Normal Approximation to Binomial Distribution Let $X$ be a binomially distributed random variable with number of trials $n$ and probability of success $p$. The mean of $X$ is $\mu=E(X) = np$ and variance of $X$ is $\sigma^2=V(X)=np(1-p)$. The general rule of thumb to use normal approximation to binomial distribution is that the sample size $n$ is sufficiently large if $np \geq 5$ and $n(1-p)\geq 5$. For sufficiently large $n$, $X\sim N(\mu, \sigma^2)$.

Confidence interval for difference in means when variances are unknown and equal Let $x_1, x_2, \cdots, x_{n_1}$ be a random sample of size $n_1$ from a population with mean $\mu_1$ and standard deviation $\sigma_1$. Let $y_1, y_2, \cdots, y_{n_2}$ be a random sample of size $n_2$ from a population with mean $\mu_2$ and standard deviation $\sigma_2$. And the two sample are independent. Let $\overline{x} = \frac{1}{n_1}\sum x_i$ and $\overline{y} = \frac{1}{n_2}\sum y_i$ be the sample means of first and second sample respectively.

Beta Type I Distribution A continuous random variable $X$ is said to have a beta type I distribution with parameters $\alpha$ and $\beta$ if its p.d.f. is given by $$ \begin{equation*} f(x)=\left\{ \begin{array}{ll} \frac{1}{B(\alpha,\beta)}x^{\alpha-1}(1-x)^{\beta-1}, & \hbox{$0\leq x\leq 1$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$ where, $B(\alpha,\beta) =\frac{\Gamma \alpha \Gamma \beta}{\Gamma (\alpha+\beta)}=\int_0^1 x^{\alpha-1}(1-x)^{\beta-1}\; dx$ is a beta function and $\Gamma \alpha$ is a gamma function. In notation, it can be written as $X\sim \beta_1(\alpha,\beta)$.

Assignment Problem The assignment problem is a special case of transportation problem. Suppose there are $n$ jobs to be performed and $n$ persons are available for these jobs. Assume that each person can do each job at a time, though with varying degree of efficiency. Let $c_{ij}$ be the cost if the $i^{th}$ person is assigned the $j^{th}$ job. Then the problem is to find an assignment so that the total cost of performing all jobs is minimum.

Bayes’ Theorem One of the important uses of conditional probability is to find out the inverse probability. If an event $B$ can happen with mutually exclusive and exhaustive events (partitions) $A_1, A_2, \cdots, A_n$ whose probabilities are known and also the conditional probabilities $P(B/A_1)$, $P(B/A_2)$, $\cdots, P(B/A_n)$ are known, then with the help of these probabilities the inverse probabilities $P(A_1/B)$, $P(A_1/B)$, $\cdots, P(A_n/B)$ can be determined. This particular rule was given by British Mathematician Bayes.

Bayes’ Theorem One of the important uses of conditional probability is to find out the inverse probability. If an event $B$ can happen with mutually exclusive and exhaustive events (partitions) $A_1, A_2, \cdots, A_n$ whose probabilities are known and also the conditional probabilities $P(B/A_1)$, $P(B/A_2)$, $\cdots, P(B/A_n)$ are known, then with the help of these probabilities the inverse probabilities $P(A_1/B)$, $P(A_1/B)$, $\cdots, P(A_n/B)$ can be determined. This particular rule was given by British Mathematician Bayes.

Introduction In the theory of probability and statistics, a Bernoulli trial or Bernoulli Experiment is a random experiment with exactly two mutually exclusive outcomes, “Success” and “Failure” with the probability of success remains same every time the experiment is conducted. The name Bernoulli trial or Bernoulli distribution named after a Swiss scientist Jacob Bernoulli. Bernoulli Trials Trials of random experiment are called Bernoulli trials, if they satisfy the following conditions

Definition of Beta Type I Distribution A continuous random variable $X$ is said to have a beta type I distribution with parameters $\alpha$ and $\beta$ if its p.d.f. is given by $$ \begin{equation*} f(x)=\left\{ \begin{array}{ll} \frac{1}{B(\alpha,\beta)}x^{\alpha-1}(1-x)^{\beta-1}, & \hbox{$0\leq x\leq 1$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$ where, $B(\alpha,\beta) =\frac{\Gamma \alpha \Gamma \beta}{\Gamma (\alpha+\beta)}=\int_0^1 x^{\alpha-1}(1-x)^{\beta-1}\; dx$ is a beta function and $\Gamma (\alpha)$ is a gamma function. Mean of Beta Type I Distribution The mean of beta type I distribution is $E(X) = \dfrac{\alpha}{\alpha+\beta}$.

Beta Type II Distribution A continuous random variable $X$ is said to have a beta type II distribution with parameters $\alpha$ and $\beta$ if its p.d.f. is given by $$ \begin{equation*} f(x)=\left\{ \begin{array}{ll} \frac{1}{B(\alpha,\beta)}\cdot\frac{x^{\alpha-1}}{(1+x)^{\alpha+\beta}}, & \hbox{$0\leq x\leq\infty$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$ where, $B(\alpha,\beta) =\frac{\Gamma \alpha \Gamma \beta}{\Gamma (\alpha+\beta)}=\int_0^1 x^{\alpha-1}(1-x)^{\beta-1}\; dx$ is a beta function and $\Gamma \alpha$ is a gamma function. In notation, it can be written as $X\sim \beta_2(\alpha,\beta)$.

Beta Type II Distribution A continuous random variable $X$ is said to have a beta type II distribution with parameters $\alpha$ and $\beta$ if its p.d.f. is given by $$ \begin{equation*} f(x)=\left\{ \begin{array}{ll} \frac{1}{B(\alpha,\beta)}\cdot\frac{x^{\alpha-1}}{(1+x)^{\alpha+\beta}}, & \hbox{$0\leq x\leq\infty$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$ where, $B(\alpha,\beta) =\frac{\Gamma \alpha \Gamma \beta}{\Gamma (\alpha+\beta)}=\int_0^1 x^{\alpha-1}(1-x)^{\beta-1}\; dx$ is a beta function and $\Gamma \alpha$ is a gamma function. Mean of Beta Type II Distribution The mean of beta type II distribution is $E(X) = \dfrac{\alpha}{\beta-1}$.

Introduction Binomial distribution is one of the most important discrete distribution in statistics. In this tutorial we will discuss about theory of Binomial distribution along with proof of some important results related to binomial distribution. Binomial Experiment Binomial experiment is a random experiment that has following properties: The random experiment consists of n repeated Bernoulli trial. Each trial has only two possible outcomes like success ($S$) and failure ($F$). All the trials are independent, i.

Bowley’s Coefficient of Skewness for grouped data Skewness is a measure of symmetry. The meaning of skewness is “lack of symmetry”. Skewness gives us an idea about the concentration of higher or lower data values around the central value of the data. For a symmetric distribution, the two quartiles namely $Q_1$ and $Q_3$ are equidistance from the median (i.e. $Q_2$). That is for symmetric distribution $Q_3 - Q_2 = Q_2 -Q_1$.

Bowley’s Coefficient of Skewness for Ungrouped data Skewness is a measure of symmetry. The meaning of skewness is “lack of symmetry”. Skewness gives us an idea about the concentration of higher or lower data values around the central value of the data. For a symmetric distribution, the two quartiles namely $Q_1$ and $Q_3$ are equidistance from the median (i.e. $Q_2$). That is for symmetric distribution $Q_3 - Q_2 = Q_2 -Q_1$.

Cauchy Distribution A continuous random variable $X$ is said to follow Cauchy distribution with parameters $\mu$ and $\lambda$ if its probability density function is given by $$ \begin{equation*} f(x) =\left\{ \begin{array}{ll} \frac{\lambda}{\pi}\cdot \frac{1}{\lambda^2+(x-\mu)^2}, & \hbox{$-\infty < x< \infty$;} \\ & \hbox{$-\infty < \mu< \infty$, $\lambda>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$ In notation it can be written as $X\sim C(\mu, \lambda)$. The parameter $\mu$ and $\lambda$ are location and scale parameters respectively.

Cauchy Distribution A continuous random variable $X$ is said to follow Cauchy distribution with parameters $\mu$ and $\lambda$ if its probability density function is given by $$ \begin{equation*} f(x; \mu, \lambda) =\left\{ \begin{array}{ll} \frac{\lambda}{\pi}\cdot \frac{1}{\lambda^2+(x-\mu)^2}, & \hbox{$-\infty < x< \infty$;} \\ & \hbox{$-\infty < \mu< \infty$, $\lambda>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$ Distribution Function of Cauchy Distribution The distribution function of Cauchy distribution is $$ \begin{equation*} F(x) =\frac{1}{\pi}\tan^{-1}\bigg(\frac{x-\mu}{\lambda}\bigg) + \frac{1}{2}.

Central moments for grouped data Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution. The mean of $X$ is denoted by $\overline{x}$ and is given by $$ \begin{eqnarray*} \overline{x}& =\frac{1}{N}\sum_{i=1}^{n}f_ix_i \end{eqnarray*} $$ Formula The first four central moments are as follows $m_1=0$ $m_2 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^2$ $m_3 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^3$ $m_4 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^4$ where, $N$ total number of observations $\overline{x}$ sample mean Example 1 Following tables shows a frequency distribution of daily number of car accidents at a particular cross road during a month of April.

Central moments for ungrouped data Let $x_1, x_2,\cdots, x_n$ be $n$ observations. The mean of $X$ is denoted by $\overline{x}$ and is given by $$ \begin{eqnarray*} \overline{x}& =\frac{1}{n}\sum_{i=1}^{n}x_i \end{eqnarray*} $$ Formula The first four sample central moments are as follows $m_1=0$ (always) $m_2 =\dfrac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^2$ $m_3 =\dfrac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^3$ $m_4 =\dfrac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^4$ where, $n$ total number of observations $\overline{x}$ sample mean Example The hourly earning (in dollars) of sample of 7 workers are : 26,21,24,22,25,24,23.

Chebyshev’s Inequality Chebyshev’s Theorem If $g(x)$ is a non-negative function and $f(x)$ be p.m.f. or p.d.f. of a random variable $X$, having finite expectation and if $k$ is any positive real constant, then $$ \begin{equation*} P[g(x)\geq k] \leq \frac{E[g(x)]}{k}\quad \text{ and }\quad P[g(x) < k] \geq 1-\frac{E[g(x)]}{k} \end{equation*} $$ Proof Discrete Case Let $X$ be a discrete random variable with p.m.f. $f(x)$. Let $g(x)$ be a non-negative function of $X$. Then $E[g(x)] = \sum g(x)f(x)$.

Chebyshev’s Inequality Let $X$ be a random variable with mean $\mu$ and finite variance $\sigma^2$. Then for any real constant $k>0$, $$ \begin{equation*} P[|X-\mu| \geq k\sigma] \leq \frac{1}{k^2}\quad \text{ and }\quad P[|X-\mu|< k\sigma] \geq 1-\frac{1}{k^2} \end{equation*} $$ OR If $\mu$ and $\sigma$ are the mean and the standard deviation of a random variable $X,$ then for any positive constant $k$, the probability is at least $1-\dfrac{1}{k^2}$ that $X$ will take on a value within $k$ standard deviations of the mean.

Testing variance or standard deviation In this tutorial we will discuss a method for testing a claim made about the population variance $\sigma^2$ or population standard deviation $\sigma$. To test the claim about the population variance or population standard deviation we use chi-square test. In this tutorial we will explain the six steps approach used in hypothesis testing to test hypothesis about the population variance or population standard deviation. Chi-square Test for Variance Let $X_1, X_2, \cdots, X_n$ be a random sample from a normal population with mean $\mu$ and variance $\sigma^2$.

Chi-square Test for the Variance In this tutorial we will discuss a method for testing a claim made about the population variance $\sigma^2$ or population standard deviation $\sigma$. To test the claim about the population variance or population standard deviation we use chi-square test. In this tutorial we will discuss some numerical examples using six steps approach used in hypothesis testing to test hypothesis about the population variance or population standard deviation.

Coefficient of variation for grouped data Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution. Formula Coefficient of variation is given by $CV =\dfrac{s_x}{\overline{x}}\times 100$ where, $\overline{x} =\dfrac{1}{N}\sum_{i=1}^{n}f_ix_i$ is the sample mean of $X$, $N$ total number of observations, $s_x=\sqrt{V(x)}$ is the standard deviation of $X$, $s_x^2=V(x)=\dfrac{1}{N}\sum_{i=1}^{n}f_ix_i^2 -(\overline{x})^2$ is the variance of $X$ Coefficient of variation of one data set is lower than the coefficient of variation of other data set, then the data set with lower coefficient of variation is more consistent than the other.

Coefficient of variation for ungrouped data Coefficient of variation is an absolute measure of variation and is used for the comparison of variability between two or more frequency distribution. Let $x_i, i=1,2, \cdots , n$ be $n$ observations. Formula Coefficient of variation is given by $CV =\dfrac{s_x}{\overline{x}}\times 100$ where, $\overline{x} =\dfrac{1}{n}\sum_{i=1}^{n}x_i$ is the sample mean of $X$, $n$ total number of observations, $s_x=\sqrt{V(x)}$ is the standard deviation of $X$, $s_x^2=V(x)=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2 -\dfrac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)$ is the variance of $X$ Example 1 The following data gives the hourly wage rates (in dollars) of 10 employees of a company.

Column Minima Method Step 1 Select the smallest cost in the first column of the transportation table. Let it be $c_{i1}$. Allocate as much as possible amount $x_{i1} = min_i(a_i, b_1)$ in the cell $(i,1)$, so that either the capacity of origin $O_i$ is exhausted or the requirement at destination $D_1$ is satisfied or both. Step 2 If $x_{i1} = b_1$, the requirement at destination $D_1$ is completely exhausted, cross-out the first column of the table and move down to the second column.

Introduction Confidence interval can be used to estimate the population parameter with the help of an interval with some degree of confidence. One such a parameter that can be estimated is a population mean. In this article we will discuss step by step procedure to construct a confidence interval for population mean when the population standard deviation is known. Confidence Interval for Mean Let $X_1, X_2, \cdots , X_{n}$ be a random sample of size $n$ from $N(\mu, \sigma^2)$ with known variance $\sigma^2$.

Confidence Interval for Mean In this tutorial we will discuss some numerical examples to understand how to construct a confidence interval for population mean when the population standard deviation is known. Example 1 Among those who take the Graduate Record Examination (GRE), 67 persons are randomly selected. This sample group has a mean score of 558 on the quantitative portion of the GRE. The population standard deviation is 139 (from the Educational Testing Service).

Introduction Confidence interval can be used to estimate the population parameter with the help of an interval with some degree of confidence. One such a parameter that can be estimated is a population mean. In this article we will discuss step by step procedure to construct a confidence interval for population mean when the population standard deviation is unknown. Confidence Interval for the mean ($\sigma$ unknown) Let $X_1, X_2, \cdots , X_{n}$ be a random sample of size $n$ from $N(\mu, \sigma^2)$ with unknown variance $\sigma^2$.

Confidence Interval for the population mean when standard deviation is unknown This tutorial covers examples on confidence interval for the population mean when the population standard deviation is unknown. Example 1 A new brand of laptop battery is produced by a company. The company claims that the battery will last for an extended period of time before a recharge is necessary. A sample of 40 batteries is tested for the length of usage time to recharge.

Introduction In this article we will discuss step by step procedure to construct a confidence interval for difference between two population means when the population standard deviations are known. CI for difference between two population means (variances are known) Let $X_1, X_2, \cdots, X_{n_1}$ be a random sample of size $n_1$ from a population with mean $\mu_1$ and standard deviation $\sigma_1$. Let $Y_1, Y_2, \cdots, Y_{n_2}$ be a random sample of size $n_2$ from a population with mean $\mu_2$ and standard deviation $\sigma_2$.

CI for difference between two population means (variances are known) In this tutorial we will discuss some examples on confidence interval for difference in means when the population variances are known. Example 1 Two kinds of thread are being compared for tensile strength. Fourty pieces of each type of thread are tested under similar conditions. Brand A has an average tensile strength of 81.6 kilograms with a standard deviation of 4.

Confidence interval for difference in means when variances are unknown and equal In this tutorial we will discuss step by step solution to some numerical problems on confidence interval for difference in means when the population variances are unknown and equal. Example 1 A high school language course is given in two sections, each using a different teaching method. The first section has $21$ students, and the grades in that section have a mean of $82.

Confidence Interval for paired t-test Let $X_1, X_2, \cdots, X_n$ and $Y_1, Y_2, \cdots, Y_n$ be two dependent samples each of size $n$ with respective means $\mu_1$ and $\mu_2$. Define $d_i = X_i - Y_i$, $i=1,2,\cdots, n$. Let $\overline{d}=\frac{1}{n} \sum d_i$ be the mean of the difference and $s_d=\sqrt{\frac{1}{n-1}\sum (d_i - \overline{d})^2}$ be the sample standard deviation of the difference. Let $C=1-\alpha$ be the confidence coefficient. Our aim is to construct $100(1-\alpha)$% confidence interval estimate of the mean of the differences from dependent samples.

Confidence Interval for paired t-test In this tutorial we will discuss how to determine confidence interval for the difference in means for dependent samples. Example 1 An experiment ws designed to estimate the mean difference in weight gain for pigs fed ration A as compared with those fed ration B. Eight pairs of pigs were used. The pigs within each pair were littermates. The rations were assigned at random to the two animals within each pair.

Introduction Confidence interval can be used to estimate the population parameter with the help of an interval with some degree of confidence. One such a parameter that can be estimated is a population proportion. In this article we will discuss step by step procedure to construct a confidence interval for population proportion. Confidence Interval for Proportion Let $X$ be the observed number of individuals possessing certain attributes (number of successes) in a random sample of size $n$ from a large population with population proportion $p$.

Confidence Interval for Proportion In this tutorial we will discuss some numerical examples to understand how to construct a confidence interval for population proportion. Example 1 In a simple random sample of 50 animated children movies, 19 show tobacco use by main characters. Construct a 90% confidence interval to estimate the proportion of all animated children movies that show tobacco use by main characters. Assume the population is normally distributed.

CI for difference between two population proportions Let $X_1$ be the observed number of individuals possessing certain attributes (number of successes) in a random sample of size $n_1$ from a large population with population proportion $p_1$ and let $X_2$ be the observed number of individuals possessing certain attributes (number of successes) in a random sample of size $n_2$ from a large population with population proportion $p_2$. The two sample are independent.

CI for difference between two population proportions In this tutorial we will discuss some examples on confidence interval for difference between two population proportions. Example 1 A random sample of 85 parts manufactured by machine A yields 10 defective and a random sample of 110 parts manufactured by machine B shows 28 defective. Compute 95% confidence interval for difference between the proportions of defectives. Are the two machine differ significantly with respect to the proportion of defectives?

Confidence Interval for ratio of variances Let $X_1, X_2, \cdots , X_{n_1}$ be a random sample of size $n_1$ from $N(\mu_1, \sigma_1^2)$ and $Y_1, Y_2, \cdots , Y_{n_2}$ be a random sample of size $n_2$ from $N(\mu_2, \sigma_2^2)$. Moreover, $X$ and $Y$ are independently distributed. $100(1-\alpha)$% confidence interval estimate for the ratio of variances is $$ \begin{aligned} \bigg(\frac{s_1^2}{s_2^2}\frac{1}{F_{(\alpha/2,n_1-1,n_2-1)}}, \frac{s_1^2}{s_2^2}\frac{1}{F_{(1-\alpha/2,n_1-1,n_2-1)}}\bigg) \end{aligned} $$ Assumptions a. The two populations are independent. b. The two samples are simple random samples.

Confidence Interval for ratio of variances In this tutorial we will discuss some examples on confidence interval for ratio of variances. Example 1 The same capacity of hard drive is manufactured on two different machines, Machine A and Machine B. Samples are taken from both machines and sample mean manufacturing times and sample variances are recorded as follows: Machine A: $n_1=13$, $\overline{x}= 127.4$, $s_1^2= 384.16$ Machine B: $n_2=9$, $\overline{y}= 108.3$, $s_2^2 =106.

Introduction Confidence interval can be used to estimate the population parameter with the help of an interval with some degree of confidence. One such a parameter that can be estimated is a population variance or population standard deviation. In this article we will discuss step by step procedure to construct a confidence interval for population variance or population standard deviation. Confidence Interval for Variance Let $X_1, X_2, \cdots , X_n$ be a random sample of size $n$ from $N(\mu, \sigma^2)$.

Confidence Interval for Variance In this tutorial we will discuss some numerical examples to understand how to construct a confidence interval for population variance or population standard deviation. Example 1 The mean replacement time for a random sample of 12 microwaves is 8.6 years with a standard deviation of 3.6 years. Construct a 95% confidence interval for the population standard deviation. Solution Here the sample size is $n=27$, sample standard deviation is $s=6.

Continuous Uniform Distribution A random variable has a uniform distribution if each value of the random variable is equally likely and the values of the random variable are uniformly distributed throughout some specified interval. A uniform distribution is a distribution with constant probability. This tutorial will help you understand the theory and proof of theoretical results of uniform distribution. Uniform distribution is also known as rectangular distribution (see the graph below).

Covariance between X and Y Let $(x_i, y_i), i=1,2, \cdots , n$ be $n$ pairs of observations. Covariance measures the simultaneous variability between the two variables. It indicates how the two variables are related. A positive value of covariance indicate that the two variables moves in the same direction, whereas a negative value of covariance indicate that the two variables moves on opposite direction. Formula The sample covariance between $x$ and $y$ is denoted by $Cov(x,y)$ or $s_{xy}$ and is defined as

Deciles for grouped data Deciles are the values of arranged data which divide whole data into ten equal parts. They are 9 in numbers namely $D_1,D_2, \cdots, D_9$. Here $D_1$ is first decile, $D_2$ is second decile, $D_3$ is third decile and so on. Decile formula for grouped data For discrete frequency distribution, the formula for $i^{th}$ decile is $D_i =\bigg(\dfrac{i(N)}{10}\bigg)^{th}$ value, $i=1,2,\cdots, 9$ where, $N$ is total number of observations.

Deciles for ungrouped data Deciles are the values of arranged data which divide whole data into ten equal parts. They are 9 in numbers namely $D_1,D_2, \cdots, D_9$. Here $D_1$ is first decile, $D_2$ is second decile, $D_3$ is third decile and so on. Formula The formula for $i^{th}$ decile is $D_i =$ Value of $\bigg(\dfrac{i(n+1)}{10}\bigg)^{th}$ observation, $i=1,2,3,\cdots, 9$ where, $n$ is the total number of observations. Example 1 The marks obtained by a sample of 20 students in a class test are as follows:

Discrete Uniform Distribution A discrete random variable has a discrete uniform distribution if each value of the random variable is equally likely and the values of the random variable are uniformly distributed throughout some specified interval. In this article, I will walk you through discrete uniform distribution and proof related to discrete uniform. You can use discrete uniform distribution Calculator. Discrete Uniform Distribution Definition A discrete random variable $X$ is said to have a uniform distribution if its probability mass function (pmf) is given by

Empirical rule for grouped data Empirical rule is the general rule of thumb that applies to bell shaped (symmetrical) distribution. The empirical rule can be stated as : $68$% of the data will fall within one standard deviation of the mean, $95$% of the data will fall within two standard deviations of the mean, $99.7$% of the data will fall within three standard deviations of the mean. Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution.

Empirical Rule for ungrouped data Empirical rule is the general rule of thumb that applies to bell shaped (symmetrical) distribution. The empirical rule can be stated as : $68$% of the data will fall within one standard deviation of the mean, $95$% of the data will fall within two standard deviations of the mean, $99.7$% of the data will fall within three standard deviations of the mean. Formula Let $x_1,x_2,\cdots, x_n$ be $n$ sample observations.

Example Find the expected value and variance of $X$ for the following probability distribution. $x$ 1 2 3 4 $P(X=x)$ 0.0625 0.1875 0.3125 0.4375 Solution The Expected value of $X$ is $$ \begin{aligned} E(X) &= \sum_x x*P(X=x) \end{aligned} $$ The expected value of $X^2$ is $$ \begin{aligned} E(X^2) &= \sum_x x^2*P(X=x) \end{aligned} $$ The variance of $X$ is $$ \begin{aligned} V(X) &= E[X-E(X)]^2 \\ &= E(X^2) - [E(X)]^2 \end{aligned} $$

Exponential distribution Exponential Distribution Definition A continuous random variable $X$ is said to have an exponential distribution with parameter $\theta$ if its p.d.f. is given by $$ \begin{equation*} f(x)=\left\{ \begin{array}{ll} \theta e^{-\theta x}, & \hbox{$x\geq 0;\theta>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$ In notation, it can be written as $X\sim \exp(\theta)$. Graph of pdf of exponential distribution Following is the graph of probability density function of exponential distribution with parameter $\theta=0.

Testing equality of two variances Many times it is desirable to compare two variances rather than comparing two means. F test is used to compare two population variances or population standard deviations. In this tutorial we will discuss six steps approach used in hypothesis testing to test whether two population variances are same or not. F-Test for equality of two variances Let $X_1, X_2, \cdots, X_{n_1}$ be a random sample of size $n_1$ from a population with variance $\sigma^2_1$ and $Y_1,Y_2, \cdots, Y_{n_2}$ be a random samples of sizes $n_2$ from a population with variance $\sigma^2_2$.

F test for two variances In this tutorial we will discuss some examples on F test for comparing two variances or standard deviations. Example 1 A professor from a graduate school claims that there is less variability in the final exam scores of students taking the statistics major than the students taking the mathematics major. Random samples of 13 mathematics majors and 10 statistics majors are selected from the his large class, and the following results are computed based on the final exam scores:

Five number summary for grouped data A five number summary is a quick and easy way to determine the the center, the spread and outliers (if any) of a data set. Five number summary includes five values, namely, minimum value ($\min$), first quartile ($Q_1$), $\text{median }$ ($Q_2$), third quartile ($Q_3$), maximum value ($\max$). Formula $\min$= lower limit of the first class, $\max$= upper limit of the last class, $Q_i=l + \bigg(\dfrac{\dfrac{iN}{4} - F_<}{f}\bigg)\times h$ ; $i=1,2,\cdots,3$

Five number summary for ungrouped data A five number summary is a quick and easy way to determine the the center, the spread and outliers (if any) of a data set. Five number summary includes five values, namely, minimum value ($\min$), first quartile ($Q_1$), $\text{median }$ ($Q_2$), third quartile ($Q_3$), maximum value ($\max$). Formula $\min$ : smallest observation of a data set, $Q_1$ :first quartile of a data set, $\text{median} = Q_2$ : median of a data set, $Q_3$ : third quartile of a data set, $\max$ : largest observation of a data set.

Five number summary for ungrouped data A five number summary is a quick and easy way to determine the the center, the spread and outliers (if any) of a data set. Five number summary includes five values, namely, minimum value ($\min$), first quartile ($Q_1$), $\text{median }$ ($Q_2$), third quartile ($Q_3$), maximum value ($\max$). Formula $\min$, $Q_1$, $\text{median}$, $Q_3$ and $\max$ The five number summary of a data gives you a rough idea about the range of the data, center of the data, variation in the data and symmetry of the data.

Gamma Distribution Gamma distribution is used to model a continuous random variable which takes positive values. Gamma distribution is widely used in science and engineering to model a skewed distribution. In this tutorial, we are going to discuss various important statistical properties of gamma distribution like graph of gamma distribution for various parameter combination, derivation of mean, variance, harmonic mean, mode, moment generating function and cumulant generating function. Gamma Distribution Definition A continuous random variable $X$ is said to have an gamma distribution with parameters $\alpha$ and $\beta$ if its p.

Definition of Gamma Distribution A continuous random variable $X$ is said to have an gamma distribution with parameters $\alpha$ and $\beta$ if its p.d.f. is given by $$ \begin{equation*} f(x)=\left\{ \begin{array}{ll} \frac{\alpha^\beta}{\Gamma(\beta)}x^{\beta -1}e^{-\alpha x}, & \hbox{$x>0;\alpha, \beta >0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$ Mean of Gamma Distribution The mean or expected value of gamma random variable is $E(X)= \dfrac{\beta}{\alpha}$ Variance of Gamma distribution The variance of gamma random variable is $V(X) = \dfrac{\beta}{\alpha^2}$.

Introduction Geometric distribution is used to model the situation where we are interested in finding the probability of number failures before first success or number of trials (attempts) to get first success in a repeated mutually independent Beronulli’s trials, each with probability of success $p$. There are two different definitions of geometric distributions one based on number of failures before first success and other based on number of trials (attempts) to get first success.

Geometric mean for ungrouped data Let $x_i, i=1,2, \cdots , n$ be $n$ positive observations then the geometric mean of $X$ is denoted by $GM$. Formula The geometric mean is given by $GM =\big(\prod_{i=1}^n x_i\big)^{1/n}=\sqrt[n]{x_1\times x_2\times \cdots \times x_n}$ Example The marks obtained by a sample of 5 students in a class test are as follows: 20,30,21,10 and 15. Find the geometric mean of the marks. Solution The geometric mean of $X$ is

Harmonic mean for grouped data Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution. Then the harmonic mean of $X$ is denoted by $HM$ and is given by $$ \begin{aligned} HM &= \dfrac{N}{\sum_{i=1}^{n} \dfrac{f_i}{x_i}} \end{aligned} $$ where $N=\sum f$ total no. of observations Example 1 Compute harmonic mean for the following frequency distribution. x 10 15 20 25 30 f 3 12 25 10 5 Solution $x$ Freq ($f$) $f/x$ 10 3 0.

Harmonic mean for ungrouped data Let $x_i, i=1,2, \cdots , n$ be $n$ positive observations then the harmonic mean of $X$ is denoted by $HM$. Formula The harmonic mean is given by $HM = \dfrac{n}{\sum_{i=1}^{n} \dfrac{1}{x_i}}$ Example A bus driver travels total 30 miles. He travel first 10 miles with speed of 30 miles per hour, second 10 miles with the speed of 35 miles per hour and last 10 miles with the speed of 45 miles per hour.

Hungarian Method to solve Assignment Problem For obtaining an optimal assignment, Hungarian method involves following steps : Step 1 Subtract the minimum of each row of the cost matrix,from all the elements of respective rows. Step 2 Subtract the minimum of each column of the modified cost matrix, from all the elements of respective columns. Step 3 Then draw the minimum number of horizontal and vertical line to cover all the zeros in the modified cost matrix.

Hypergeometric Experiment A hypergeometric experiment is an experiment which satisfies each of the following conditions: The population or set to be sampled consists of $N$ individuals, objects, or elements (a finite population). Each object can be characterized as a “defective” or “non-defective”, and there are $M$ defectives in the population. A sample of $n$ individuals is drawn in such a way that each subset of size $n$ is equally likely to be chosen.

Inter Quartile Range for grouped data Inter quartile range is given by $IQR = Q_3-Q_1$ where, $Q_1$ is the first quartile $Q_3$ is the third quartile The formula for $i^{th}$ quartile is $$ \begin{aligned} Q_i=l + \bigg(\frac{\frac{iN}{4} - F_<}{f}\bigg)\times h; \quad i=1,2,3 \end{aligned} $$ where, $l :$ the lower limit of the $i^{th}$ quartile class $N=\sum f :$ total number of observations $f :$ frequency of the $i^{th}$ quartile class $F_< :$ cumulative frequency of the class previous to $i^{th}$ quartile class $h :$ the class width Example 1 A class teacher has the following data about the number of absences of 35 students of a class.

Inter Quartile Range for ungrouped data Inter quartile range is given by $IQR = Q_3-Q_1$ where $Q_1$ is the first quartile $Q_3$ is the third quartile The formula for $i^{th}$ quartile is $Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$ where $n$ is the total number of observations. Example 1 A random sample of 15 patients yielded the following data on the length of stay (in days) in the hospital. 5, 6, 9, 10, 15, 10, 14, 12, 10, 13, 13, 9, 8, 10, 12.

Karl Pearson coefficient of skewness for grouped data Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution. Karl Pearson coefficient of skewness formula The Karl Pearson’s coefficient skewness for grouped data is given by $S_k =\dfrac{Mean-Mode)}{sd}=\dfrac{\overline{x}-\text{Mode}}{s_x}$ OR $S_k =\dfrac{3(Mean-Median)}{sd}=\dfrac{\overline{x}-M}{s_x}$ where, $\overline{x}$ is the sample mean, $M$ is the median, $s_x$ is the sample standard deviation. Sample mean The sample mean $\overline{x}$ is given by $$ \begin{eqnarray*} \overline{x}& =\frac{1}{N}\sum_{i=1}^{n}f_ix_i \end{eqnarray*} $$

Karl Pearson coefficient of skewness for ungrouped data Let $x_i, i=1,2, \cdots , n$ be $n$ observations. Formula The Karl Pearson’s coefficient skewness is given by $S_K =\dfrac{3(Mean-Median)}{sd}=\dfrac{3(\overline{x}-M)}{s_x}$ where, $\overline{x}$ is the sample mean, $M$ is the median, $s_x$ is the sample standard deviation. Sample mean The sample mean $\overline{x}$ is given by $$ \begin{eqnarray*} \overline{x}& =\frac{1}{n}\sum_{i=1}^{n}x_i \end{eqnarray*} $$ Sample Median Arrange the data in ascending order of magnitude. Median of $X$ is given by

Karl Pearson’s Correlation Coefficient Let $(x_i, y_i), i=1,2, \cdots , n$ be $n$ pairs of observations then the Karl Pearson’s coefficient of correlation between two variables $X$ and $Y$ is denoted by $r_{xy}$ or $r$ and is given by $r = \dfrac{Cov(X,Y)}{\sqrt{Var(X) Var(Y)}}$ where, the sample covariance between $x$ and $y$ is $$ \begin{aligned} Cov(x,y) =s_{xy}&=\frac{1}{n-1}\sum_{i=1}^{n}(x_i -\overline{x})(y_i-\overline{y})\\ &= \frac{1}{n-1}\bigg(\sum_{i=1}^n x_iy_i - \frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}\bigg) \end{aligned} $$ the sample variance of $x$ is $$ \begin{aligned} V(x) =s_{x}^2 &=\frac{1}{n-1}\sum_{i=1}^{n}(x_i -\overline{x})^2\\ &= \frac{1}{n-1}\bigg(\sum_{i=1}^n x_i^2 - \frac{(\sum_{i=1}^n x_i)^2}{n}\bigg) \end{aligned} $$

Kelly’s coefficient of skewness for grouped data For a symmetric distribution, the first decile namely $D_1$ and nineth decile $D_9$ are equidistance from the median i.e. $D_5$. Thus, $D_9 - D_5 = D_5 -D_1$. Kelly’s coefficient of skewness is based on deciles $D_{1}$, $1^{st}$ decile, $D_{5}$, $5^{th}$ decile, and $D_{9}$, $9^{th}$ decile). Only 20% of the observations are excluded from the measure. Formula $S_k =\dfrac{D_9+D_1 -2*Median}{D_9-D_1}$ OR $S_k =\dfrac{P_{90}+P_{10} -2*Median}{P_{90}-P_{10}}$

Kelly’s Coefficient of Skewness for ungrouped data For a symmetric distribution, the first decile namely $D_1$ and nineth decile $D_9$ are equidistance from the median i.e. $D_5$. Thus, $D_9 - D_5 = D_5 -D_1$. The coefficient of skewness based on deciles is defined as $S_k = \dfrac{D_9+D_1 - 2D_5}{D_9 -D_1}$ OR $S_k =\dfrac{P_{90}+P_{10} -2*Median}{P_{90}-P_{10}}$ where, $D_1 (=P_{10})$ is the first decile (or tenth percentile) of the data $Median=D_5 (=P_{50})$ is the median of the data $D_9 (=P_{90})$ is the ninth decile (or nineteenth percentile) of the data.

Laplace Distribution A continuous random variable $X$ is said to have a Laplace distribution (Double exponential distribution or bilateral exponential distribution), if its p.d.f. is given by $$ \begin{equation*} f(x;\mu, \lambda)=\left\{ \begin{array}{ll} \frac{1}{2\lambda}e^{-\frac{|x-\mu|}{\lambda}}, & \hbox{$-\infty < x< \infty$;} \\ & \hbox{$-\infty < \mu < \infty$, $\lambda >0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$ In Laplace distribution $\mu$ is called location parameter, since it locates the curve of the distribution, and $\lambda$ is called scale parameter, since the shape of the curve depends on the value of $\lambda$.

Least Cost Entry Method This method is also known as Matrix Minima Method. Step 1 Select the smallest cost in the cost matrix of the transportation table. Let it be $c_{ij}$. Allocate $x_{ij} = min_{i,j}(a_i, b_j)$ in the cell $(i,j)$. Step 2 If $x_{ij} = a_i$, then cross-out the $i^{th}$ row of the transportation table and decrease $b_j$ by $a_i$ and goto Step 3. If $x_{ij} = b_j$, then cross-out the $j^{th}$ column of the transportation table and decrease $a_i$ by $b_j$ and goto Step 3.

Mean absolute deviation for grouped data Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution. The mean of $X$ is denoted by $\overline{x}$ and is given by $$ \begin{eqnarray*} \overline{x}& =\frac{1}{N}\sum_{i=1}^{n}f_ix_i \end{eqnarray*} $$ Formula The mean absolute deviation about mean is given by $MAD =\dfrac{1}{N}\sum_{i=1}^{n}f_i|x_i -\overline{x}|$ where, $N$ total number of observations $\overline{x}$ sample mean Example 1 A librarian keeps the records about the amount of time spent (in minutes) in a library by college students.

Mean absolute deviation for ungrouped data Let $x_i, i=1,2, \cdots , n$ be $n$ observations. The mean of $X$ is denoted by $\overline{x}$ and is given by $$ \begin{eqnarray*} \overline{x}& =\frac{1}{n}\sum_{i=1}^{n}x_i \end{eqnarray*} $$ Formula: $MAD =\dfrac{1}{n}\sum_{i=1}^{n}|x_i -\overline{x}|$ where, $n$ total number of observations $\overline{x}$ sample mean Example 1 The age (in years) of 6 randomly selected students from a class are : 22,25,24,23,24,20. Compute mean absolute deviation about mean. Solution $x_i$ $(x_i-xb)$ $|x_i-xb|$ 22 -1 1 25 2 2 24 1 1 23 0 0 24 1 1 20 -3 3 Total 138 8 The sample mean of $X$ is

Mean, median and mode for grouped data Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution. Formula Sample mean The mean of $X$ is denoted by $\overline{x}$ and is given by $\overline{x} =\dfrac{1}{N}\sum_{i=1}^{n}f_ix_i$ In case of continuous frequency distribution, $x_i$’s are the mid-values of the respective classes. Sample median The median is given by $\text{Median } = l + \bigg(\dfrac{\frac{N}{2} - F_<}{f}\bigg)\times h$ where, $N$, total number of observations $l$, the lower limit of the median class $f$, frequency of the median class $F_<$, cumulative frequency of the pre median class $h$, the class width Sample mode The mode of the distribution is given by

Mean, median and mode for ungrouped data Let $x_i, i=1,2, \cdots , n$ be $n$ observations. Formula The mean of $X$ is denoted by $\overline{x}$ and is given by $\overline{x} =\dfrac{1}{n}\sum_{i=1}^{n}x_i$ Arrange the data in ascending order of magnitude. Median of $X$ is given by Median $$ \begin{equation*} M= \left\{ \begin{array}{ll} \text{value of }\big(\frac{n+1}{2}\big)^{th}\text{ observation}, & \hbox{if $n$ is odd;} \\ \text{average of }\big(\frac{n}{2}\big)^{th}\text{ and }\big(\frac{n}{2}+1\big)^{th} \text{ observation}, & \hbox{if $n$ is even.

Moment coefficient of kurtosis for grouped data Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution. The mean of $X$ is denoted by $\overline{x}$ and is given by $$ \begin{eqnarray*} \overline{x}& =\frac{1}{N}\sum_{i=1}^{n}f_ix_i \end{eqnarray*} $$ Formula The moment coefficient of kurtosis $\beta_2$ is defined as $\beta_2=\dfrac{m_4}{m_2^2}$ The moment coefficient of kurtosis $\gamma_2$ is defined as $\gamma_2=\beta_2-3$ where $N$ total number of observations $\overline{x}$ sample mean $m_2 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^2$ is second central moment $m_4 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^4$ is fourth central moment Example 1 Following tables shows a frequency distribution of daily number of car accidents at a particular cross road during a month of April.

Moment coefficient of kurtosis for ungrouped data Let $x_1, x_2,\cdots, x_n$ be $n$ observations. The mean of $X$ is denoted by $\overline{x}$ and is given by $$ \begin{eqnarray*} \overline{x}& =\frac{1}{n}\sum_{i=1}^{n}x_i \end{eqnarray*} $$ Formula The moment coefficient of kurtosis $\beta_2$ is defined as $\beta_2=\dfrac{m_4}{m_2^2}$ The moment coefficient of kurtosis $\gamma_2$ is defined as $\gamma_2=\beta_2-3$ where $n$ total number of observations $\overline{x}$ sample mean $m_2 =\dfrac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^2$ is second sample central moment $m_4 =\dfrac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^4$ is fourth sample central moment Example The hourly earning (in dollars) of sample of 7 workers are :

Moment coefficient of skewness for grouped data Let $(x_i,f_i), i=1,2, \cdots , n$ be given frequency distribution. The mean of $X$ is denoted by $\overline{x}$ and is given by $$ \begin{eqnarray*} \overline{x}& =\frac{1}{N}\sum_{i=1}^{n}f_ix_i \end{eqnarray*} $$ Formula The moment coefficient of skewness $\beta_1$ is defined as $\beta_1=\dfrac{m_3^2}{m_2^3}$ The moment coefficient of skewness $\gamma_1$ is defined as $\gamma_1=\sqrt{\beta_1}=\dfrac{m_3}{m_2^{3/2}}$ where $n$ total number of observations $\overline{x}$ sample mean $m_2 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^2$ is second central moment $m_3 =\frac{1}{N}\sum_{i=1}^n f_i(x_i-\overline{x})^3$ is third central moment Example 1 Following tables shows a frequency distribution of daily number of car accidents at a particular cross road during a month of April.

Moment coefficient of skewness for ungrouped data Let $x_1, x_2,\cdots, x_n$ be $n$ observations. The mean of $X$ is denoted by $\overline{x}$ and is given by $$ \begin{eqnarray*} \overline{x}& =\frac{1}{n}\sum_{i=1}^{n}x_i \end{eqnarray*} $$ Formula The moment coefficient of skewness $\beta_1$ is defined as $\beta_1=\dfrac{m_3^2}{m_2^3}$ The moment coefficient of skewness $\gamma_1$ is defined as $\gamma_1=\sqrt{\beta_1}=\dfrac{m_3}{m_2^{3/2}}$ where $n$ total number of observations $\overline{x}$ sample mean $m_2 =\dfrac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^2$ is second sample central moment $m_3 =\dfrac{1}{n}\sum_{i=1}^n (x_i-\overline{x})^3$ is third sample central moment Example The hourly earning (in dollars) of sample of 7 workers are :

Multiplication Rule in Probability If $A$ and $B$ are two events in a probability experiment such that $P(A)\neq 0$ and $P(B)\neq 0$, then the probability that both the events occurs simultaneously is $$ \begin{aligned} P(A\cap B)& = P(A|B)\times P(B)\\ &=P(B|A)\times P(A) \end{aligned} $$ In case of independent events, the probability that both the events occurs simultaneously is $$ \begin{aligned} P(A\cap B)& = P(A)\times P(B) \end{aligned} $$

Introduction A negative binomial distribution is based on an experiment which satisfies the following three conditions: An experiment consists of q sequence of independent Bernoulli’s trials (i.e., each trial can result in a success ($S$) or a failure ($F$)), The probability of success is constant from trial to trial, i.e., $P(S\text{ on trial } i) = p$, for $i=1,2,3,\cdots$, The experiment continues until a total of $r$ successes have been observed, where $r$ is a specified positive integer.

Normal approximation to Poisson distribution In this tutorial we will discuss some numerical examples on Poisson distribution where normal approximation is applicable. For large value of the $\lambda$ (mean of Poisson variate), the Poisson distribution can be well approximated by a normal distribution with the same mean and variance. Let $X$ be a Poisson distributed random variable with mean $\lambda$. The mean of $X$ is $\mu=E(X) = \lambda$ and variance of $X$ is $\sigma^2=V(X)=\lambda$.

Normal Distribution Normal distribution is one of the most fundamental distribution in Statistics. It is also known as Gaussian distribution. Definition A continuous random variable $X$ is said to have a normal distribution with parameters $\mu$ and $\sigma^2$ if its probability density function is given by $$ \begin{equation*} f(x;\mu, \sigma^2) = \left\{ \begin{array}{ll} \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2\sigma^2}(x-\mu)^2}, & \hbox{$-\infty< x<\infty$,} \\ & \hbox{$-\infty<\mu<\infty$, $\sigma^2>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$ where $e= 2.

North-West Corner Method The North-West cornet method starts at the northwest corner cell of the transportation problem table. Step 1 Allocate as much as possible to the selected cell, and adjust the associated amounts of supply and demand by subtracting the allocated amount. The maximum possible amount is allocated in in (1,1) cell, i.e., $x_{11} = \min(a_1, b_1)$. Step 2 If $b_1 > a_1$, move vertically downwards to the second row and make the second allocation of amount $x_{21} = \min(a_2,b_1-x_{11})$ in the cell (2,1).

Octiles for grouped data Octiles are the values of arranged data which divide whole data into eight equal parts. They are 7 in numbers namely $O_1,O_2, \cdots, O_7$. Here $O_1$ is first octile, $O_2$ is second octile, $O_3$ is third octile and so on. Formula For discrete frequency distribution, the formula for $i^{th}$ octile is $O_i =\bigg(\dfrac{i(N)}{8}\bigg)^{th}$ value, $i=1,2,\cdots, 7$ where, $N$ is total number of observations. For continuous frequency distribution, the formula for $i^{th}$ octile is

Octiles for ungrouped data Octiles are the values of arranged data which divide whole data into eight equal parts. They are 7 in numbers namely $O_1,O_2, \cdots, O_7$. Here $O_1$ is first octile, $O_2$ is second octile, $O_3$ is third octile and so on. The formula for $i^{th}$ octile is $O_i =$ Value of $\bigg(\dfrac{i(n+1)}{8}\bigg)^{th}$ observation, $i=1,2,3,\cdots, 7$ where, $n$ is the total number of observations. Example Diastolic blood pressure (in mmHg) of a sample of 18 patients admitted to the hospitals are as follows:

Outliers for ungrouped data An outlier is any observation that is $1.5(IQR)$ away from the first quartile ($Q_1$) or third quartile ($Q_3$). Formula $x$ is an outlier if $x$ is below $Q_1 -1.5\times IQR$ or above $Q_3+1.5\times IQR$, where, $Q_1$ is the first quartile, $Q_3$ is the third quartile, $IQR = Q_3-Q_1$ is an inter-quartile range. $Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$ where $n$ is the total number of observations.

Paired $t$-test (Dependent Sample) Let $X_1, X_2, \cdots, X_n$ and $Y_1, Y_2, \cdots, Y_n$ be two dependent samples each of size $n$ with respective means $\mu_1$ and $\mu_2$. Define $d_i = X_i - Y_i$, $i=1,2,\cdots, n$. Then $\mu_d= \mu_1 - \mu_2$. Let $\overline{d}=\frac{1}{n} \sum d_i$ be the mean of the difference and $s_d=\sqrt{\frac{1}{n-1}\sum (d_i - \overline{d})^2}$ be the sample standard deviation of the difference. Assumptions a. The samples are dependent (matched pairs).

Paired $t$-test (Dependent Sample) In this tutorial we will discuss some numerical examples on paired t test. Example 1 A new prep class was designed to improve AP statistics test scores. Five students were selected at random. The numbers of correct answers on two practice exams were recorded; one before the class and one after. The data recorded in the table below. We want to test if the numbers of correct answers, on average, are higher after the class.

Percentiles for grouped data Percentiles are the values which divide whole distriution into hundred equal parts. They are 99 in numbers namely $P_1, P_2, \cdots, P_{99}$. Here $P_1$ is first percentile, $P_2$ is second percentile and so on. Formula For discrete frequency distribution, the formula for $i^{th}$ percentile is $P_i =\bigg(\dfrac{i(N)}{100}\bigg)^{th}$ value, $i=1,2,\cdots, 99$ where, $N$ is total number of observations. For continuous frequency distribution, the formula for $i^{th}$ percentile is

Percentiles for ungrouped data Percentiles are the values of arranged data which divide whole data into hundred equal parts. They are 9 in numbers namely $P_1,P_2, \cdots, P_{99}$. Here $P_1$ is first percentile, $P_2$ is second percentile, $P_3$ is third percentile and so on. Formula The formula for $i^{th}$ percentile is $P_i =$ Value of $\bigg(\dfrac{i(n+1)}{100}\bigg)^{th}$ observation, $i=1,2,3,\cdots,99$ where $n$ is the total number of observations. Example 1 The test score of a sample of 20 students in a class are as follows:

Introduction In this article we will discuss step by step procedure to construct a plus four confidence interval for population proportion. The plus four confidence interval for the population proportion can be used when the confidence coefficient is more than 90% and the sample size of the population is at least 10. Plus Four Confidence Interval for Proportion Let $X$ be the observed number of individuals possessing certain attributes (number of successes) in a random sample of size $n$ from a large population with population proportion $p$.

Plus Four Confidence Interval for Proportion In this article we will discuss step by step examples to construct a plus four confidence interval for population proportion. Example 1 Using Plus Four Confidence Interval for Proportion In a random sample of 60 students from a college, 32 opted mathematics as major subject. Using plus four method find a 95% confidence interval for the proportion of students who opted mathematics as a major subject.

Plus Four CI for difference between two population proportions The plus four method produces more accurate confidence interval for difference between two population proportions. We assume that we had four additional observations. One success and one failure in each of the two samples. The plus four method when the confidence coefficient is at least 90% and both the sample sizes are at least 5. Let $X_1$ be the observed number of individuals possessing certain attributes (number of successes) in a random sample of size $n_1$ from a large population with population proportion $p_1$ and let $X_2$ be the observed number of individuals possessing certain attributes (number of successes) in a random sample of size $n_2$ from a large population with population proportion $p_2$.

Plus Four CI for difference between two population proportions In this tutorial we will discuss some step by step numerical examples to estimate plus four confidence interval for the difference between two population proportions. Example For two independent samples A and B following information is available: Sample A has 790 number of successes from a sample size of 805. Sample B has 798 number of successes from a sample size of 811.

Poisson approximation to binomial distribution Let $X$ be a binomially distributed random variable with number of trials $n$ and probability of success $p$. The mean of $X$ is $\mu=E(X) = np$ and variance of $X$ is $\sigma^2=V(X)=np(1-p)$. The general rule of thumb to use Poisson approximation to binomial distribution is that the sample size $n$ is sufficiently large and $p$ is sufficiently small such that $\lambda=np$ (finite). For sufficiently large $n$ and small $p$, $X\sim P(\lambda)$.

Poisson Distribution Poisson distribution helps to describe the probability of occurrence of a number of events in some given time interval or in a specified region. The time interval may be of any length, such as a minutes, a day, a week etc. Definition of Poisson Distribution A discrete random variable $X$ is said to have Poisson distribution with parameter $\lambda$ if its probability mass function is $$ \begin{equation*} P(X=x)= \left\{ \begin{array}{ll} \frac{e^{-\lambda}\lambda^x}{x!

Quartile Deviation for ungrouped data Quartile Deviation is given by $QD = \dfrac{Q_3-Q_1}{2}$ Coefficient of Quartile Deviation is given by Coefficient of $QD = \dfrac{Q_3-Q_1}{Q_3+Q_1}$ where $Q_1$ is the first quartile $Q_3$ is the third quartile The formula for $i^{th}$ quartile is $Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$ where $n$ is the total number of observations. Example 1 A random sample of 15 patients yielded the following data on the length of stay (in days) in the hospital.

Quartile for grouped data Quartiles are the values which divide whole distriution into four equal parts. They are 3 in numbers namely $Q_1$, $Q_2$ and $Q_3$. Here $Q_1$ is first quartile, $Q_2$ is second quartile and $Q_3$ is third quartile. Quartile formula for grouped data For discrete frequency distribution, the formula for $i^{th}$ quartile is $Q_i =\bigg(\dfrac{i(N)}{4}\bigg)^{th}$ value, $i=1,2,3$ where, $N$ is total number of observations. For continuous frequency distribution, the formula for $i^{th}$ quartile is

Quartiles for ungrouped data Quartiles are the values of arranged data which divide whole data into four equal parts. They are 3 in numbers namely $Q_1$, $Q_2$ and $Q_3$. Here $Q_1$ is first quartile, $Q_2$ is second quartile and $Q_3$ is third quartile. The formula for $i^{th}$ quartile is $Q_i =$ Value of $\bigg(\dfrac{i(n+1)}{4}\bigg)^{th}$ observation, $i=1,2,3$ where $n$ is the total number of observations. Example 1 A random sample of 15 patients yielded the following data on the length of stay (in days) in the hospital.

Row Minima Method Step 1 Select the smallest cost in the first row of the transportation table. Let it be $c_{1j}$. Allocate as much as possible amount $x_{1j} = min_j(a_1, b_j)$ in the cell $(1,j)$, so that either the capacity of origin $O_1$ is exhausted or the requirement at destination $D_j$ is satisfied or both. Step 2 If $x_{1j} = a_1$, the availability at origin $O_1$ is completely exhausted, cross-out the first row of the table and move down to the second row.

Sample size to estimate mean The minimum sample size required to estimate the population mean $\mu$ is $$ \begin{aligned} n &= \bigg(\frac{z*\sigma}{E}\bigg)^2 \end{aligned} $$ where $\sigma$ is the population standard deviation $z=z_{\alpha/2}$ is the critical value of $Z$ $E$ is the margin of error. Example 1 An efficiency expert wishes to determine the average time that it takes to drill three holes in a certain metal clamp. How large a sample will she need to be 90% confident that her sample mean will be within 15 seconds of the true mean?

Sample size required to estimate means independent samples The formula to estimate the sample size required to estimate the mean of two independent samples is $$ n_i =2\bigg(\frac{z\sigma}{E}\bigg)^2; i=1,2 $$ where $n_i$ is the sample size for $i^{th}$ group, $z$ is the $Z_{\alpha/2}$, $\sigma$ is the population standard deviation and $E$ is the margin of error.

Sample size required to estimate means dependent samples The formula to estimate the sample size required to estimate the difference between means of dependent samples is $$ n =\bigg(\frac{z\sigma_d}{E}\bigg)^2 $$ where, $z$ is the $Z_{\alpha/2}$ (critical value of $Z$), $\sigma_d$ is the population standard deviation of difference between scores, $E$ is the margin of error.

Sample size to estimate proportion The minimum sample size required to estimate the population proportion is $$ n =p*(1-p)\bigg(\frac{z}{E}\bigg)^2 $$ where, $p$ is the proportion of success, $z$ is the $Z_{\alpha/2}$ (critical value of $Z$), $E$ is the margin of error. Example 1 A survey of 1000 randomly selected registered U.S. voters found that 37% were in favor of the recent tax reform plan. A new survey is being proposed to estimate the true proportion in favor of the recent tax reform plan.

Sample size required to estimate proportions $p_1-p_2$ The minimum sample size required to estimate the proportions $p_1-p_2$ is $$ n =\big[p_1*(1-p_1)+p_2*(1-p_2)\big]\bigg(\frac{z}{E}\bigg)^2 $$ where $p_1$ is the proportion of successes in first group, $p_2$ is the proportion of successes in second group, $z$ is the $Z_{\alpha/2}$ and $E$ is the margin of error.

Simple linear regression from raw data Let $(x_i, y_i), i=1,2, \cdots , n$ be $n$ pairs of observations. The simple linear regression model of $Y$ on $X$ is $$y_i=\beta_0 + \beta_1x_i +e_i$$ where, $y$ is a dependent variable, $x$ is an independent variable, $\beta_0$ is an intercept, $\beta_1$ is the slope, $e$ is the error term. Formula By the method of least square, the model parameters $\beta_0$ and $\beta_1$ can be estimated as

Simple linear regression from sum and sum of squares Let $(x_i, y_i), i=1,2, \cdots , n$ be $n$ pairs of observations. The simple linear regression model of $Y$ on $X$ is $$y_i=\beta_0 + \beta_1x_i +e_i$$ where, $y$ is a dependent variable, $x$ is an independent variable, $\beta_0$ is an intercept, $\beta_1$ is the slope, $e$ is the error term. Formula By the method of least square, the model parameters $\beta_0$ and $\beta_1$ can be estimated as

Simple linear regression from summarize data Let $(x_i, y_i), i=1,2, \cdots , n$ be $n$ pairs of observations. The simple linear regression model of $Y$ on $X$ is $$y_i=\beta_0 + \beta_1x_i +e_i$$ where, $y$ is a dependent variable, $x$ is an independent variable, $\beta_0$ is an intercept, $\beta_1$ is the slope, $e$ is the error term. Formula The simple linear regression model parameters $\beta_0$ and $\beta_1$ can be estimated using the method of least square.

Spearman’s Rank correlation Coefficient Let $(x_1, y_1), (x_2, y_2), \cdots , (x_n, y_n)$ be the ranks of $n$ individuals in two characteristics $A$ and $B$ respectively. Formula Then the Spearman’s rank correlation coefficient is denoted by $\varrho$ and is given by $\varrho = 1- \dfrac{6 \sum_{i=1}^{n}d_i^2}{n(n^2-1)}$ where, $d_i = x_i - y_i$ is the difference between the pairs of ranks of the $i^{th}$ individual in the two characteristics and $n$ is the number of pairs.

Summary statistic for grouped data Summary statistic summarize and provide information about the sample data. It includes the minimum value of the data, first quartile ($Q_1$), median (i.e., $Q_2$), mean ($\overline{x}$), third quartile ($Q_3$) and the minimum value of the data. Summary statistic includes minimum value ($\min$), first quartile ($Q_1$), $\text{median }$ ($Q_2$), sample mean ($\overline{x}$), third quartile ($Q_3$), maximum value ($\max$). Formula $\min$, $Q_1$, $\text{median}$, $\overline{x}$, $Q_3$ and $\max$

Summary statistic for ungrouped data Summary statistic summarize and provide information about the sample data. It includes the minimum value of the data, first quartile ($Q_1$), median (i.e., $Q_2$), mean ($\overline{x}$), third quartile ($Q_3$) and the minimum value of the data. Summary statistic includes minimum value ($\min$), first quartile ($Q_1$), $\text{median }$ ($Q_2$), sample mean ($\overline{x}$), third quartile ($Q_3$), maximum value ($\max$). Formula $\min$, $Q_1$, $\text{median}$, $\overline{x}$, $Q_3$ and $\max$

p value of the test p-value of the test is the probability that the test statistic under null hypothesis will take on values as extreme as or more extreme than the observed value of test statistic. The smaller p-value of the test indicate strong evidence against the null hypothesis $H_0$. In this tutorial we discuss about how to find the $p$-value of the t-test. The p-value of the $t$-test depends on the direction of the alternative hypothesis.

One Sample t test for mean In this tutorial we will explain the six steps approach used in hypothesis testing to test hypothesis about the population mean when the population standard deviation is unknown. One sample t test for mean Let $X_1, X_2, \cdots, X_n$ be a random sample from a normal population with mean $\mu$ and unknown variance $\sigma^2$. Let $\overline{x}=\frac{1}{n} \sum x_i$ be the sample mean and $s^2 =\frac{1}{n-1} \sum (x_i -\overline{x})^2$ be the sample variance.

One Sample $t$-Test for mean In this tutorial we will discuss some numerical examples on one sample t-test for testing population mean when population standard deviation is unknown. This tutorial will show you step by step procedure to solve numerical examples on one sample t-test for population mean. Example 1 The Healthy Food Company claims that its cereal boxes contain, on average, 453 grams of cereal. We suspect that the cereal boxes contain, on average, less than claimed.

Two sample t test with unknown and equal variances In this tutorial we will discuss two sample t-test for testing difference between two population means when the population variances are unknown but equal. Two sample t test with unknown and equal variances Let $\overline{x}_1$ be the sample mean and $s_1$ be the sample standard deviation of a random sample of size $n_1$ from a population with mean $\mu_1$ and variance $\sigma^2_1$.

Two sample t test for means with unknown but equal variances In this tutorial we will discuss some numerical examples on two sample t test for difference between two population means when the population variances are unknown but equal. Example 1 A high school language course is given in two sections, each using a different teaching method. The first section has $21$ students, and the grades in that section have a mean of $82.

Two sample t test for means with unknown and unequal variances In this tutorial we will discuss two sample t test for testing difference between two population means when the population variances are unknown and unequal. Two sample t test for means with unknown and unequal variances Let $\overline{x}_1$ be the sample mean and $s_1$ be the sample standard deviation of a random sample of size $n_1$ from a population with mean $\mu_1$ and variance $\sigma^2_1$.

Two Sample $t$-Test (Independent Sample with Unequal Variances) In this tutorial we will discuss some numerical examples on two sample t test for difference between two population means when the population variances are unknown and unequal. Example 1 A statistician claims that the average score on logical reasoning test taken by students who major in Physics is less than that of students who major in English. The result of the exams, given to 22 Physics students and 33 English students, is shown here.

Testing Correlation Coefficient Let $(x_i, y_i), i=1,2,\cdots, n$ be a random sample of $n$ pairs of observations drawn from a bivariate normal population with correlation coeffficient $\rho$. Let $r_{xy}$ or $r$ be the observed correlation coefficient between $x$ and $y$. A test of significance for a linear relationship between the variables $x$ and $y$ can be performed using the sample correlation coefficient $r_{xy}$. We wish to test the hypothesis $H_0 : \rho =0$ (there is no significant linear relationship between $x$ and $y$) against $H_a : \rho \neq 0$ (there is a significant linear relationship between $x$ and $y$).

Testing Significance of Linear Relationship A test of significance for a linear relationship between the variables $x$ and $y$ can be performed using the sample correlation coefficient $r_{xy}$. Example 1 For a sample of eight bears, researchers measured the distances around the bears’ chests and weighed the bears. The Sample correlation coefficient between the chest size and weight of bears is $r=0.744$ for $n=8$ bears. Using $\alpha=0.05$, determine if there is a positive linear correlation between chest size and weight.

Testing Correlation Coefficient Let $(X_i, Y_i), i=1,2,\cdots, n$ be a random sample of $n$ pairs of observations drawn from a bivariate population with correlation coeffficient $\rho$. Let $r$ be the observed correlation coefficient between $X$ and $Y$. We wish to test the hypothesis $H_0 : \rho =\rho_0$ against $H_a : \rho \neq \rho_0$. Assumptions The population from which, the samples drawn, is a bivariate normal. The relationship between $X$ and $Y$ is linear.

Testing Correlation Coefficient In this tutorial we will discuss step by step solution of numerical problems on testing whether the population correlation coefficient is $\rho_0$ or not. Example 1 The median records shows that the correlation between the age of the mother and the birth weight of their first child is less than -0.34. A random sample of 8 mother’s age and the birth weight of their first child are as follows:

Testing Homogeneity of Two Correlation Coefficient We wish to test the null hypothesis that the correlation $\rho_1$ between $X$ and $Y$ in one population is the same as the correlation $\rho_2$ between $X$ and $Y$ in another population. Let $n_1$ be the sample pairs of observation from first population with sample correlation coefficient $r_1$ and $n_2$ be the sample pairs of observation from second population with sample correlation coefficient $r_2$.

Testing Homogeneity of Two Correlation Coefficient In this tutorial we will discuss step by step solution to some problems on testing homogeneity of two correlation coefficients. Example 1 The correlation between the mathematics scores and statistics scores of group of 50 male students is 0.358 and the correlation between the mathematics scores and statistics scores of groups of 40 female students is 0.569. Is the relationship between mathematics scores and statistics scores for male students is weaker than for female students?

Transhipment Problem The transhipment problem is more general than that of regular transportation problem, where direct shipments only are allowed between a source and destination. In transhipment problem available commodity frequently moves from one source to another source or destination before reaching its actual destination. In Transhipment Network, the nodes of the network that acts as both source and destination are called transhipment nodes. The nodes of the network which acts as source only are called pure supply nodes and the nodes which acts as destination only are called pure demand nodes.

Transportation Problem Transportation problem is a special class of linear programming problem that deals with transporting (or shipping) a commodity from various origins or sources (e.g. factories) to various destinations or sinks (e.g., warehouses). In this type of problem the objective is to determine the transportation schedule that minimizes the total transportation cost while satisfying supply and demand conditions. The general Transportation Problem (TP) can be defined as follows: Suppose that there are $m$ origins $O_1, O_2, \ldots, O_m$ and $n$ destinations $D_1, D_2, \ldots, D_n$.

Travelling Salesman Problem Suppose a salesman wants to visit certain number of cities, say, $n$. Let $c_{ij}$ be the distance from city $i$ to city $j$. Then the problem of salesman is to select such a route that starts from his home city, passes through each city once and only once, and returns to his home city in the shortest possible distance. Such a problem is known as Travelling Salesman Problem.

UV Method For Finding Optimal Solution of TP Step 1 First, construct a transportation table entering the origin capacities $a_i$, the destination requirements $b_j$ and the cost $c_{ij}$. (If the TP is unbalanced convert it into a balanced TP by adding a dummy row or dummy column as per the requirement taking zero costs). Step 2 Find an initial basic feasible solution by any one method (preferably Vogel’s Approximation Method).

Variance and standard deviation for grouped data Let $(x_i,f_i), i=1,2, \cdots , n$ be the observed frequency distribution. Formula Sample variance The sample variance of $X$ is denoted by $s_x^2$ and is given by $s_x^2 =\dfrac{1}{N-1}\sum_{i=1}^{n}f_i(x_i -\overline{x})^2$ OR $s_x^2 =\dfrac{1}{N-1}\bigg(\sum_{i=1}^{n}f_ix_i^2-\frac{\big(\sum_{i=1}^n f_ix_i\big)^2}{N}\bigg)$ where, $N=\sum_{i=1}^n f_i$ is the total number of observations, $\overline{x}$ is the sample mean. Sample standard deviation The sample standard deviation of $X$ is defined as the positive square root of sample variance.

Variance and standard deviation for ungrouped data Let $x_i, i=1,2, \cdots , n$ be $n$ observations. Formula Sample variance of $X$ is denoted by $s_{x}^2$ and is given by $s_x^2 =\dfrac{1}{n-1}\sum_{i=1}^{n}(x_i -\overline{x})^2=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\dfrac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)$ where, $\overline{x}=\dfrac{1}{n}\sum_{i=1}^{n}x_i$ is the sample mean The sample standard deviation of $X$ is defined as the positive square root of the sample variance. The sample standard deviation of $X$ is given by $s_x =\sqrt{s_x^2}$ Example 1 The age (in years) of 6 randomly selected students from a class are

Vogel’s Approximation Method (VAM) Vogel’s approximation method is an improved version of the least cost entry method. It gives better starting solution as compared to any other method. Step 1 For each row (column), determine the penalty measure by subtracting the smallest unit cost element in the row (column) from the next smallest unit cost element in the same row (column). Step 2 Select the row or column with the largest penalty.

Weibull Distribution A continuous random variable $X$ is said to have a Weibull distribution with three parameters $\mu$, $\alpha$ and $\beta$ if the random variable $Y =\big(\frac{X-\mu}{\beta}\big)^\alpha$ has the exponential distribution with p.d.f. $$ \begin{equation*} f(y) = e^{-y}, y>0. \end{equation*} $$ The probability density function of Weibull random variable $X$ is $$ \begin{equation*} f(x;\alpha, \beta)=\left\{ \begin{array}{ll} \frac{\alpha}{\beta} \big(\frac{x-\mu}{\beta}\big)^{\alpha-1}e^{-\big(\frac{x-\mu}{\beta}\big)^\alpha}, & \hbox{$x>\mu$, $\alpha, \beta>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$

p value of the test p-value of the test is the probability that the test statistic under null hypothesis will take on values as extreme as or more extreme than the observed value of test statistic. The smaller p-value of the test indicate strong evidence against the null hypothesis $H_0$. In this tutorial we discuss about how to find the $p$-value of the Z-test. The p-value of the $Z$-test depends on the direction of the alternative hypothesis.

One sample Z test for mean In this tutorial we will explain the six steps approach used in hypothesis testing to test hypothesis about the population mean when the population standard deviation is known. One Sample Z Test For Population Mean Let $X_1, X_2, \cdots, X_n$ be a random sample from a normal population with mean $\mu$ and known variance $\sigma^2$. Let $\overline{x}=\frac{1}{n} \sum X_i$ be the sample mean. Assumptions a.

Z-test for testing population mean In this tutorial we will discuss some numerical examples on one sample z test for population mean. Example 1 The average cost of tuition and room and board at a small private liberal arts college is reported to be $8,500 per term, but a financial administrator believes that the average cost is lower. A study conducted using 350 small liberal arts colleges showed that the average cost per term is $8,445.

$Z$-Test for Proportion Let $X$ be the observed number of individuals possessing certain attributes (say, number of successes) in a random sample of size $n$ from a large population, then $\hat{p}=\frac{X}{n}$ be the observed proportion of successes. Let $p$ be the population proportion of successes and $q = 1- p$ be the population proportion of failures. Assumptions Assumptions for testing a proportion are as follows: a. The sample is a random sample.

Z-Test for Proportion In this tutorial we will discuss some numerical examples on one sample Z test for testing population proportion. Example 1 An insurance company states that 90% of its claims are settled within 30 days. A consumer group selected a random sample of 75 of the company’s claims to test this statement. If the consumer group found that 55 of the claims were settled within 30 days, do they have sufficient reason to support their contention that fewer than 90% of the claims are settled within 30 days?

Two sample Z test for means In this tutorial we will discuss two sample Z test for testing the difference between means of two populations when the population standard deviations are known. In practice two sample Z test is not used often, because in reality the population standard deviations $\sigma_1$ and $\sigma_2$ are unknown. In such a situation, the sample standard deviations are used and two sample t test (Equal variances) or two sample t-test (Unequal variances) is used.

Two sample Z test for means In this tutorial we will discuss some numerical examples on two sample z test for testing difference between the means. Example 1 An industrial engineer would like to determine whether there are more units produced on the night shift than on the day shift. Assume that the population standard deviation for the number of units produced on the day shift is 21 and is 28 on the night shift.

Two sample proportion test Suppose we want to compare two distinct populations $A$ and $B$ with respect to possessions of certain attribute among their members. Suppose take samples of sizes $n_1$ and $n_2$ from the population A and B respectively. Let $X_1$ and $X_2$ be the observed number of successes i.e., number of units possessing the attributes, from the two samples respectively. Then, $\hat{p}_1=\frac{X_1}{n_1}$ be the observed proportion of successes in the sample from population $A$.

$Z$-Test for difference of Proportions In this tutorial we will discuss some numerical examples on two sample Z test for proportions using traditional approach and p value approach. Example 1 A survey indicate that of 900 women randomly sampled, 345 use smartphones. For the men, 450 of the 1025 who were randomly sampled use smartphones. Test whether a percentage of women who uses smartphone is less than men. Use $\alpha=0.05$.

Bernoulli Distribution In this tutorial, we will discuss example on Bernoulli’s distribution. The discrete random variable $X$ is said to have Bernoulli distribution if its probability mass function is given by $$ \begin{equation*} P(X=x) = p^x q^{1-x}, \; x=0,1; 0<p,q<1; q=1-p. \end{equation*} $$ Example The battery from a lot of batteries has 85% chances of non-defective. We select a batter at random from a lot. a. What is the probability distribution of non-defective battery?

Binomial Distribution In this tutorial, we will provide you step by step solution to some numerical examples on Binomial distribution to make sure you understand the Binomial distribution clearly and correctly. Upon successful completion of this tutorial, you will be able to understand how to calculate binomial probabilities. Definition of Binomial Distribution A discrete random variable $X$ is said to have Binomial distribution with parameters $n$ and $p$ if the probability mass function of $X$ is $$ \begin{eqnarray*} P(X=x) & =& \binom{n}{x} p^x q^{n-x},\\ & & \qquad \; x = 0,1,2, \cdots, n; \;\\ & & 0 \leq p \leq 1, q = 1-p \end{eqnarray*} $$ where,

Continuous Uniform Distribution The continuous uniform distribution is the simplest probability distribution where all the values belonging to its support have the same probability density. It is also known as rectangular distribution. This tutorial will help you understand how to solve the numerical examples based on continuous uniform distribution. Definition of Uniform Distribution A continuous random variable $X$ is said to have a Uniform distribution (or rectangular distribution) with parameters $\alpha$ and $\beta$ if its p.

Discrete uniform distribution In this tutorial we will discuss some examples on discrete uniform distribution and learn how to compute mean of uniform distribution, variance of uniform distribution and probabilities related to uniform distribution. Definition of Discrete Uniform Distribution A discrete random variable $X$ is said to have a uniform distribution if its probability mass function (pmf) is given by $$ \begin{aligned} P(X=x)&=\frac{1}{N},\;\; x=1,2, \cdots, N. \end{aligned} $$ The expected value of discrete uniform random variable is $E(X) =\dfrac{N+1}{2}$.

Exponential Distribution In this tutorial, we will provide you step by step solution to some numerical examples on exponential distribution to make sure you understand the exponential distribution clearly and correctly. Definition of Exponential Distribution A continuous random variable $X$ is said to have an exponential distribution with parameter $\theta$ if its p.d.f. is given by $$ \begin{equation*} f(x)=\left\{ \begin{array}{ll} \theta e^{-\theta x}, & \hbox{$x\geq 0;\theta>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right.

Geometric Distribution In this tutorial, we will provide you step by step solution to some numerical examples on geometric distribution to make sure you understand the geometric distribution clearly and correctly. Definition of Geometric Distribution A discrete random variable $X$ is said to have geometric distribution with parameter $p$ if its probability mass function is given by $$ \begin{equation*} P(X=x) =\left\{ \begin{array}{ll} q^x p, & \hbox{$x=0,1,2,\ldots$} \\ & \hbox{$0<p<1$, $q=1-p$} \\ 0, & \hbox{Otherwise.

Laplace Distribution A continuous random variable $X$ is said to have a Laplace distribution, if its p.d.f. is given by $$ \begin{equation*} f(x;\mu, \lambda)=\left\{ \begin{array}{ll} \frac{1}{2\lambda}e^{-\frac{|x-\mu|}{\lambda}}, & \hbox{$-\infty < x< \infty$;} \\ & \hbox{$-\infty < \mu < \infty$, $\lambda >0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$ Distribution Function Distribution function of $L(\mu,\lambda)$ distribution is $$ \begin{equation*} F(x) = \left\{ \begin{array}{ll} \frac{1}{2}e^{\frac{(x-\mu)}{\lambda}}, & \hbox{$x< \mu$;} \\ &\\ 1-\frac{1}{2}e^{-\frac{(x-\mu)}{\lambda} }, & \hbox{$x\geq \mu$;} \end{array} \right.

Definition of Log-normal distribution The continuous random variable $X$ has a log-normal distribution if the random variable $Y=\ln (X)$ has a normal distribution with mean $\mu$ and standard deviation $\sigma$. The probability density function of $X$ is $$ \begin{aligned} f(x) & = \frac{1}{\sqrt{2\pi}\sigma x}e^{-\frac{1}{2\sigma^2}(\ln x -\mu)^2},x\geq 0 \end{aligned} $$ In Log-normal distribution $\mu$ is called location parameter, since it locates the curve of the distribution, and $\sigma$ is called scale parameter, since the shape of the curve depends on the value of $\sigma$.

Log-normal distribution Normal distribution is not suitable when the data are highly skewed or data contains outliers. In such a situation, log-normal distribution is often a good choice. If $Y\sim N(\mu,\sigma^2)$ then $X= e^Y$ follows a log-normal distribution with parameter $\mu$ and $\sigma^2$. Definition of Log-normal Distribution The continuous random variable $X$ has a Log-normal Distribution if the random variable $Y=\ln (X)$ has a normal distribution with mean $\mu$ and standard deviation $\sigma$.

Negative Binomial Distribution In this tutorial, we will provide you step by step solution to some numerical examples on negative binomial distribution to make sure you understand the negative binomial distribution clearly and correctly. Definition of Negative Binomial Distribution A discrete random variable $X$ is said to have negative binomial distribution if its p.m.f. is given by $$ \begin{aligned} P(X=x)&= \binom{x+r-1}{r-1} p^{r} q^{x},\\ & \quad \quad x=0,1,2,\ldots; r=1,2,\ldots\\ & \quad\quad \qquad 0<p, q<1, p+q=1.

Poisson distribution In this tutorial, we will provide you step by step solution to some numerical examples on Poisson distribution to make sure you understand the Poisson distribution clearly and correctly. Definition of Poisson Distribution A discrete random variable $X$ is said to have Poisson distribution with parameter $\lambda$ if its probability mass function is $$ \begin{equation*} P(X=x)= \left\{ \begin{array}{ll} \frac{e^{-\lambda}\lambda^x}{x!} , & \hbox{$x=0,1,2,\cdots; \lambda>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right.

Confidence interval for means (variances are unknown and unequal) Let $x_1, x_2, \cdots, x_{n_1}$ be a random sample of size $n_1$ from a population with mean $\mu_1$ and standard deviation $\sigma_1$. Let $y_1, y_2, \cdots, y_{n_2}$ be a random sample of size $n_2$ from a population with mean $\mu_2$ and standard deviation $\sigma_2$. And the two sample are independent. Let $\overline{x} = \frac{1}{n_1}\sum x_i$ and $\overline{y} = \frac{1}{n_2}\sum y_i$ be the sample means of first and second sample respectively.

Hypergeometric Experiment In this tutorial, we will provide you step by step solution to some numerical examples on hypergeometric distribution to make sure you understand the hypergeometric distribution clearly and correctly. Definition of Hypergeometric Distribution Suppose we have an hypergeometric experiment. That is, suppose there are $N$ units in the population and $M$ out of $N$ are defective, so $N-M$ units are non-defective. Let $X$ denote the number of defective in a completely random sample of size $n$ drawn from a population consisting of total $N$ units.

Sample size to test mean The minimum sample size required to test the hypothesis about the population mean $\mu$ is $$ \begin{aligned} n &= \bigg(\frac{z_{1-\alpha/2}+Z_{1-\beta}}{ES}\bigg)^2 \end{aligned} $$ where $ES=\dfrac{|\mu_1-\mu_0|}{\sigma}$ is the effect size, $\sigma$ is the population standard deviation, $\alpha$ is the level of significance, $1-\beta$ is the power of the test. Example A researcher wishes to test the hypothesis $H_0:\mu=94$ against $H_a:\mu=97$ at $0.05$ level of significance. The population standard deviation is $\sigma= 4.

Sample size required to test means independent samples The $ES$ is defined as $$ ES=\frac{|\mu_1-\mu_2|}{\sigma} $$ where $\mu_1$ is the mean of the first group, $\mu_2$ is the mean of the second group, $\sigma$ is the standard deviation. The formula for determining the sample size required in each group to ensure that the test has a specified power is $$ n =2\bigg(\frac{Z_{1-\alpha/2}+Z_{1-\beta}}{ES}\bigg)^2 $$ where $\alpha$ is the selected level of significance, $1-\beta$ is the selected power and $ES$ is the effect size.

Sample size required to test means dependent samples The $ES$ is defined as $$ ES=\frac{|\mu_d|}{\sigma_d} $$ where $\mu_d$ is the mean of the difference, $\sigma_d$ is the standard deviation of the difference. The formula for determining the sample size required in each group to ensure that the test has a specified power is $$ n =\bigg(\frac{Z_{1-\alpha/2}+Z_{1-\beta}}{ES}\bigg)^2 $$ where $\alpha$ is the selected level of significance, $1-\beta$ is the selected power and $ES$ is the effect size.

Sample size required to test proportion Use this calculator to find the minimum sample size required to test hypothesis about proportion $p$. Sample size required to test proportion The $ES$ is defined as $$ ES=\frac{|p_1-p_0|}{\sqrt{p_0*(1-p_0)}} $$ where $p_0$ is the proportion under null hypothesis, $p_1$ is the proportion under alternative hypothesis. The formula for determining the sample size required to ensure that the test has a specified power is $$ n =\bigg(\frac{Z_{1-\alpha/2}+Z_{1-\beta}}{ES}\bigg)^2 $$

Sample size required to test proportions $p_1-p_2$ The formula for determining the sample size required in each group t ensure that the test has a specified power is $$ n_i =2\bigg(\frac{Z_{1-\alpha/2}+Z_{1-\beta}}{ES}\bigg)^2; i=1,2. $$ where $\alpha$ is the selected level of significance, $1-\beta$ is the selected power and $ES$ is the effect size. The $ES$ is defined as $$ ES=\frac{|p_1-p_2|}{\sqrt{p*(1-p)}} $$ where $p_1$ is the proportion for first group, $p_2$ is the proportion for second group, $p$ is the overall proportion, based on pooling the data from the two comparison groups, that is $p=\frac{p_1+p_2}{2}$.

Normal Distribution A continuous random variable $X$ is said to have a normal distribution if its probability density function is given by $$ \begin{equation*} f(x;\mu, \sigma^2) = \left\{ \begin{array}{ll} \frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2\sigma^2}(x-\mu)^2}, & \hbox{$-\infty< x<\infty$, $-\infty<\mu<\infty$, $\sigma^2>0$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$ where $\mu$ and $\sigma^2$ are the parameters of normal distribution. In notation it can be written as $X\sim N(\mu,\sigma^2)$. Mean Normal distribution The mean of $N(\mu,\sigma^2)$ distribution is $E(X)=\mu$.

CI for difference between two population means with unknown and unequal variances In this tutorial we will discuss step by step solution to some numerical problems on confidence interval for difference in means when the population variances are unknown and unequal. Example 1 To compare the mean lifespans of African elephants in the wild and in a zoo, you randomly select several lifespans from both locations. The results are given below:

Weibull Distribution In this tutorial we will discuss about the Weibull distribution and examples. Weibull distribution is a continuous probability distribution. Weibull distribution is one of the most widely used probability distribution in reliability engineering. This tutorial help you to understand how to calculate probabilities related to Weibull distribution and step by step guide on Weibuill Distribution Examples for different numerical problems. Three parameter Weibull Distribution A continuous random variable $X$ is said to have a Weibull distribution with three parameters $\mu$, $\alpha$ and $\beta$ if the probability density function of Weibull random variable $X$ is

Geometric mean for grouped data Let $(x_i,f_i), i=1,2, \cdots , n$ be the given frequency distribution then the geometric mean of $X$ is denoted by $GM$. Formula The geometric mean of grouped data is given by $GM=\bigg(\prod_{i=1}^n x_i^{f_i}\bigg)^{1/N}$ OR $\log (GM) =\frac{1}{N}\sum_{i=1}^n f_i\log(x_i)$ where, $N=\sum_i f_i$ total number of observations Example Compute geometric mean for the following frequency distribution. x 5-8 9-12 13-16 17-20 21-24 f 2 13 21 14 5 Solution Class Interval Class Boundries mid-value ($x_i$) Freq ($f_i$) $log(x_i)$ $f_i*log(x_i)$ 5-8 4.

Quartile Deviation for grouped data Quartile deviation is half the difference between the thrid quartile $Q_3$ and the first quartile $Q_1$ of a frequency distribution. It is also known as Semi interquartile range. Quartile Deviation is given by $QD = \dfrac{Q_3-Q_1}{2}$ where, $Q_1$ is the first quartile $Q_3$ is the third quartile Quartile Deviation Formula for Grouped Data The formula for $i^{th}$ quartile is $$ \begin{aligned} Q_i=l + \bigg(\frac{\frac{iN}{4} - F_<}{f}\bigg)\times h; \quad i=1,2,3 \end{aligned} $$