Confidence Interval for mean-t examples

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. The sample results are as follow:

Sample size: 40,

Sample Mean: 6.5 hrs,

Sample Standard Deviation: 1.3 hrs.

Construct a 99% confidence interval for the average length of usage time to recharge.

Solution

Given that sample size $n = 40$, sample mean $\overline{X}= 6.5$, sample standard deviation $s = 1.3$. The confidence level is $1-\alpha = 0.99$.

Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.99$. Thus, the level of significance is $\alpha = 0.01$.

Step 2 Given information

Given that sample size $n =40$, sample mean $\overline{X}=6.5$, sample standard deviation $s=1.3$.

Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for the population mean $\mu$ is

$$ \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned} $$ where $E = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}$ , and $t_{\alpha/2, n-1}$ is the $t$ value providing an area of $\alpha/2$ in the upper tail of the students’ $t$ distribution.

Step 4 Determine the critical value

The critical value of $t$ for given level of significance and $n-1$ degrees of freedom is $t_{\alpha/2,n-1}$.

Thus $t_{\alpha/2,n-1} = t_{0.005,40-1}= 2.708$ .

Step 5 Compute the margin of error

The margin of error for mean is $$ \begin{aligned} E & = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}\\ & = 2.708 \frac{1.3}{\sqrt{40}} \\ & = 0.557. \end{aligned} $$

Step 6 Determine the confidence interval

$99$% confidence interval estimate for population mean is $$ \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 6.5 - 0.557 & \leq \mu \leq 6.5 + 0.557\\ 5.943 &\leq \mu \leq 7.057. \end{aligned} $$ Thus, $99$% confidence interval estimate for population mean is $(5.943,7.057)$.

Interpretation

We can be $99$% confident that the average length of usage time to recharge is between $5.943$ and $7.057$.

Example 2

Find the 95% confidence interval for the mean number of ounces of coffee that a machine dispenses in 12 ounce cups. Assume the variable is normally distributed. The data is shown below:

12.03, 12.10, 12.02, 11.98, 12.00, 12.05, 11.97.

Solution

	$x_i$	$x_i^2$
	12.03	144.72
	12.10	146.41
	12.02	144.48
	11.98	143.52
	12.00	144.00
	12.05	145.20
	11.97	143.28
Total	84.15	1011.62

Sample mean

The sample mean of $X$ is

$$ \begin{aligned} \overline{x} &=\frac{1}{n}\sum_{i=1}^n x_i\\ &=\frac{84.15}{7}\\ &=12.0214\text{ ounce} \end{aligned} $$

The average of ounces of coffee is $12.0214$ ounce.

Sample variance

Sample variance of $X$ is

$$ \begin{aligned} s_x^2 &=\dfrac{1}{n-1}\bigg(\sum_{i=1}^{n}x_i^2-\frac{\big(\sum_{i=1}^n x_i\big)^2}{n}\bigg)\\ &=\dfrac{1}{6}\bigg(1011.6151-\frac{(84.15)^2}{7}\bigg)\\ &=\dfrac{1}{6}\big(1011.6151-\frac{7081.2225}{7}\big)\\ &=\dfrac{1}{6}\big(1011.6151-1011.60321\big)\\ &= \frac{0.01189}{6}\\ &=0.002 \end{aligned} $$

Sample standard deviation

The sample standard deviation is

$$ \begin{aligned} s_x &=\sqrt{s_x^2}\\ &=\sqrt{0.002}\\ &=0.0445 \text{ ounce} \end{aligned} $$

Sample size $n = 20$, sample mean $\overline{X}= 12.0214$, sample standard deviation $s = 0.0445$.

The confidence level is $1-\alpha = 0.95$.

Step 1 Specify the confidence level $(1-\alpha)$

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

Step 2 Given information

Sample size $n =20$, sample mean $\overline{X}=12.0214$, sample standard deviation $s=0.0445$.

Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for the population mean $\mu$ is

$$ \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned} $$ where $E = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}$ , and $t_{\alpha/2, n-1}$ is the $t$ value providing an area of $\alpha/2$ in the upper tail of the students’ $t$ distribution.

Step 4 Determine the critical value

The critical value of $t$ for given level of significance and $n-1$ degrees of freedom is $t_{\alpha/2,n-1}$ .

Thus $t_{\alpha/2,n-1} = t_{0.025,20-1}= 2.093$ .

Step 5 Compute the margin of error

The margin of error for mean is $$ \begin{aligned} E & = t_{(\alpha/2,n-1)} \frac{s}{\sqrt{n}}\\ & = 2.093 \frac{0.0445}{\sqrt{20}} \\ & = 0.0208. \end{aligned} $$

Step 6 Determine the confidence interval

$95$% confidence interval estimate for population mean is $$ \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 12.0214 - 0.021 & \leq \mu \leq 12.0214 + 0.021\\ 12.0006 &\leq \mu \leq 12.0422. \end{aligned} $$ Thus, $95$% confidence interval estimate for population mean is $(12.0006,12.0422)$.