## 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.

## Example 1

A statistician wishes to test the claim that the standard deviation of the weights of firemen is less than 25 pounds. She selected a random sample of 20 firemen and found $s = 23.2$ pounds. Assuming that the weights of firemen are normally distributed, test the claim of the statistician at the 0.05 level of significance.

### Solution

Given that the sample size $n = 20$ and sample standard deviation $s = 23.2$.

### Step 1 Hypothesis Problem

The hypothesis testing problem is $H_0 : \sigma = 25$ against $H_1 : \sigma < 25$ ($\text{left-tailed}$).

### Step 2 Test Statistic

The test statistic for testing above hypothesis testing problem is

`$$ \chi^2 =\frac{(n-1)s^2}{\sigma^2} $$`

The test statistic follows a chi-square distribution with $n-1$ degrees of freedom.

### Step 3 Level of Significance

The significance level is $\alpha = 0.05$.

### Step 4 Critical Value

As the alternative hypothesis is **$\text{left-tailed}$**, the critical value of $\chi^2$ for $\alpha=0.05$ level of significance and $n-1 = 19$ degrees of freedom $\text{ is }$ $\text{10.117}$ (from $\chi^2$ statistical table).

The rejection region (i.e. critical region) is $\chi^2 < 10.117$.

### Step 5 Test Statistic

The test statistic under the null hypothesis is

`$$ \begin{aligned} \chi^2 &=\frac{(n-1)s^2}{\sigma^2_0}\\ &= \frac{(20-1)*(23.2)^2}{(25)^2}\\ &= 16.362 \end{aligned} $$`

### Step 6 Decision (Traditional approach)

The test statistic is $\chi^2 =16.362$ which falls $outside$ the critical region, we **$\text{fail to reject}$** the null hypothesis.

OR

### Step 6 Decision ($p$-value Approach)

This is a **$\text{left-tailed}$** test, so the p-value is the area to the left of the test statistic ($\chi^2=16.362$) is p-value = $0.367$.

The p-value is $0.367$ which is **$\text{greater than}$** the significance level of $\alpha = 0.05$, we **$\text{fail to reject}$** the null hypothesis.

## Example 2

An engineer is investing the amount of standard deviation in the time it takes a 3D printer to make a particular part. The engineer believes that the standard deviation in the time it takes to make the part is more than 2. Test this at $\alpha = 0.01$ level of significance, using 11 sample times taken while the printer was making these parts. The sample standard deviation is 2.3.

### Solution

Given that the sample size $n = 11$ and sample standard deviation $s = 2.3$.

### Step 1 Hypothesis Problem

The hypothesis testing problem is $H_0 : \sigma = 2$ against $H_1 : \sigma > 2$ ($\text{right-tailed}$)

### Step 2 Test Statistic

The test statistic for testing above hypothesis testing problem is

`$$ \chi^2 =\frac{(n-1)s^2}{\sigma^2} $$`

The test statistic follows chi-square distribution with $n-1$ degrees of freedom.

### Step 3 Level of Significance

The significance level is $\alpha = 0.01$.

### Step 4 Critical Value

As the alternative hypothesis is **$\text{right-tailed}$**, the critical value of $\chi^2$ for $\alpha=0.01$ level of significance and $n-1 = 10$ degrees of freedom $\text{ is }$ $\text{23.209}$ (from $\chi^2$ statistical table).

The rejection region (i.e. critical region) is $\chi^2 > 23.209$.

### Step 5 Test Statistic

The test statistic under the null hypothesis is

`$$ \begin{aligned} \chi^2& =\frac{(n-1)s^2}{\sigma^2_0}\\ &= \frac{(11-1)*(2.3)^2}{(2)^2}\\ &= 13.225 \end{aligned} $$`

### Step 6 Decision (Traditional approach)

The test statistic is $\chi^2 =13.225$ which falls $outside$ the critical region, we **$\text{fail to reject}$** the null hypothesis.

OR

### Step 6 Decision ($p$-value Approach)

This is a **$\text{right-tailed}$** test, so the p-value is the area to the left of the test statistic ($\chi^2=13.225$) is p-value = $0.2114$.

The p-value is $0.2114$ which is **$\text{greater than}$** the significance level of $\alpha = 0.01$, we **$\text{fail to reject}$** the null hypothesis.

## Example 3

A cigarette manufacturer wishes to test the claim that the variance of nicotine content of its cigarettes is 0.644. Nicotine content is measured in milligrams and is assumed normally distributed. A sample of 20 cigarettes has a standard deviation of 1.00 milligram. At $\alpha = 0.01$, is there enough evidence to reject the manufacturer's claim?

### Solution

Given that the sample size $n = 20$ and sample standard deviation $s = 1$.

### Step 1 Hypothesis Problem

The hypothesis testing problem is $H_0 : \sigma^2 = 0.644$ against $H_1 : \sigma^2 \neq 0.644$ ($\text{two-tailed}$)

### Step 2 Test Statistic

The test statistic for testing above hypothesis testing problem is

`$$ \chi^2 =\frac{(n-1)s^2}{\sigma^2} $$`

The test statistic follows chi-square distribution with $n-1$ degrees of freedom.

### Step 3 Level of Significance

The significance level is $\alpha = 0.01$.

### Step 4 Critical Value

As the alternative hypothesis is **$\text{two-tailed}$**, the critical values of $\chi^2$ for $\alpha=0.01$ level of significance and $n-1 = 19$ degrees of freedom $\text{ are }$ $\text{6.844 and 38.582}$ (from $\chi^2$ statistical table).

The rejection region (i.e. critical region) is $\chi^2 < 6.844$ or $\chi^2 > 38.582$.

### Step 5 Test Statistic

The test statistic under the null hypothesis is

`$$ \begin{aligned} \chi^2 &=\frac{(n-1)s^2}{\sigma^2_0}\\ &= \frac{(20-1)*(1)^2}{(0.8024961)^2}\\ &= 29.503 \end{aligned} $$`

### Step 6 Decision (Traditional approach)

The test statistic is $\chi^2 =29.503$ which falls $outside$ the critical region, we **$\text{fail to reject}$** the null hypothesis.

OR

### Step 6 Decision ($p$-value Approach)

This is a **$\text{two-tailed}$** test, so the p-value is the area to the left of the test statistic ($\chi^2=29.503$) is p-value = $0.0585$.

The p-value is $0.0585$ which is **$\text{greater than}$** the significance level of $\alpha = 0.01$, we **$\text{fail to reject}$** the null hypothesis.