CHI SQUARE TEST FOR THE VARIANCE OR STANDARD DEVIATION:

Assuming that the data are normally distributed, the chi-square test can be used for the variance or standard deviation: Is the population variance or standard deviation is equal to a specified value?

The test statistic is:

$\chi^2_{\text{stat}} = \frac{(n-1)\,S^2}{\sigma^2}$

where

$$n = \text{sample size}$$ $$S^2 = \text{sample variance}$$ $$\sigma^2 = \text{hypothetical population variance}$$

We want to know if the standard deviation has changed from a hypthetical $$5$$ mL in a bottling soda company Thus, you use a two-tail test with the following null and alternative hypotheses:

$$H_0: \sigma^2 = 25$$ ($$\sigma = 5$$ mL) $$H_1: \sigma^2 \neq 25$$.

With a sample of 30 bottles, you reject the null hypothesis if the computed test statistic falls into either the lower or upper tail of a chi-square distribution with $$n-1 = 30 - 1 =29$$ degrees of freedom,

If the sample standard deviation is $$S = 8$$ mL, the test statistic would be:

$$\chi^2 = \frac{(30-1)\,(8)^2}{5^2}=74.24$$

The theshold values with a risk alpha of $$0.05$$ are:

qchisq(c(0.025, 1 - 0.025), 29)
## [1] 16.04707 45.72229

Hence we reject $$H_0$$, in favor of the alternative hypthesis that the variance has increased with a p value of:

(p = 1 - pchisq(74.24, 29))
## [1] 7.803442e-06