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

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