A random sample of 41 observations from a normal population possessed a mean \(\bar x = 88\) and a standard deviation s = 6.9.

a. Test \({H_0}:{\sigma ^2} = 30\) against \({H_a}:{\sigma ^2} > 30\). Use\(\alpha = 0.05.\)

The number of samples is, n=41.

The sample mean,\(\bar x = 88.\)

The sample standard deviation is, s=6.9.

The test statistic for the large sample test of Hypothesis about \({\sigma ^2}\) is,

\(\chi _c^2 = \frac{{\left( {n - 1} \right){s^2}}}{{\sigma _0^2}}\)

Where n is the number of sample size, s is the sample variance, and \(\sigma _0^2\) is the population variance.\(\)

a.

The null and alternative hypothesis are:

\(\begin{aligned}{H_0}:{\sigma ^2} = 30\\{H_0}:{\sigma ^2} > 30\end{aligned}\)

The significance level is, \(\alpha = 0.05.\)

The test statistic is computed as:

\(\begin{aligned}\chi _c^2 &= \frac{{\left( {n - 1} \right){s^2}}}{{\sigma _0^2}}\\ &= \frac{{\left( {41 - 1} \right){{\left( {6.9} \right)}^2}}}{{30}}\\ &= \frac{{40 \times 47.61}}{{30}}\\ &= 63.48\end{aligned}\)

This is a one-tailed test. The rejection region is, \(\chi _c^2 > \chi _\alpha ^2.\)

The degrees of freedom is computed as:

\(\begin{aligned}\left( {n - 1} \right) &= \left( {41 - 1} \right)\\ &= 40\end{aligned}\)

The \(\chi _\alpha ^2\) value with \(\alpha = 0.05\)and degree of freedom 40 obtained from the Chi-square table is,

\(\begin{aligned}\chi _\alpha ^2 &= \chi _{0.05}^2\\ &= 55.76\end{aligned}\)

Here, \(\chi _c^2 > \chi _\alpha ^2 \Rightarrow 63.48 > 55.76\).The test statistic \(\chi _c^2\)falls in a rejection region. So, we reject the null hypothesis.

Hence, there is no sufficient evidence to conclude that the true value of the variance is 30.