Answer to Question #263102 in Statistics and Probability for John

Question

Test Statistic for σ2 is given by: χ2 = (n −1)s 2 σ2 ; which is Chi-squared distribbuted with ν = n −1 degrees of freedom when the population random variable is normally distributed with variance equal to σ2 3 (a) A random sample of n = 100 observations was drawn from a normal population. The sample variance was calculated to be s 2 = 220. Test with α = 0.05 to determine whether we can infer that the population variance differs from 300. (b) Repeat part (a) changing the sample size to n = 50 (c) What is the effect of decreassing the sample size?

Answer 1

From the information we are given,

$n=100,\space s^2=220$

a.

The hypotheses to be tested are,

$H_0:\sigma^2=300,\space against\space H_1:\sigma\not=300$

As given above, the test statistic is,

$\chi^2=(n-1)*s^2/\sigma^2=99*220/300=72.6$

$\chi^2$ is compared with a chi-squared table value at $\alpha=5\%$ with $v=n-1=100-1=99$ degrees of freedom.

The table value is given as,

$\chi^2_{\alpha/2,v}=\chi^2_{0.05/2,99}=\chi^2_{0.025,99}=128.422$ and the null hypothesis is rejected if $\chi^2\gt \chi^2_{0.025,99}$

Since $\chi^2=72.6\lt\chi^2_{0.025,99}=128.422,$ we fail to reject the null hypothesis and conclude that there is insufficient evidence to show that the population variance differs from 300 at 5% level of significance.

b.

when $n=50$,

The test statistic is given as,

$\chi^2=(50-1)*220/300=35.93333$ which is compared with the table value at $\alpha=0.05$ with $v=n-1=50-1=49$ degrees of freedom.

The table value is given as,

$\chi^2_{0.05/2,49}=\chi^2_{0.025,49}=70.22241$ and the null hypothesis is rejected if $\chi^2\gt\chi^2_{0.05,49}$

Since $\chi^2=35.93333\lt\chi^2_{0.025,49}=70.22241$, we fail to reject the null hypothesis and conclude that there is insufficient evidence to show that the population variance differs from 300 at 5% level of significance.

c.

Clearly, decreasing the sample size decreases the value of the test statistic as well as the table value.