Two independent samples of sizes 8 and 7 contained the following values: Sample I : 19 17 15 21 16 18 16 14 Sample II : 16 14 15 19 15 18 16 Do the estimates of the population variance differ significantly at 5% level?

$s^2=\frac{{\sum }_i(x_i-\bar{x}{)}^2}{n-1}$

$\bar{x}_1=\dfrac{19+ 17+ 15 +21+ 16+ 18 +16 +14}{8}=17$

$s_1^2=\dfrac{\sum_i(x_i-17)^2}{8-1}=5.142857$

$\bar{x}_2=\dfrac{16 +14 +15 +19+ 15 +18 +16}{7}=\dfrac{113}{7}$

$s_2^2=\dfrac{\sum_i(x_i-\dfrac{113}{7})^2}{7-1}=3.142857$

The following null and alternative hypotheses need to be tested:

$H_0:\sigma_1^2=\sigma_2^2$

$H_1:\sigma_1^2\not=\sigma_2^2$

This corresponds to a two-tailed test, for which a F-test for two population variances needs to be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ $df_1=8-1=7, df_2=7-1=6$ degrees of freedom, and the the rejection region for this two-tailed test test is 

$R = \{F: F < 0.1954 \text{ or } F > 5.6955\}$

The F-statistic is computed as follows:

$F=\dfrac{s_2^2}{s_1^2}=\dfrac{5.142857}{3.142857}=1.6363$

Since from the sample information we get that

$F_L=0.1954\le F=1.6363\le5.6955=F_R,$

it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population variance $\sigma_1^2$  is different than the population variance $\sigma_2^2,$ at the $\alpha = 0.05$ significance level.