Answer to Question #265820 in Statistics and Probability for Diwakar

Question

Two samples of 12 and 18 concrete cubes are weighed. the sample averages are found to be 8.3kg and 8.1kg respectively, with standard deviations of 0.2kg and 0.3kg the population variances are not known. test at 1% significance level whether the 2 samples could be from the same population

Answer 1

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 $\alpha = 0.01,$ and the critical values are $F_L = 0.202$ and $F_U = 4.05.$

The rejection region for this two-tailed test test is $R = \{F: F < 0.305 \text{ or } F > 2.87\}.$

The F-statistic is computed as follows:

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{0.2^2}{0.3^2}=\dfrac{4}{9}\approx0.444$$

Since from the sample information we get that 

$$F_L = 0.202 \le F = 0.444 \le F_U =4.05,$$

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 $\alpha = 0.01$ significance level.

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_1:\mu_1\not=\mu_2$

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ and the degrees of freedom are $df_{Total} = df_1 + df_2 = 12-1 + 18-1 = 28,$ assuming that the population variances are equal.

Hence, it is found that the critical value for this two-tailed test is $t_c = 2.763262,$ for $\alpha = 0.01$ and $df = 28.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.763262\}.$

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t=\dfrac{\bar{X_1}-\bar{X_2}}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{8.3-8.1}{\sqrt{\dfrac{(12-1)0.2^2+(18-1)0.3^2}{12+18-2}(\dfrac{1}{12}+\dfrac{1}{18})}}$$

$$\approx2.0232$$

Since it is observed that $|t| = 2.0232 \le t_c = 2.763262,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for two-tailed $\alpha=0.01, df=28, t=2.0232$ is $p=0.052694,$ and since $p = 0.052694 \ge 0.01=\alpha,$ it is concluded that the null hypothes is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2,$ at the $\alpha = 0.01$ significance level.

Therefore, there is enough evidence to claim that 2 samples could be from the same population, at the $\alpha = 0.01$ significance level.