Answer to Question #242394 in Statistics and Probability for Walid Abdulbagi

Question

IV. Solve the problem.

In a certain college in Metro Manila, a study was conducted to determine whether or not the IQ scores of students who came from provincial high schools differ significantly from those students who came from city high schools. An IQ test was given to 200 (100 from each group) college freshmen and the results are as follows:

Students from Prov. H.S.: x1 = 99 and s1= 5

Students from City H.S.: x2 = 102 and s2 = 8

Set up and test the appropriate statistical hypotheses using 0.05 level of significance.

Answer 1

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 $\alpha=0.05,$ the degrees of freedom are computed as follows, assuming that the population variances are equal:

$$df=n_1+n_2-2=100+100-2=198.$$

The critical value for this right-tailed test $\alpha=0.05, df=198$ degrees of freedom is $t_c=1.972017.$

The rejection region for this right-tailed test is $R=\{t:|t|>1.972017\}.$

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

$$t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{99-102}{\sqrt{\dfrac{(100-1)(5)^2+(100-1)(8)^2}{100+100-2}(\dfrac{1}{100}+\dfrac{1}{100})}}$$

$$\approx-3.1800$$

Since it is observed that $|t|=3.18>1.972017=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for two-tailed test $\alpha=0.05,$ $df=198$ degrees of freedom is $p=0.00171,$ and since $p=0.00171<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$

is different than $\mu_2,$ at the $\alpha = 0.05$ significance level.

The degrees of freedom are computed as follows, assuming that the population variances are unequal:

$$df=\dfrac{(\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2})^2}{\dfrac{(s_1^2/n_1)^2}{n_1-1}+\dfrac{(s_2^2/n_2)^2}{n_2-1}}$$

$$=\dfrac{(\dfrac{(5)^2}{100}+\dfrac{(8)^2}{100})^2}{\dfrac{((5)^2/100)^2}{100-1}+\dfrac{((8)^2/100)^2}{100-1}}\approx166.104427$$

The critical value for this right-tailed test $\alpha=0.05, df=166.104427$ degrees of freedom is $t_c=1.9743.$

The rejection region for this right-tailed test is $R=\{t:|t|>1.9743\}.$

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

$$t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s_2^2}{n_2}}}$$

$$=\dfrac{99-102}{\sqrt{\dfrac{(5)^2}{100}+\dfrac{(8)^2}{100}}}\approx-3.1800$$

Since it is observed that $|t|=3.18>1.9743=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for two-tailed test $\alpha=0.05,$ $df=166.104427$ degrees of freedom is $p=0.001757,$ and since $p=0.001757<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$

is different than $\mu_2,$ at the $\alpha = 0.05$ significance level.