Answer to Question #155873 in Statistics and Probability for Cyrille

Question

All freshmen in a particular school were found to have variation in grades expressed as a standard deviation. Two samples among these freshmen, made up of 30 and 50 students each, were found to have means of 90 and 86 with a standard deviation of 2.75 and 1.50 , respectively. Based on their grades, is the first group really brighter than the second group at .05 level of significance? 

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0: \mu_1\leq\mu_2$

$H_1: \mu_1>\mu_2$

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

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{2.75^2}{1.5^2}=3.361$$

$df_1=n_1-1=29, df_2=n_2=1=49, \alpha=0.05$

The critical values are $F_L=0.502$ and $F_U=1.881.$

Since $F=3.361,$ then the null hypothesis of equal variances is rejected.

This t-statistic follows a t-distribution approximately, with estimated degrees of freedom

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

$$=\dfrac{(\dfrac{2.75^2}{30}+\dfrac{1.50^2}{50})^2}{\dfrac{1}{30-1}(\dfrac{2.75^2}{30})^2+\dfrac{1}{50-1}(\dfrac{1.50^2}{50})^2}=39.532,$$

Based on the information provided, the significance level is $\alpha=0.05,$ and the degrees of freedom are $df=39.532,$ , assuming that the population variances are unequal.

It is found that the critical value for this right-tailed test is $t_c=1.6843,$ for $\alpha=0.05$ and $df=39.532.$

The rejection region for this right-tailed test is $R=\{t: t>1.6843\}.$ 

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

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

$$=\dfrac{90-86}{\sqrt{\dfrac{2.75^2}{30}+\dfrac{1.5^2}{50}}}=7.3387$$

Since it is observed that $t=7.3387>1.6843=t_c,$ it is then concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that population mean $\mu_1$ is greater than $\mu_2,$ at the 0.05 significance level.

Using the P-value approach: The p-value is $p=0,$ and since $p=0<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that population mean $\mu_1$ is greater than $\mu_2,$ at the 0.05 significance level.

Therefore, there is enough evidence to claim that the first group really brighter than the second group, at the 0.05 significance level.