Question

A researcher wishes to determine the performance level of students with pre-school education and with no preschool education. Fifteen senior high school students with pre-school education and fifteen senior high school students with no pre-school education were randomly selected and the results are tabulated below.

Education

 (mean)

 (standard deviation)

 (sample size)

With pre-school education

89.4

3.25

15

With no pre-school education

85.2

2.85

15

Is it safe to conclude that those with pre-school education have a better performance than those with no pre-school education? Use 99% level of confidence

Answer 1

Testing for Equality of Variances

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

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{3.25^2}{2.85^2}=1.3004$$

The critical values for $\alpha=0.01, df_1-n_1=1=14$degrees o freedom, $df_2=n_2-1=14$ degrees o freedom, are $F_L = 0.2326$ and $F_U = 4.2993,$ and since $F = 1.3004,$ then the null hypothesis of equal variances is not rejected.

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1\le\mu_2$

$H_a:\mu_1>\mu_2$

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.

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

$$df=n_1+n_2=15+15-2=28$$

Based on the information provided, the significance level is $\alpha = 0.01,$

and the degrees of freedom are $df = 28$ degrees of freedom.

Hence, it is found that the critical value for this right-tailed test is $t_c = 2.46714,$ for $\alpha = 0.01$ and $df = 28.$

The rejection region for this right-tailed test is $R = \{t: t > 2.46714\}.$

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

$$t=\dfrac{\bar{X}_1-\bar{X}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$ $$=\dfrac{89.4-85.2}{\sqrt{\dfrac{(15-1)(3.25)^2+(15-1)(2.85)^2}{15+15-2}(\dfrac{1}{15}+\dfrac{1}{15})}}$$ $$\approx3.7631$$

Since it is observed that $t = 3.7631> t_c = 2.46714,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for right-tailed test, $df=28$ degrees of freedom, $t=3.7631,$ is $p=0.000395,$ and since $p = 0.000395< 0.01=\alpha,$ it is concluded that the null hypothes is rejected.

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

is greater than $\mu_2,$ at the $\alpha = 0.01$ significance level.