Question

In an attempt to compare the performance of students with more than one electronic gadget and those with only one or none, the mean grades of students and standard deviations were taken and shown in the table below.

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c} & mean & standard & sample \\ & &deviation &\\ \hline students & & & \\ with & \bar{x}_1=83 & s_1=10 & n_1=12\\ 0-1\ gadget & & & \\ \hdashline students & & & \\ with\ more & \bar{x}_2=79 & s_2=14 & n_2=12\\ than\ 1\ gadget & & & \end{array}$$

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.

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

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{10^2}{14^2}=\dfrac{25}{49}\approx0.51$$

The critical values are $F_L=0.288$  and $F_U=3.474,$ and since $F=0.51,$

then the null hypothesis of equal variances is not rejected.

The significance level is $\alpha=0.05,$ and the degrees of freedom are $df=12+12-2=22.$  

It is found that the critical value for this two-tailed test is $t_c=2.073873,$for is $\alpha=0.05,$ and $df=22.$  

The rejection region for this two-tailed test is $R=\{t:|t|>2.073873\}.$

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

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

$$=\dfrac{83-79}{\sqrt{(\dfrac{(12-1)10^2+(12-1)14^2}{12+12-2})(\dfrac{1}{12}+\dfrac{1}{12})}}$$

$$\approx0.8054$$

Since it is observed that $|t|=0.8054<2.073873=t_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu_1$

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

Using the P-value approach: the p-value for two-tailed, $t=0.8054, df=22,$ $\alpha=0.05$ is $p=0.4292,$ and since $p=0.4292>0.05=\alpha,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu_1$

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