Answer to Question #141414 in Statistics and Probability for Paul

Question

With using Independent Samples T-test formula, work on the problem.

Suppose that we want to know the effect of XYZ memory enhancer on studen'ts memory level. So we conducted a study with 20 students. 10 of them were given XYZ memory enhancer and the other 10 did not receive the treatment. We measured the memory level of two groups after the study, Using the .05 alpha, was the XYZ enhancer effective? Use the data below for computation.
Students with , Students without
XYZ XYZ
9 4
10 5
7 2
9 5
8 2
7 2
10 5
8 3
9 4
7 5

Answer 1

Using ti84 sample mean and sd of two samples given are calculated

he provided sample means are shown below:

$\bar X_1 = 8.4$ ,$\bar X_2 ==3.7$

Also, the provided sample standard deviations are:

$s_1 = 1.174$ $s_2 = 1.337$ and the sample sizes are /

 $n_1 = 10\\ and\\ n_2 = 10 .$

The following null and alternative hypotheses need to be tested:

$Ho: \mu_1 ​ = \mu_2 \\ ​ Ha: \mu_1 ​ > \mu_2$

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.

Hence, it is found that the critical value for this right-tailed test$t_c = 1.736, for ,\alpha = 0.05 ,$using $df=\frac{\frac{s_{1}^{2}}{n_1}+\frac{s_{_2}^{2}}{n_2}}{\frac{\frac{s_{1}^{2}}{n_1}}{n_1-1}+\frac{\frac{s_{2}^{2}}{n_2}}{n_2-1}}$ = 17.704

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

t=$\frac{\overline{X1}-\overline{X2}}{\sqrt{(\frac{s1^2}{n1})+\frac{s2^2}{n2}}}$ =$\frac{8.4-3.7}{\sqrt{\frac{1.174^2}{10}+\frac{1.337^2}{10}}}$ =8.353

Since it is observed that

t=8.353>tc​=1.736, it is then concluded that the null hypothesis is rejected.

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean

μ1 is greater than μ2 at the 0.05 significance level.

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