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 XYZ Students without XYZ
9 4
10 5
7 2
9 5
8 2
7 2
10 5
8 3
9 4
7 5

Answer 1

$X_1=8.4$

$s_1^2=1.378$

$X_2=3.7$

$s_2^2=1.789$

Two independent sample t-test for unequal variances is used since it works when variances are not equal.

The test statistic for unequal variances is given by

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

which has a t distribution with m degrees of freedom.

$m=\frac{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}$

$t=\frac{8.4-3.7}{\sqrt{\frac{1.378}{10}+\frac{1.789}{10}}}=8.352$

$m=\frac{(\frac{1.378}{10}+\frac{1.789}{10})^2}{\frac{(\frac{1.378}{10})^2}{9}+\frac{(\frac{1.789}{10})^2}{9}}=17.7=18$

$Cv=t_{0.025,18}=2.101$

Since the test statistic t=8.352 is greater than the critical value cv=2.101, we reject the null hypothesis and conclude that there is enough evidence to support the claim that the XYZ enhancer is effective.