Answer to Question #117631 in Statistics and Probability for gly

Question

In an Ergonomics lab activity, specifically that for memory recall, 12 students were given ten minutes to try to memorize a list of 20 nonsense words. Each student was then asked to list as many of the words he or she could remember both one hour and twenty-four hours later. The number of words recalled correctly for each student is shown below:
Student A B C D E F G H I J K L
1-hour later 14 9 18 12 13 17 16 16 19 8 15 7
24-hours later 10 6 14 6 8 10 12 10 14 5 10 5
Determine, at the 95% confidence level, that for all such students, the mean number of words recalled after one hour exceeds that recalled after twenty-four hours by 5 words.

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c: } Student & 1-hour \ later & 24-hours\ later & Difference \\ \hline A & 14 & 10 & 4\\ B & 9 & 6 & 3 \\ C & 18 & 14 & 4 \\ D & 12 & 6 & 6 \\ E & 13 & 8 & 5 \\ F & 17 & 10 & 7 \\ G & 16 & 12 & 4 \\ H & 16 & 10 & 6 \\ I & 19 & 14 & 5 \\ J & 8 & 5 & 3 \\ K & 15 & 10 & 5 \\ L & 7 & 5 & 2 \end{array}$$

$\bar{x_d}=4.5, s_d=1.446, n=12, df=n-1=11$

The following null and alternative hypotheses need to be tested:

$H_0:\mu_d\leq5$

$H_1:\mu_d>5$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a two-tailed test is $t_c=1.796.$

The rejection region for this one-tailed test is $R=\{t:t>1.796\}$.  

The t-statistic is computed as follows:

$$t={\bar{x_d}-\mu_d\over s_d/\sqrt{n}}={4.5-5\over 1.446/\sqrt{12}}\approx-1.198$$

Since it is observed that $t=-1.198<1.796=t_c,$ it is then concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is greater than 5, at the 0.05 significance level.

Using the P-value approach: The p-value is $p=0.8719,$ and since $p=0.8719>0.05,$ it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the population mean $\mu$ is greater than 5, at the 0.05 significance level.