Question

When people learn a new task, their performance usually improves when they are tested the next day, but only if they get 6 hours sleep (Sticckgold, et al., 2000). The following data demonstrate this phenomenon. The participants learned a visual discrimination task on one day. Half of the participants were allowed to have at least 6 hours of sleep and the other half were kept awake all night.

6 hours sleep No sleep

n =14 n = 14

M = 72 M =65

SS = 932 SS = 706

Is there a significant difference between the two conditions? Use a two-tailed test with LaTeX: \alpha

α

= .01.

Answer 1

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: ​

Ha:

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.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the degrees of freedom are df=26. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal.

Hence, it is found that the critical value for this two-tailed test is tc​=2.779, for α=0.01 and df=26.

The rejection region for this two-tailed test is R={t:∣t∣>2.779}.

(3) Test Statistics

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

(4) The decision about the null hypothesis

Since it is observed that ∣t∣=0.647≤tc​=2.779, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.5232, and since p=0.5232≥0.01, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is different than μ2​, at the 0.01 significance level.

Graphically

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