Question

A sleep center hypothesizes that people who sleep only four hours will score lower than people who sleep for eight hours on a cognitive skills test. The center recruited 20 participants and split them into two groups, giving one group 8 hours of sleep and the other only 4 hours. The following morning, the CAT (Cognitive Ability Test) was conducted, with scores ranging from 1-9, 9 being the best score. Use this information to answer questions . CAT Scores Group X: Eight hrs sleep 4 7 9 4 3 3 8 6 3 7 Group Y: Four hrs sleep 7 8 1 4 2 3 5 2 7 4 Conduct the following hypothesis test: - A one-tail T-test for a two-sample difference in means at the 95% confidence level - with Null Hypothesis that the Group X mean CAT score is equal to the Group Y mean CAT score - and with Alternate Hypothesis that the Group X mean CAT score is greater than the Group Y mean CAT score a). Calculate the mean and standard deviation of the scores for each group. (10%)

b)Using the correct degrees of freedom (df = group X size + group Y size ̶ # of groups), the correct number of tails, and at the correct confidence level, determine the critical value of t. (10%)

c). Explain under which scenarios using a pooled variance be inadvisable, then, calculate the pooled variance (formula for S2 is on page 379) for the groups. (10%)

d). Calculate the test statistic, Ttest (formula for t is on page 380). (10%)

e). The sleep center’s statistician tells you that the p-value for the test is 0.1535. Summarize the result of the study. Compare the mean scores in each group. Compare the test statistic to the critical value. Compare the p-value to alpha. Do you find a statistically significant difference between Group X and Group Y on cognitive test performance? Is there a meaningful/practical difference? Explain your decisions and Justify your claims

Answer 1

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS HA: ( Right tailed test)

a)

b) degrees of freedom= n1+n2-2= 10+10-2= 20-2=18

c) We can not use pooled variance when data are paired.

sp^2​=(n1​−1)s1^2​+(n2​−1)s2^2​​/n1​+n2​−2

Plugging in the corresponding values in the formula above, we get that

sp^2​=(10−1)(2.27)^2+(10−1)(2.41)^2​/10+10−2=5.4805

Now that we have the pooled variance, we can compute the pooled standard deviation by simply taking the squared root of the value obtained for the variance.

The following pooled standard deviation is obtained:

sp​= sqrt(5.4805) ​=2.341

d) test statistic t=

t= 1.05

e) Critical value of t = 1.73 with 18 d.f at 0.05 level of significance.

Since critical value is greater than calculated value of t we therefore do not reject null hypothesis at 0.05 level of significance.

Also P value= 0.1535>0.05 hence not significant we therefore do not reject null hypothesis H0 at 0.05 level of significance.

No there is no a statistically significant difference between Group X and Group Y on cognitive test performance.