A college student is interested in whether there is a difference between male and female students in the amount of time they spend studying each week. The student gathers information from a random sample of male and female students on campus. The amounts of time spent studying are normally distributed. The data are: Males Emirates 2? 25 25 29 19 18 10 23 16 20 22 15 14 19 a. What statistical test should be used to analyze these data? b. Identify Ho and Ha for this study. c. Conduct the appropriate analysis. d. Should Ha be rejected? What should the researcher conclude? e. If significant, compute and interpret the effect size. f. If significant, draw a graph representing the data. g. Determine the 95% confidence interval.

Given the two samples of male and female collected on the basis of study hours we state the Hypothesis as

Null: H0 : There is no difference in study hours between male and female.

Alternative: H1 : There is difference in study hours between male and female.

Now since the data is normally distributed we can use the 2-sample t-test to test the

Hypothesis. Here we can take our level of significance as 0.05 or a confidence level of 95%.

Thus below are the results of the testing using Minitab software.

Two-Sample T-Test and CI: Males, Females

Two-sample T for Males vs Females

N Mean StDev SE Mean

Males 7 19.00 6.11 2.3

Females 7 21.29 4.72 1.8

Difference = mu (Males) - mu (Females)

Estimate for difference: -2.29

95% CI for difference: (-8.71, 4.14)

T-Test of difference = 0 (vs not =): T-Value = -0.78 P-Value = 0.450 DF = 11

Individual Value Plot of Males, Females

Difference is not significant, thus no graph needed, because of interpretation

Interpretation: Since the P-value(0.450) is greater than the significance level of (0.05) thus we can accept the null hypothesis and say that there is no difference in study hours for male and female.