The null hypothesis is that the program did not improve the students' time management skills. The alternate hypothesis is that the program did improve their time management.

I think a better way to test these hypotheses than the one given would be to test the time management skills of students who went through the program against those who didn't (i.e have a control group). Since there isn't a control group, let's see if significantly more than half the students improved.

Because this is a question regarding a proportion (42/60 = 0.7 = 70% of the students responded yes), we use a one-proportion z-test.

The test statistic is calculated as:

Where p-hat is the proportion you're testing against (0.50), p is the proportion you have (0.70), q is 1 - p (0.30), and n is your sample size (60).

(a)

The null hypothesis is p < 0.50 (less than 50% of the students improved).
The alternate hypothesis is p > 0.50 (more than 50% of the students improved).

(b) The critical value for a one-sided z-test is at the 0.05 level is 1.65. Therefore, our decision rule is that we will reject the null hypothesis if the observed value of z is greater than 1.65.

(c) Using the formula above,

z = 0.70 - 0.50 = 0.20/0.059 = 3.38
z = (0.70*0.30/60)

(d) Because the test statistic is larger than the critical value, we can reject the null hypothesis and assume that more than half of the students improved following the program.

------------------------------------------------------------------------------------------------------------

You could also calculate a 95% confidence interval using the following formula:

where p-hat is your percentage (0.70), n is the number of cases (60), and 1.96 is the critical value of the z-distribution.

If the confidence interval did not include 0, that would be evidence that at least some of the students improved, and the program is effective.