Quick Stats question

1. Oct 30, 2014

xsgx

• OP warned about not using the homework template
Leisure Time In a Gallup poll: 1010 adults were randomly selected and asked if they were satisfied or dissatisfied with the amount of leisure time
that they had. Of this sample 657 said that they were satisfied and 353 said that they were dissatisfied. Use a 0.01 significance level to test the claim that 2/3 of adults are satisfied with the amount of leisure time that they have.

In this case is the null hypotheses 2/3 adults or the calculated proportion (.65) from the sample size?

2. Oct 30, 2014

xsgx

My guess is that the null hypothesis is Ho=.65 and the alternative hypothesis is h1>.65. But I want to be sure.

3. Oct 30, 2014

xsgx

For the first step I calculated P by dividing the amount of people who responded that they do have enough leisure time by the sample size: this gave me a P value of .65 after rounding. Since there is no previously known value of P this becomes the null hypothesis . The ⅔ of adults claim then leads to the alternative hypothesis H1 >.65 because ⅔ = .67 after rounding. I then plugged in the values of .65 for the “previously known proportion”, .67 for P^(the claimed proportion), and the sample size of 1010 into the test statistic for a sample proportion formula from the book: . I used the traditional method of using a critical value, and since the alternative hypothesis contains a “greater than” sign the hypothesis test will be right tailed. Since the significance is a=.01 I calculated the critical value by using statdisk and plugging in .99 into the “cumlative area from the left” tab which gave me a critical value of 2.33, and because the test statistic 1.333 is far below that critical value 2.33 I failed to reject the null hypothesis.

My solution to the problem. Please let me know whether I have made any errors.

4. Oct 30, 2014

RUber

Your null hypothesis should be the claim, i.e. 2/3. Use that to calculate your variance.
Also, if the question is test the validity of the claim, generally you will use a 2-tailed test, so your critical value would be for .995.
I can't imagine that your decision will change, since the proportions are so close anyway, but it might be worth another look.

5. Oct 30, 2014

xsgx

^^ In that case would this be a satisfactory answer? For the first step I calculated P^ by dividing the amount of people who responded that they do have enough leisure time by the sample size: this gave me a P^ value of .65 after rounding. Since there is already an established claim that ⅔ of adults say they are satisfied with the amount of leisure time that becomes the null hypothesis Ho: P=.67 after rounding .Therefore the alternative hypothesis would be H1: P =/ .67 . I then plugged in the values of .65 for the P^, .67 for established/claimed proportion, and the sample size of 1010 into the test statistic for a sample proportion formula from the book: . I used the traditional method of using a critical value,and since the alternative hypothesis contains a =/ sign the hypothesis test will be two tailed. Since the significance is a=.01 I calculated the critical value by using statdisk and plugging in .005 into the “cumlative area from the left” tab which gave me a critical value of -2.58 for the left tail and 2.58, for the right tail, and because the test statistic -1.352 is far above the left critical value -2.58 I failed to reject the null hypothesis.

6. Oct 31, 2014

RUber

All looks right to me.