Question

From the 2008 General Social Survey, females and males were asked about the number of hours a day that the subject watched TV. Females (n = 698) reported a mean of 3.08 hours with a standard deviation of 2.70 hours. Males (n = 626) reported a mean of 2.87 hours with a standard deviation of 2.61 hours. Test that the mean hours of TV watched by men and women is different from zero at the 5% significance level.

(A) What are the null and alternative hypotheses?

(B) Based on the significance level at which you are testing, what is (are) the critical value(s) for the test?

a) it is a two-sided test. Thus, the tcritis ±1.960.

b) it is a two-sided test. Thus, the tcrit is ±1.645.

c) A one-sided-test with tcrit = 1.96

d) A one-sided test with tcrit = -1.96

(C.1) Calculate the appropriate test statistic. What is the standard error you calculated?

(C.2)Calculate the appropriate test statistic. What is the test statistic you calculated?

(D) Calculate the corresponding p-value from the appropriate table or online calculator.

(E) What conclusions can you draw from the hypothesis test? Be sure to comment on evidence from both the test statistic and p-value.

(F) Which conclusion is true according to your results?

a) |tcalc|<|tcrit|andp>α

Thus I cannot reject the plausibility of zero hours for the true mean difference in hours of TV watched per week by males and females in the population and cannot conclude that the true mean difference in hours of TV watched per week by males and females in the population is statistically significantly different from zero hours at the 5% significance level.

b) |tcalc|>|tcrit|andp<α

Thus I can reject the plausibility of zero hours for the true mean difference in hours of TV watched per week by males and females in the population and cannot conclude that the true mean difference in hours of TV watched per week by males and females in the population is statistically significantly different from zero hours at the 5% significance level.

c) |tcalc|>|tcrit|andp>α

Thus I cannot reject the plausibility of zero hours for the true mean difference in hours of TV watched per week by males and females in the population and cannot conclude that the true mean difference in hours of TV watched per week by males and females in the population is statistically significantly different from zero hours at the 5% significance level.

d)|tcalc|<|tcrit|andp<α

Thus I can reject the plausibility of zero hours for the true mean difference in hours of TV watched per week by males and females in the population and cannot conclude that the true mean difference in hours of TV watched per week by males and females in the population is statistically significantly different from zero hours at the 5% significance level.

(G.1) Construct a 95% confidence interval around the difference-in-means estimate. Find the lower bound of the interval you calculated. (In this case, be sure to use the standard error you calculated when determining the test statistic that uses information about the population proportion.)

(G.2) Construct a 95% confidence interval around the difference-in-means estimate. Find the upper bound of the interval you calculated. (In this case, be sure to use the standard error you calculated when determining the test statistic that uses information about the population proportion.)

(H) How would you interpret the confidence interval?

a) We can be 95% confident that the true mean difference in hours of TV watched per week by males and females in the population falls between lower bound and upper bound.

b) We can not be 95% confident that the true mean difference in hours of TV watched per week by males and females in the population falls between lower bound and upper bound.

(G.2) What connections can you draw between the confidence interval and the hypothesis test?

a) Because zero falls within the confidence interval, we fail to reject the null hypothesis.

b) Because zero does not fall within the confidence interval, we have another piece of evidence to reject the null hypothesis.

c) No answer text provided.

d) No answer text provided.

Answer 1

NULL HYPOTHESIS H0:

ALTERNATIVE HYPOTHESIS Ha:

a) it is a two-sided test. Thus, the tcritis ±1.960.

STANDARD ERROR=0.146

Z TEST STATISTIC= 1.438

The P-Value is .1504. The result is not significant at p > .05.

SINCE P VALUE> 0.05 THEREFORE NOT SIGNIFICANT

WE DO NOT HAVE SUFFICIENT EVIDENCE TO CONCLUDE THAT THERE IS DIFFERENCE IN MEANS.

F] |tcalc|>|tcrit|andp>α

Thus I cannot reject the plausibility of zero hours for the true mean difference in hours of TV watched per week by males and females in the population and cannot conclude that the true mean difference in hours of TV watched per week by males and females in the population is statistically significantly different from zero hours at the 5% significance level.

G] 95% CONFIDENCE INTERVAL

MARGIN OF ERROR= ZC*SE= 1.96*0.146= 0.286

DIFFERENCE OF MEANS MOE

0.210.29

(-0.08 , 0.5)

Because zero falls within the confidence interval, we fail to reject the null hypothesis.