Question

13. A study published in 2008 by researchers at UT Austin found that 124 out of 1,923 U of T females had over $6,000 in credit card debt while 65 out of 1,236 males had over $6,000 in credit card debt. Test using 0.05, if there is evidence that the proportion of female students at U of T with more than $6,000 credit card debt.

a. Verify that the sample size is large enough in each group to use the normal distribution to perform a hypothesis test for a difference in two groups.

b. Write out the null and alternative hypotheses.

c. Find the value of the pooled standard error. Round to 4 decimal places.

d. The test statistic for this sample is z=1.77, find the p-value. Round to 4 decimal places.

e. Make a formal decision for the hypothesis test based on your p-value in part d. Interpret your decision in the context of the original question.

Answer 1

a.

To use the normal distribution to perform a hypothesis test for a difference in two groups, the condition np > 10 and n(1-p) > 10 should be satisfied. That each sample includes at least 10 which had over $6,000 in credit card debt  and 10 which do not had over $6,000 in credit card debt. The conditions are satisfied for both males and females, thus normal distribution to perform a hypothesis test for a difference in two groups can be used.

b.

H0: pf = pm; H1: pf > pm

c.

p-hat pooled = Total number of students with more than $6000 credit card debt / Total number of students in sample
= (124 + 65) / (1923 + 1236) = 0.05982906

Pooled standard error, SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

= sqrt{ 0.05982906 * ( 1 - 0.05982906 ) * [ (1/1923) + (1/1236) ] }
= 0.0086

d.

p-value = P(z > 1.77) = 0.0384

e.

As, p-value is less than the significance level 0.05, we reject H0 and conclude that there is significant evidence that proportion of females had over $6,000 in credit card debt is greater than proportion of males had over $6,000 in credit card debt.