Question

A​ study, which randomly surveyed 3,700 households and drew on this information from the CRA​ (Canada Revenue​ Agency), found that 76​% of households have conducted at least one pension rollover from an​ employer-sponsored retirement plan. Suppose a recent random sample of 110 households in a certain county was taken and respondents were asked whether they had ever funded an RRSP account with a rollover from an​ employer-sponsored retirement plan. Based on the sample data​ below, can you conclude at the 0.01 level of significance that the proportion of households in the county that have funded an RRSP with a rollover is different from the proportion for all households reported in the​ study?

77 respondents said they had funded an​ account; 33 respondents said they had not
A) Determine the null and alternative hypotheses. Choose the correct answer below.

a. H0​: ​π≥0.76

HA​: ​π<0.76

b. H0​: ​π≤0.76

HA​: ​π>0.76

c. H0​: ​π=0.76

HA​: ​π≠0.76

d. H0​: ​π≠0.76

HA​: ​π= 0.76

B) Calculate the test statistic.
Z = ? (Round to two decimal places as needed)

C) Find p-value.

p = ? (Round to four decimal places as​ needed)

D) Determine a conclusion. (choose between reject/do not reject and between insufficient/sufficient)

(Reject/Do not reject) H0. There is (insufficient/sufficient) evidence to conclude that the proportion of households in the county that have funded an RRSP with a rollover is different from the proportion for all households in the study.

Answer 1

A) The correct hypotheses are: H0​: π=0.76 and HA​: π≠0.76.

B) Test statistic Z = 1.44

C) p-value ≈ 0.1496

D) Do not reject H0. There is insufficient evidence to conclude that the proportion of households funding an RRSP with a rollover differs from the reported proportion.

Step-by-step explanation:

A hypothesis test is conducted to compare the sample proportion of households funding an RRSP with a rollover to the reported proportion in the study. The null hypothesis (H0) assumes no difference, stating that the population proportion (π) equals the reported value (0.76), while the alternative hypothesis (HA) suggests a difference in proportions (π ≠ 0.76).

Calculating the test statistic Z using the sample data yields Z = 1.44. The p-value, representing the probability of observing the sample proportion given H0 is true, is approximately 0.1496. Since the p-value is higher than the significance level of 0.01, we do not have enough evidence to reject the null hypothesis. Therefore, we fail to conclude that the proportion of households funding an RRSP with a rollover differs from the reported proportion of 76% in the study.