4 . Individuals filing federal income tax returns prior to March 31 had an avera

Question

4. Individuals filing federal income tax returns prior to March 31 had an average refund of $1102. Consider the population of “last-minute” filers who mail their returns during the last five days of the income tax period (typically April 10 to April 15).

a. A researcher suggests that one of the reasons that individuals wait until the last five days to file their returns is that on average those individuals have a lower refund than early filers. Develop appropriate hypotheses such that rejection of H0 will support the researcher’s contention.

b. For a sample of 600 individuals who filed a return between April 10 and April 15, the sample mean refund was $1050 and the standard deviation was $500. Compute the p-value.

c. Using =.05, what is your conclusion?

d. Test the hypotheses using the critical value approach( = 0.025).

5. Media Metrix, Inc. tracks Internet users in seven countries: Australia, Great Britain, Canada, France, Germany, Japan, and the United State. According to resent measurement figures, American home users rank first in Internet usage with a mean of 14 hours per month. Assume that in a following-up study involving a sample of 205 Canadian Internet users, the sample mean was 12.8 hour per month and the sample standard deviation was 6.2 hours.

a. Formulate the null and alterative hypotheses that can be used to determine whether the sample data support the conclusion that Canadian Internet users have a population mean less than the U.S. mean of 14 hours per month.

b. What is the p-value?

c.   Using =0.01, what is your conclusion?

d. Test the hypotheses using the critical value approach ( =0.01).

Explanation / Answer

4) a) H0: µ = 1102

H1: µ < 1102

b) p value = p [Z < 1050 - 1102 * 600 / 500]

= p [Z < -2.5475]

= 0.0054

c) Reject H0

d) Decision rule : If Z statistic < -1.96, then reject H0

Z test statistic = xbar - µ / [ s/n]

= 1050 - 1102 / [500/600]

= -2.55

Z < -1.96, Reject H0

5)

a) H0: µ = 14

H1: µ < 14

b) p value = p [Z < 12.8 - 14 * 205 / 6.2]

= p [Z < -2.77]

= 0.0028

c) Reject H0, since p value < 0.01

d) Decision rule : If Z statistic < -2.326, then reject H0

Z test statistic = xbar - µ / [ s/n]

= 1050 - 1102 / [500/600]

= -2.55

Z < -2.326, Reject H0

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