A recent study focused on the number of times men and women who live alone buy t

Question

A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. Assume that the distributions follow the normal probability distribution and the population standard deviations are equal. The information is summarized below.

Statistic: Men Women

Sample mean: 24.51 22.69

Sample standard deviation: 4.48 3.86

Sample size: 35 40

At the 0.01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month?

a. State the decision rule for 0.01 significance level: H0: Men= Women   H1: Men Women.(Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)

Reject H0 if t < _____ or t > ______

b. Compute the value of the test statistic. (Round your answer to 3 decimal places.)

Value of the test statistic is _______

c. What is your decision regarding the null hypothesis? (Answered)

The decision is do not reject the null hypothesis that the means are the same.

d. What is the p-value? (Round your answer to 3 decimal places.)

p-value is ______

Explanation / Answer

Solution:

From the given information we have:
x1 = 24.51, x2 = 22.69
s1 = 4.48, s2 = 3.86
n1 = 35, n2 = 40
= 0.01
a) Test hypothesis:
H0: Men= Women
H1: Men Women

b) Test statistic:

Z = (x1 - x2)/sqrt[s12/n1 + s22/n2]

= (24.51-22.69)/ sqrt(4.482/35 + 3.862/ 40)
=1.87

d) The p-value is 2*P(Z>1.87) =0.0614 (check standard normal table)

c) Since p-value is larger than =0.01, we do not reject Ho. So we can not conclude that there is a difference in the mean number of times men and women order take-out dinners in a month.

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