A researcher wants to estimate the mean difference in the amount of television t

Question

A researcher wants to estimate the mean difference in the amount of television that married men and women watch. Six randomly selected married couples are studied for a month, and the daily average number of hours spent watching television for each person is recorded. (See below. If the table is not displayed properly below, please see Section 5.4 Exercise #12 in the textbook ) Assume that the differences between the men's and women's times are approximately normally distributed. Construct a 90% confidence interval for the mean of the differences between the men's and women's average daily number of hours spent watching television. Married Couple Kimball Linge McClelland Nguyen Oliver Pontus Husband Wife 3.84.6 3.2 5.2 2.6 3.5 2.43.54.1 1.8 3.6 3.9 a. (-4509, 0.3842) b. (1.0710, 1.0042) . (-1.065, 0.9980) d. (0.5449, 0.4782)

Explanation / Answer

Ans:

n=6

df=6-1=5

critical t value=tinv(0.1,5)=2.015

90% confidence interval for mean difference

=-0.0333+/-2.015*(0.6218/sqrt(6))

=-0.0333+/-0.5115

=(-0.5449, 0.4782)

Option d is correct.

Husband wife d 1 3.8 3.2 0.6 2 4.6 5.2 -0.6 3 2.6 3.5 -0.9 4 2.4 1.8 0.6 5 3.5 3.6 -0.1 6 4.1 3.9 0.2 mean= -0.0333 std. dev= 0.6218

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