Question

Random and independent samples of 95 recent prime time airings from each of two major networks have been considered. The first network aired a mean of 110.6 commercials during prime time, with a standard deviation of 4.3 commercials. The second network aired a mean of 109.4 commercials, with a standard deviation of 4.7 commercials. As the sample sizes are quite large, the population standard deviations can be estimated using the sample standard deviations. Construct a 95% confidence interval for Hly, the difference between the mean number of commercials, aired during prime time by the first network and the mean number of commercials paired during prime time by the second network. Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. What is the lower limit of the 95% confidence interval? What is the upper limit of the 95% confidence interval?

Answer 1

(21 – 22) + ZC* ((sl)2/n1 + (s2)2/n2)

LOWER LIMIT =

(110.6 - 109.4) - 1.96 * |(4.3)2/95+ (4.7)2/95)

(1.2) – 1.96* ((18.49/95 + 22.09/95)

(1.2) – 1.96* (0.4271)

(1.2-1.96 +0.6535

(1.2-1.281

LOWER LIMIT= -0.08

UPPER LIMIT=

(1.2) +1.281

UPPER LIMIT=