Question

Random and independent samples of

100

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

4.6

commercials. The second network aired a mean of

109.4

commercials, with a standard deviation of

4.5

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

−μ1μ2

, the difference between the mean number of commercials

μ1

aired during prime time by the first network and the mean number of commercials

μ2

aired 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. (If necessary, consult a list of formulas.)

What is the lower limit of the 95% confidence interval?
What is the upper limit of the 95% confidence interval?

Answer 1

Given:

s₁ - estimate of population standard deviation = 4.6

s₂ - an estimate of population standard deviation = 4.5

c - confidence level = 95%

The formula to find the confidence interval for the difference of population means is,

The sample sizes are sufficiently large, so use z instead of t.

To find z, the area is 1 - (alpha/2)

alpha = 1 - c = 1 - 0.95 = 0.05

alpha/2 = 0.025

1 - (alpha/2) = 0.975

Using z table the z score for area 0.975 is 1.96

The lower limit of the 95% confidence interval = -0.06

The upper limit of the 95% confidence interval = 2.46