Question

II. Please construct the confidence interval for the following example. First, compute the confidence interval. Second, point out the margin of error. Third, point out the width of the confidence interval. Finally, , interpret the final result in the context. (Formula and computation)

A random sample of 178 households reported an average of 2.1 TV sets per household. The sample standard deviation is 0.1. What would be the confidence interval? Please construct the confidence interval at the 99% confidence level, and interpret the result in the context of the example.

Answer 1

t critical value at 0.01 significance level with 177 df = 2.604

99% confidence interval for is

- t * S / sqrt(n) < < + t * S / sqrt(n)

2.1 - 2.604 * 0.1 / sqrt(178) < < 2.1 + 2.604 * 0.1 / sqrt(178)

2.08 < < 2.12

99% CI Is ( 2.08 , 2.12 )

Margin of error = t * S / sqrt(n) = 2.604 * 0.1 / sqrt(178) = 0.02

Width of CI = 2.12 - 2.08 = 0.04

Interpretation = We are 99% confident that true average TV sets per household are between

2.08 and 2.12 .