Question

A recently televised broadcast of 60 minutes had a 15 share, meaning that among 5000 monitored households with TV sets in use, 15% of them were tuned to 60 minutes. Use a 0.05 significance level to test the claim of an advertiser that among the household with TV sets in use, less than 20% were tuned to 60 minutes.

Answer 1

Explanation:

n = 5000, p = 0.15, q = 0.85

μ = p = 0.15

σ = √(pq/n) = 0.005

At 0.05 significance, z = 1.96.

0.15 ± 1.96 × 0.005

(0.140, 0.160)

0.20 is outside of the confidence interval, so we can conclude with 95% confidence that the claim is true.