A newspaper reporter reports that the governors approval rat

A newspaper reporter reports that the governor\'s approval rating stands at 65%. The article adds that the poll is based on a random sample of 972 adults and has a margin of of 2.5%. What level of confidence did the pollsters use?

Solution

Governer\'s approval rating = 65%

n = 972

Margin of Error = 2.5% = 0.025

Standard Error = sqrt[p * (1 - p)/n] = sqrt[0.65 * (1 - 0.65)/972] = 0.015

Margin of Error = Critical value * Standard Error

0.025 = Critical value * 0.015

Critical value = 1.634

Area under the Normal table corresponding to Z value of 1.634 is 0.95

1 - alpha/2 = 0.95

alpha = 0.10

Confidence Level = 1 - alpha = 1 - 0.1 = 0.90 = 90%