Question

Confidence Intervals. A newspaper publishes a survey that estimates that 70% of Americans try to avoid soda. The newspaper reports that margin of error is 5%. Assume that the survey used a simple random sample and that the margin of error published corresponds to a 95% confidence level. Calculate the sample size in the survey.

Answer 1

Solution :

Given that,

= 0.70

1 - = 1 - 0.70= 0.30

margin of error = E = 5% = 0.05

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.96 / 0.05)2 * 0.70 * 0.30

= 322.69

Sample size =323