In a survey conducted with 600 participants across the United States, 450 were found to have studied science in college. If we were to predict the population proportion with a 99.7% confidence, what would the confidence interval be? A. 69.7% to 80.3% B. 73.23% to 76.76% C. 71.46% to 78.53% D. 73.5% to 76.5%

Confidence interval of a population proportion is given by p^ + or - sqrt (p^ (1 - p^) / n); where p^ = 450/600 = 0.75 and n = 600

99.7% convidence interval = 0.75 + or - 2.96 x sqrt (0.75 (1 - 0.75) / 600) = 0.75 + or - 2.96 x sqrt (0.75 (0.25) / 600) = 0.75 + or - 2.96 x 0.0177 = 0.75 + or - 0.0524 = 0.697 to 0.803 = 69.7% to 80.3%

You need to find the confidence interval of proportion

The proportion of participants studying science=p^=450/600=0.75

The confidence interval for proportion is given by,

p^ ± Z (alpha/2) * √ (p^ (1-p^) / n)

For 99.7% confidence we have alpha = 1-0.997=0.003

From the normal table we get the critical value of Z = 2.97

So the confidence interval will be

=0.75±2.97*√ (0.75 (1-0.75) / 600)

=0.69749,0.8025

Hence answer is A.