Construct a 99% confidence interval of the population proportion using the given information. X= 105, n= 150

150

Explanation:n =

x = 106

\hat p = x / n = 106 / 150 = 0.707

1 -\hat p = 1 - 0.707 = 0.293

At 99% confidence level the z is ,

\alpha  = 1 - 99% = 1 - 0.99 = 0.01

\alpha / 2 = 0.01 / 2 = 0.005

Z\alpha/2 = Z0.005 = 2.576

Margin of error = E = Z\alpha / 2 * \sqrt((\hat p * (1 - \hat p)) / n)

= 2.576 * (\sqrt((0.707 * 0.293) / 150)

=0.096

A 99% confidence interval for population proportion p is ,

\hat p- E < P <\hat p + E

0.707 - 0.096 < p < 0.707 + 0.096

0.611 < p < 0.803

answer; mixture;