Question

In: Statistics and Probability

A survey indicated that 28% of tenured university professors are female. Suppose the survey was based...

A survey indicated that 28% of tenured university professors are female. Suppose the survey was based on a sample of 500 professors.

The 95% confidence interval for the population proportion, p, is

Group of answer choices

0.26 to 0.30

0.24 to 0.32

0.27 to 0.29

0.28 to 0.34

Expert Solution

Solution :

Given that,

n = 500

Point estimate = sample proportion = =0.28

1 - = 0.72

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 ( Using z table )

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

= 1.96 ( $\sqrt$ ((0.28*0.72) /500 )

E = 0.04

A 95% confidence interval proportion p is ,

- E < p < + E

0.28-0.04 < p < 0.28 +0.04

0.24 to 0.32

orchestra answered 2 years ago

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