Question

Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best point estimate for p, the margin of error, and the confidence interval. Assume the results come from a random sample.

A 90% confidence interval for p given that p (with the hat)= 0.4 and n equals 475.

Round your answer for the best point estimate to two decimal places, and your answers for the margin of error and the confidence interval to three decimal places

. Best point estimate =

Margin of error

The 90% confidence interval is what to what

Answer 1

Solution :

Given that,

n = 475

Point estimate = sample proportion = = 0.40

1 - = 1 - 0.40 = 0.60

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 * (((0.40 * 0.60) / 475)

= 0.037

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.4 - 0.037 < p < 0.4 + 0.037

0.636 < p < 0.437

The 95% confidence interval for the population proportion p is : (0.363 , 0.437)