Question

A city is investigating the chances of getting public support for a new bill. In a survey of 1250 families, 400 families were in favor of the new bill.

a. What is the point estimate for the population proportion in favor of the bill?

b. Write a 95% confidence interval for the population proportion. (Use the t table to obtain the critical value and round to two decimal places). (  ,  )

c. Write a 90% confidence interval for the population proportion. (Use the t table to obtain the critical value and round to two decimal places). (  ,  )

Answer 1

Solution :

n = 1250

x = 400

a ) = x / n = 400 / 1250 = 0.320

1 - = 1 - 0.320 = 0.680

b )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

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

= 1.96 * (((0.320 * 0.680) / 1250)

= 0.026

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.320 - 0.026 < p < 0.320 + 0.026

0.294 < p < 0.346

c ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.64

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

= 1.64 * (((0.320 * 0.680) / 1250)

= 0.022

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.320 - 0.022 < p < 0.320 + 0.022

0.298 < p < 0.342