Question

A survey was run by a high school student in order to determine what proportion of mortgage-holders in his town expect to own their house within 10 years. He surveyed 133 mortgage holders and found that the proportion of these that did expect to own their house within 10 years is 0.67.

The student decides to construct a 95% confidence interval for the population proportion.

a)Calculate the margin of error that the high school student will have. Give your answer as a decimal to 3 decimal places.

Margin of error =

A different high school student sees this survey and decides to try and repeat it. This second high school student also surveys 133 mortgage holders, and this student also finds that the proportion of these that do expect to own their house within 10 years is 0.67. However, this student wants to construct a 99% confidence interval for the population proportion.

b)Calculate the margin of error that the second student will have. Give your answer as a decimal to 3 decimal places.

Margin of error =

Answer 1

Solution :

Given that,

n = 133

Point estimate = sample proportion = = 0.67

1 - = 1 - 0.67 = 0.33

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z0.025 = 1.960

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

E = 1.96 (((0.67 * 0.33) / 133)

E = 0.080

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

Z/2 = Z0.005 = 2.576

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

E = 2.576 (((0.67 * 0.33) / 133)

E = 0.105