Question

In a survey of 356 randomly selected gun owners, it was found that 82 of them said they owned a gun primarily for protection.

Find the margin of error and 95% confidence interval for the percentage of all gun owners who would say that they own a gun primarily for protection. Round all answers to 3 decimal places.

Margin of Error (as a percentage): %
Confidence Interval: % to %

Answer 1

Solution :

Given that,

n = 356

x = 82

Point estimate = sample proportion = = x / n = 0.230

1 - = 0.770

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

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

= 1.96 * (((0.230 * 0.770) / 356)

= 0.044

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.230 - 0.044 < p < 0.230 + 0.044

0.186 < p < 0.274

Margin of Error = 0.044 = 4.4%

Confidence Interval: 18.6% to 27.4%