Question

Suppose that a survey reported that 45% of respondents favored gun control, with a margin of error of ±4 percentage points. (SHOW YOUR WORK)

(a) What was the approximate size of the sample, n? (Round answer to the nearest whole number.)

(b) What is an approximate 95% confidence interval for the percentage of the corresponding population who favor gun control?

(c) Write a statement interpreting the interval you found in part (b) that could be understood by someone with no training in statistics. A survey found that 45% of those asked favor gun control. This was based on a sample of about individuals, but the results indicate that it is likely that between % and % of the population favors gun control.

Answer 1

Solution :

Given that,

= 0.45

1 - = 0.55

margin of error = E = 0.04

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

a)

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.96 /0.04 )² *0.45*0.55

= 594.25

sample size = 595

b)

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

= 1.96 * (((0.45*0.55) / 595)

= 0.040

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.45 - 0.040 < p < 0.45 + 0.040

0.41 < p < 0.44

c)

A survey found that 45% of those asked favor gun control. This was based on a sample of about individuals, but the results

indicate that it is likely that between 41% and 44% of the population favors gun control.