Question

A pollster surveyed a sample of 980 adult Americans, asking them if they own a personal firearm. 34% of the sample said yes.

1. What is a 90% confidence interval estimate for the percentage of Americans that own a firearm?

2. A gun owners’ group claims that more Americans own a firearm in 2015 than ten years ago, when the percentage of owners was 30%. At the 0.05 level of significance, has the percentage of owners increased?

3. What is the p-value for this problem? How does it tell you what conclusion to draw about the null hypothesis?

Answer 1

1) At 90% confidence interval the critical value is z_0.05 = 1.645

The 90% confidence interval for population proportion is

+/- z_0.05 * sqrt((1 - )/n)

= 0.34 +/- 1.645 * sqrt(0.34 * (1 - 0.34)/980)

= 0.34 +/- 0.025

= 0.315, 0.365

b) H₀: P = 0.3

    H₁: P > 0.3

The test statistic z = ( - P)/sqrt(P(1 - P)/n)

                             = (0.34 - 0.3)/sqrt(0.3 * 0.7/980)

                             = 2.73

At alpha = 0.05, the critical value is z_0.95 = 1.645

Since the test statistic value is greater than the critical value (2.73 > 1.645), so we should reject H₀.

So there is sufficient evidence to support the claim that the percentage of owners has increased.

3) P-value = P(Z > 2.73)

                 = 1 - P(Z < 2.73)

                 = 1 - 0.9968

                 = 0.0032

Since the P-value is less than the significance level(0.0032 < 0.05), so we should not reject H₀.