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Question

Execute these steps.

a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results.

Employ the traditional method of hypothesis testing unless otherwise specified.

The percentages of adults 25 years of age and older who have completed 4 or more years of college are 23.6%23.6 \% for females and 27.8%27.8 \% for males. A random sample of women and men who were 25 years old or older was surveyed with these results. Estimate the true difference in proportions with 95%95 \% confidence, and compare your interval with the Almanac statistics.

Women Men
Sample size 350 400
No. who completed 4 or more years 100 115

Solution

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Step 1
1 of 2

Given:

c=95%c=95\%

x1=100x_1=100

n1=350n_1=350

x2=115x_2=115

n2=400n_2=400

The sample proportion is the number of successes divided by the sample size:

p^1=x1n1=1003500.2857\hat{p}_1=\dfrac{x_1}{n_1}=\dfrac{100}{350}\approx 0.2857

p^2=x2n2=115400=0.2875\hat{p}_2=\dfrac{x_2}{n_2}=\dfrac{115}{400}=0.2875

For confidence level 1α=0.951-\alpha=0.95, determine zα/2=z0.025z_{\alpha/2}=z_{0.025} using table E (look up 0.025 in the table, the z-score is then the found z-score with opposite sign):

zα/2=1.96z_{\alpha/2}=1.96

The endpoints of the confidence interval for p1p2p_1-p_2 are then:

(p^1p^2)zα/2p^1(1p^1)n1+p^2(1p^2)n2=(0.28570.2875)1.960.2857(10.2857)350+0.2875(10.2875)4000.0667(p^1p^2)+zα/2p^1(1p^1)n1+p^2(1p^2)n2=(0.28570.2875)+1.960.2857(10.2857)350+0.2875(10.2875)4000.0631\begin{align*} &(\hat{p}_1-\hat{p}_2)-z_{\alpha/2}\cdot \sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \\ &=(0.2857-0.2875)-1.96\sqrt{\dfrac{0.2857(1-0.2857)}{350}+\dfrac{0.2875(1-0.2875)}{400}} \\ &\approx -0.0667 \\ \\ &(\hat{p}_1-\hat{p}_2)+z_{\alpha/2}\cdot \sqrt{\dfrac{\hat{p}_1(1-\hat{p}_1)}{n_1}+\dfrac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \\ &=(0.2857-0.2875)+1.96\sqrt{\dfrac{0.2857(1-0.2857)}{350}+\dfrac{0.2875(1-0.2875)}{400}} \\ &\approx 0.0631 \end{align*}

We note that the confidence interval contains the difference of the Almanac statistics: 23.6%27.8%=4.2%=0.04223.6\%-27.8\%=-4.2\%=-0.042

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