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Chapter 10: Problem 28

The article "Poll Finds Most Oppose Return to Draft, Wouldn't Encourage Children to Enlist" (Associated Press, December 18,2005 ) reports that in a random sample of 1000 American adults, 700 indicated that they oppose the reinstatement of a military draft. Is there convincing evidence that the proportion of American adults who oppose reinstatement of the draft is greater than twothirds? Use a significance level of \(.05\).

Short Answer

Expert verified
The conclusion will be based on the comparison between the calculated z-score and the critical value. If the z-score is greater than the critical value, we can say there is convincing evidence that the proportion of American adults who oppose the draft reinstatement is greater than two thirds with a significance level of 0.05. Otherwise, we cannot say this.

Step by step solution

01

State the hypotheses

The null hypothesis (H0) indicates that the proportion (p) is equal to two thirds, i.e., H0: p = 2/3. The alternative hypothesis (Ha) is that the proportion is greater than two thirds, i.e., Ha: p > 2/3.
02

Calculate test statistic

We calculate the one-sample z-score using the following formula: \(z = (\hat{p} - p_0) / sqrt[(p_0(1 - p_0))/n]\) where \(\hat{p}\) is the sample proportion, \(p_0\) is the assumed population proportion under the null hypothesis, and \(n\) is the sample size. Here, \(\hat{p} = 700/1000 = 0.7\), \(p_0 = 2/3=0.6667\), and \(n = 1000\). Substituting these values into the formula gives a z-score.
03

Calculate the critical value

We now calculate the critical value for a one-tailed test with a significance level of 0.05, we find that the critical value is 1.645.
04

Making a decision

Compare the calculated z-score with the critical value. If the z-score is greater than the critical value, we reject H0. Otherwise, we fail to reject H0.

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