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Chapter 11: Problem 76

An Associated Press article (San Luis Obispo Telegram-Tribune, September 23,1995 ) examined the changing attitudes of Catholic priests. National surveys of priests aged 26 to 35 were conducted in 1985 and again in 1993\. The priests surveyed were asked whether they agreed with the following statement: Celibacy should be a matter of personal choice for priests. In \(1985,69 \%\) of those surveyed agreed; in \(1993,38 \%\) agreed. Suppose that the samples were randomly selected and that the sample sizes were both 200 . Is there evidence that the proportion of priests who agreed that celibacy should be a matter of personal choice declined from 1985 to 1993 ? Use \(\alpha=.05\).

Short Answer

Expert verified
The decision about whether there is evidence that the proportion of priests who agree that celibacy should be a matter of personal choice declined from 1985 to 1993 would be based on the p-value calculated relative to the chosen significance level of 0.05. If the calculated p-value is less than 0.05, then there is evidence of a decline.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)) is that the proportion of priests who agree that celibacy should be a matter of personal choice did not decline from 1985 to 1993. This is stated mathematically as \(H_0: p_{1985} = p_{1993}\). The alternative hypothesis (\(H_a\)), which we want to gather evidence for, is that the proportion did decline. This is stated as \(H_a: p_{1985} > p_{1993}\).
02

Compute the Test Statistic

The test statistic is a z-score (z). It is calculated as \[ z = \frac{(p_{1985} - p_{1993}) - 0}{\sqrt{p(1-p)(\frac{1}{n_{1985}} + \frac{1}{n_{1993}})}} \]where \(p_{1985}\) and \(p_{1993}\) are the sample proportions for 1985 and 1993 respectively, \(p\) is the pooled sample proportion, and \(n_{1985}\) and \(n_{1993}\) are the sample sizes. The pooled sample proportion, assuming \(H_0\) is true, is the total number of successes divided by the total sample size: \[ p = \frac{p_{1985}*n_{1985} + p_{1993}*n_{1993}}{n_{1985} + n_{1993}} \].
03

Find the P-Value

Using the standard normal Z distribution, we can find the probability that we would get a Z score as extreme as our test statistic, given that the null hypothesis is true. This probability is the p-value. We calculate this as \(P(Z > z)\), because we're looking for evidence of a decrease from 1985 to 1993.
04

Make a Decision

If the p-value is less than \(\alpha\) (the significance level, set to 0.05), we reject the null hypothesis in favor of the alternative. If the p-value is not less than \(\alpha\), we do not reject the null hypothesis.

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