Question

a marriage counselor has traditionally seen that the proportion p of all married couples for whom her communication program can prevent divorce is 80%. after making some recent changes, the marriage counselor now claims that her program can prevent divorce in more than 80% of married couples. in a random sample of 250 married couples who completed her program, 214 of them stayed together. based on this sample, is there enough evidence to support the marriage counselors claim at the 0.05 level of significance? perform a one - tailed test. then complete the parts below. carry your intermediate computations to three or more decimal places. (if necessary, consult a list of formulas.) (a) state the null hypothesis h0 and the alternative hypothesis h1. h0:□ h1:□ (b) determine the type of test statistic to use. select (c) find the value of the test statistic. (round to three or more decimal places.) □ (d) find the p - value. (round to three or more decimal places.) □ (e) is there enough evidence to support the marriage counselors claim that the proportion of married couples for whom her program can prevent divorce is more than 80%? yes no

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is that the proportion $p$ is equal to the traditional proportion. The alternative hypothesis $H_1$ is the counselor's claim.
$H_0:p = 0.8$
$H_1:p>0.8$

Step2: Determine test - statistic type

We are testing a proportion, so we use the z - test for a proportion. The formula for the test - statistic $z$ in a one - sample proportion test is $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$, where $\hat{p}$ is the sample proportion, $p_0$ is the hypothesized proportion, and $n$ is the sample size.

Step3: Calculate sample proportion

The sample size $n = 250$ and the number of successes $x = 214$. The sample proportion $\hat{p}=\frac{x}{n}=\frac{214}{250}=0.856$.

Step4: Calculate test - statistic value

$p_0 = 0.8$, $n = 250$, $\hat{p}=0.856$.
$z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}=\frac{0.856 - 0.8}{\sqrt{\frac{0.8\times(1 - 0.8)}{250}}}=\frac{0.056}{\sqrt{\frac{0.8\times0.2}{250}}}=\frac{0.056}{\sqrt{\frac{0.16}{250}}}=\frac{0.056}{\sqrt{0.00064}}\approx2.224$

Step5: Calculate p - value

Since this is a right - tailed test, the p - value is $P(Z>z)$. Using the standard normal distribution table, $P(Z > 2.224)=1 - P(Z\leq2.224)$. From the standard normal table, $P(Z\leq2.224)\approx0.987$, so the p - value is $1 - 0.987 = 0.013$.

Step6: Make a decision

The significance level $\alpha=0.05$. Since the p - value ($0.013$) is less than $\alpha = 0.05$, we reject the null hypothesis.

Answer:

(a) $H_0:p = 0.8$; $H_1:p>0.8$
(b) z - test for a proportion
(c) $2.224$
(d) $0.013$
(e) Yes