(a) State the null hypothesis H, and the alternative hypothesis H, • Ho : 0 H, : 0 (b) Determine the type of test statistic to use. (Choose one) V (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 counselor's claim that the proportion of married couples for whom her program can prevent divorce is more than 76%? OYes ONo

Let's go through each part of the problem step-by-step:

### (a) State the null hypothesis \( H_0 \) and the alternative hypothesis \( H_a \)
- **Null Hypothesis (\( H_0 \))**: \( p = 0.76 \)
  - This states that the proportion of married couples for whom the program can prevent divorce is 76%.
  
- **Alternative Hypothesis (\( H_a \))**: \( p > 0.76 \)
  - This states that the proportion of married couples for whom the program can prevent divorce is more than 76%.

### (b) Determine the type of test statistic to use
The appropriate test statistic for this problem is the **z-statistic**. This is because we are dealing with a hypothesis test for a proportion.

### (c) Find the value of the test statistic
As calculated previously:

\[
z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}
\]
Where:
- \(\hat{p} = \frac{171}{205} \approx 0.8341\)
- \(p_0 = 0.76\)
- \(n = 205\)
- Standard Error \(SE \approx 0.0297\)

Thus:

\[
z = \frac{0.8341 - 0.76}{0.0297} \approx 2.4933
\]

### (d) Find the p-value
The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the value observed under the null hypothesis.

Using a z-table or a statistical calculator, we find the p-value corresponding to \( z = 2.4933 \).

\[
\text{p-value} \approx 0.0063
\]

### (e) Is there enough evidence to support the marriage counselor's claim that the proportion of married couples for whom her program can prevent divorce is more than 76%?

Given that the p-value \( 0.0063 \) is less than the significance level \( \alpha = 0.01 \), we reject the null hypothesis.

**Conclusion:** Yes, there is enough evidence to support the marriage counselor's claim that the proportion of married couples for whom her program can prevent divorce is more than 76%.

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Do you have any more questions, or would you like more details? 

Here are five related questions:
1. How is the p-value interpreted in hypothesis testing?
2. What is the impact of the sample size on the z-test result?
3. What would the p-value indicate if the z-score was lower than the critical value?
4. How does the choice of significance level affect the decision in hypothesis testing?
5. What are the assumptions required for using a z-test for proportions?

**Tip:** When performing hypothesis testing, it's crucial to remember that rejecting the null hypothesis does not prove the alternative hypothesis; it simply indicates that there is enough evidence to support it.

Explore how to conduct hypothesis testing for proportions using a z-test. Step-by-step calculation and interpretation of the test statistic and p-value to determine if there is enough evidence to support a claim about the proportion of married couples for whom a program can prevent divorce being more than 76%.

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