We are conducting a hypothesis test for the population proportion. In this case, the null hypothesis $H_0: p = 0.7$ states that the population proportion of successes is 0.7, and the alternative hypothesis $H_1: p \neq 0.7$ states that the proportion is different from 0.7. We can perform a two-tailed Z-test for the population proportion.

Formula for the test statistic:

The formula for the test statistic in a proportion test is:

$Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$

Where:

• $\hat{p}$ is the sample proportion of successes,
• $p_0$ is the population proportion under the null hypothesis,
• $n$ is the sample size.

Step 1: Calculate the sample proportion $\hat{p}$

The sample proportion is:

$\hat{p} = \frac{\text{number of successes}}{\text{sample size}} = \frac{441}{687} = 0.6415$

Step 2: Plug values into the formula

We know:

• $p_0 = 0.7$,
• $n = 687$,
• $\hat{p} = 0.6415$.

Now calculate the test statistic:

$Z = \frac{0.6415 - 0.7}{\sqrt{\frac{0.7(1 - 0.7)}{687}}}$

First, calculate the standard error:

$\text{SE} = \sqrt{\frac{0.7 \times 0.3}{687}} = \sqrt{\frac{0.21}{687}} \approx 0.0176$

Now calculate the Z-score:

$Z = \frac{0.6415 - 0.7}{0.0176} = \frac{-0.0585}{0.0176} \approx -3.326$

Conclusion:

The test statistic $Z$ for this sample is approximately -3.33.

Would you like to know more about how to interpret this result or conduct the hypothesis test?

Here are 5 related questions:

1. What is the p-value for this Z-score, and how do we interpret it?
2. How do we reject or fail to reject the null hypothesis using this test statistic?
3. What is a two-tailed test, and why is it relevant here?
4. How does changing the significance level $\alpha$ affect the hypothesis test result?
5. What is the effect of increasing the sample size on the test statistic?

Tip: Always verify if assumptions like sample randomness or normality (when needed) are satisfied before applying hypothesis tests.