To find the test statistic $z$ for the hypothesis test, you can use the formula:

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

where:

• $\hat{p}$ is the sample proportion
• $p_0$ is the hypothesized population proportion (from $H_0: p = 0.50$)
• $n$ is the sample size

Given:

• $\hat{p} = 0.47$
• $p_0 = 0.50$
• $n = 100$

Let's calculate the test statistic $z$:

$z = \frac{0.47 - 0.50}{\sqrt{\frac{0.50(1 - 0.50)}{100}}}$

Now, we can compute this step-by-step:

1. Calculate the numerator: $0.47 - 0.50 = -0.03$.
2. Calculate the denominator:

$\sqrt{\frac{0.50(1 - 0.50)}{100}} = \sqrt{\frac{0.50 \times 0.50}{100}} = \sqrt{\frac{0.25}{100}} = \sqrt{0.0025} = 0.05$

3. Now, divide the numerator by the denominator:

$z = \frac{-0.03}{0.05} = -0.60$

So, the test statistic $z$ is $-0.60$.

Would you like more details or have any further questions?

Here are 8 related questions you might find helpful:

1. How do you interpret the $z$-value in a hypothesis test?
2. What is the next step after finding the test statistic in hypothesis testing?
3. How do you determine the p-value from the test statistic $z$?
4. What is the critical value for a two-tailed test at a 5% significance level?
5. How does the sample size affect the $z$-value in hypothesis testing?
6. What are the assumptions for using a $z$-test for proportions?
7. How would the test statistic change if the sample proportion were 0.53 instead of 0.47?
8. What are the differences between a $z$-test and a $t$-test?

Tip: When the sample size is large (generally $n > 30$), the sampling distribution of the sample proportion can be approximated by a normal distribution, allowing the use of the $z$-test.