Question 1169897
Absolutely, let's solve this hypothesis testing problem.

**Understanding the Problem**

We're conducting a one-tailed z-test for proportions to determine if there's enough evidence to suggest that more than 50% of registered voters would vote for the Republican candidate.

**Given Information**

* Sample size (n) = 1000
* Number of voters favoring the Republican candidate (x) = 520
* Sample proportion (p̂) = x/n = 520/1000 = 0.52
* Null hypothesis (H0): p = 0.50
* Alternative hypothesis (Ha): p > 0.50

**(a) Calculating the Test Statistic (z)**

The formula for the z-test statistic for proportions is:

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

Where:

* p̂ is the sample proportion (0.52)
* p0 is the hypothesized population proportion (0.50)
* n is the sample size (1000)

Let's plug in the values:

$$z = \frac{0.52 - 0.50}{\sqrt{\frac{0.50(1 - 0.50)}{1000}}}$$

$$z = \frac{0.02}{\sqrt{\frac{0.25}{1000}}}$$

$$z = \frac{0.02}{\sqrt{0.00025}}$$

$$z = \frac{0.02}{0.015811388}$$

$$z \approx 1.265$$

So, the test statistic is approximately z = 1.265.

**(b) Calculating the p-value (using z = 1.28)**

We are given that z = 1.28. We need to find the p-value for a right-tailed test (since Ha: p > 0.50).

Using a standard normal distribution table or a calculator, we find the area to the right of z = 1.28.

* The area to the left of z = 1.28 is approximately 0.8997.
* The area to the right of z = 1.28 is 1 - 0.8997 = 0.1003.

Therefore, the p-value is approximately 0.1003.

**Answers**

(a) The test statistic is z ≈ 1.265.

(b) If z = 1.28, the p-value ≈ 0.1003.