Question 1169897: A random sample of n=1000 registered voters and found that 520 would vote for the Republican candidate in a state senate race. Let p represent the proportion of registered voters who would vote for the Republican candidate.
H0:p=.50
Ha:p>.50
(a) The test statistic is z =
(b) Regardless of what you acutally computed, suppose your answer to part (a) was z = 1.28. Using this z, p-value =
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website! 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.
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