You are conducting a study to see if the proportion of voters who prefer Candidate A is significantly less than 0.56. You use a significance level of alpha= 0.001.

Question

H_0 : p = 0.56
H_1 : p < 0.56

You obtain a sample of size n=201 in which there are 98 successes.

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic = ____

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value = ____

The p-value is...
A. less than (or equal to) alpha
B. greater than alpha

This test statistic leads to a decision to...
A. reject the null
B. accept the null
C. fail to reject the null

As such, the final conclusion is that...
A. There is sufficient evidence to warrant rejection of the claim that the proportion of voters who prefer Candidate A is less than 0.56.
B. There is not sufficient evidence to warrant rejection of the claim that the proportion of voters who prefer Candidate A is less than 0.56.
C. The sample data support the claim that the proportion of voters who prefer Candidate A is less than 0.56.
D. There is not sufficient sample evidence to support the claim that the proportion of voters who prefer Candidate A is less than 0.56.

GPT-4o mini · Answer

To conduct the hypothesis test, we will start by calculating the test statistic using the formula for the z-test for proportions:

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

Where:

\(\hat{p} = \frac{x}{n}\) is the sample proportion,
\(p_0\) is the population proportion under the null hypothesis,
\(n\) is the sample size,
\(x\) is the number of successes.

Given:

Sample size \(n = 201\)
Number of successes \(x = 98\)
Hypothesized proportion \(p_0 = 0.56\)

First, calculate the sample proportion \(\hat{p}\):

\[ \hat{p} = \frac{98}{201} \approx 0.4876 \]

Next, we compute the standard error:

\[ \text{Standard Error} = \sqrt{\frac{p_0(1 - p_0)}{n}} = \sqrt{\frac{0.56 \times (1 - 0.56)}{201}} = \sqrt{\frac{0.56 \times 0.44}{201}} \approx \sqrt{\frac{0.2464}{201}} \approx \sqrt{0.001225} \approx 0.0350 \]

Now, substituting \(\hat{p}\), \(p_0\), and the standard error into the z-test formula:

\[ z = \frac{0.4876 - 0.56}{0.0350} \approx \frac{-0.0724}{0.0350} \approx -2.069 \]

Rounding the test statistic to three decimal places:

\[ \text{test statistic} = -2.069 \]

Next, we calculate the p-value using the z-score. Since this is a one-tailed test, we find the area to the left of the calculated z-value. Using the standard normal distribution table or calculator:

\[ \text{p-value} = P(Z < -2.069) \approx 0.0191 \]

Rounding the p-value to four decimal places:

\[ \text{p-value} = 0.0191 \]

Now we compare the p-value to the significance level (\(\alpha = 0.001\)):

The p-value (0.0191) is greater than \(\alpha\) (0.001).

Decisions:

P-value Comparison:
- P-value is:
  - B. greater than alpha.
Decision based on p-value:
- Since the p-value is greater than alpha, we fail to reject the null hypothesis:
  - C. fail to reject the null.
Final Conclusion:
- Since we failed to reject the null hypothesis, we conclude:
  - B. There is not sufficient evidence to warrant rejection of the claim that the proportion of voters who prefer Candidate A is less than 0.56.

Summary of Results:

Test statistic = -2.069
P-value ≈ 0.0191
P-value is B. greater than alpha.
Decision: C. fail to reject the null.
Final conclusion: B. There is not sufficient evidence to warrant rejection of the claim that the proportion of voters who prefer Candidate A is less than 0.56.