Answer :
Let's go through the problem step-by-step to determine the test statistic and the p-value for this sample.
1. State the hypotheses:
- Null hypothesis (\(H_0\)): The proportion of voters who prefer Candidate A is \( p_0 = 0.58 \).
- Alternative hypothesis (\(H_1\)): The proportion of voters who prefer Candidate A is greater than 0.58 (\( p > 0.58 \)).
2. Given data:
- Sample size (\(n\)): 434
- Number of successes (\(k\)): 266
- Significance level (\(\alpha\)): 0.005
- Population proportion under the null hypothesis (\(p_0\)): 0.58
3. Sample proportion (\(\hat{p}\)):
[tex]\[ \hat{p} = \frac{k}{n} = \frac{266}{434} = 0.6138 \][/tex]
4. Calculate the standard error (SE) of the sample proportion:
[tex]\[ \text{SE} = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.58 \cdot (1 - 0.58)}{434}} = \sqrt{\frac{0.58 \cdot 0.42}{434}} = \sqrt{\frac{0.2436}{434}} = \sqrt{0.000561} \approx 0.0237 \][/tex]
5. Calculate the test statistic (z):
[tex]\[ z = \frac{\hat{p} - p_0}{\text{SE}} = \frac{0.6138 - 0.58}{0.0237} \approx \frac{0.0338}{0.0237} \approx 1.389 \][/tex]
So, the test statistic \(z\) is approximately \(1.389\).
6. Determine the p-value:
The p-value corresponds to the probability of getting a value as extreme as, or more extreme than, the observed value of the test statistic under the null hypothesis.
Since we are dealing with a right-tailed test:
[tex]\[ p\text{-value} = 1 - \Phi(z) \approx 1 - \Phi(1.389) \][/tex]
Looking up \( \Phi(1.389) \) in the standard normal distribution table (or using a calculator/software):
[tex]\[ \Phi(1.389) \approx 0.9176 \][/tex]
Hence:
[tex]\[ p\text{-value} = 1 - 0.9176 = 0.0824 \][/tex]
So, the results are:
- Test statistic \( \approx 1.389 \)
- \( p \)-value \( \approx 0.0824 \)
Therefore, the test statistic is [tex]\(1.389\)[/tex] (accurate to three decimal places) and the [tex]\( p \)[/tex]-value for this sample is [tex]\(0.0824\)[/tex] (accurate to four decimal places).
1. State the hypotheses:
- Null hypothesis (\(H_0\)): The proportion of voters who prefer Candidate A is \( p_0 = 0.58 \).
- Alternative hypothesis (\(H_1\)): The proportion of voters who prefer Candidate A is greater than 0.58 (\( p > 0.58 \)).
2. Given data:
- Sample size (\(n\)): 434
- Number of successes (\(k\)): 266
- Significance level (\(\alpha\)): 0.005
- Population proportion under the null hypothesis (\(p_0\)): 0.58
3. Sample proportion (\(\hat{p}\)):
[tex]\[ \hat{p} = \frac{k}{n} = \frac{266}{434} = 0.6138 \][/tex]
4. Calculate the standard error (SE) of the sample proportion:
[tex]\[ \text{SE} = \sqrt{\frac{p_0 (1 - p_0)}{n}} = \sqrt{\frac{0.58 \cdot (1 - 0.58)}{434}} = \sqrt{\frac{0.58 \cdot 0.42}{434}} = \sqrt{\frac{0.2436}{434}} = \sqrt{0.000561} \approx 0.0237 \][/tex]
5. Calculate the test statistic (z):
[tex]\[ z = \frac{\hat{p} - p_0}{\text{SE}} = \frac{0.6138 - 0.58}{0.0237} \approx \frac{0.0338}{0.0237} \approx 1.389 \][/tex]
So, the test statistic \(z\) is approximately \(1.389\).
6. Determine the p-value:
The p-value corresponds to the probability of getting a value as extreme as, or more extreme than, the observed value of the test statistic under the null hypothesis.
Since we are dealing with a right-tailed test:
[tex]\[ p\text{-value} = 1 - \Phi(z) \approx 1 - \Phi(1.389) \][/tex]
Looking up \( \Phi(1.389) \) in the standard normal distribution table (or using a calculator/software):
[tex]\[ \Phi(1.389) \approx 0.9176 \][/tex]
Hence:
[tex]\[ p\text{-value} = 1 - 0.9176 = 0.0824 \][/tex]
So, the results are:
- Test statistic \( \approx 1.389 \)
- \( p \)-value \( \approx 0.0824 \)
Therefore, the test statistic is [tex]\(1.389\)[/tex] (accurate to three decimal places) and the [tex]\( p \)[/tex]-value for this sample is [tex]\(0.0824\)[/tex] (accurate to four decimal places).
