Answer :
Let's go through the process of determining the test statistic and the p-value step-by-step.
Step 1: Define the null and alternative hypotheses
The political pollster claims that 55% of voters prefer candidate A. We want to test if the true proportion is less than 55%.
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(p = 0.55\)[/tex]
- Alternative hypothesis ([tex]\(H_1\)[/tex]): [tex]\(p < 0.55\)[/tex]
Step 2: Calculate the sample proportion
The pollster sampled 75 voters, and 39 preferred candidate A.
[tex]\[ \hat{p} = \frac{39}{75} = 0.52 \][/tex]
Step 3: Calculate the test statistic (z-score)
The test statistic for a proportion can be calculated using the formula:
[tex]\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \][/tex]
Where:
- [tex]\(\hat{p}\)[/tex] is the sample proportion (0.52).
- [tex]\(p_0\)[/tex] is the claimed proportion under the null hypothesis (0.55).
- [tex]\(n\)[/tex] is the sample size (75).
Now we substitute these values into the formula:
[tex]\[ z = \frac{0.52 - 0.55}{\sqrt{\frac{0.55(1 - 0.55)}{75}}} \][/tex]
First, calculate the standard error:
[tex]\[ \text{Standard error} = \sqrt{\frac{0.55 \times 0.45}{75}} = \sqrt{\frac{0.2475}{75}} = \sqrt{0.0033} \approx 0.0574 \][/tex]
Then calculate the z-score:
[tex]\[ z = \frac{0.52 - 0.55}{0.0574} \approx \frac{-0.03}{0.0574} \approx -0.5222 \][/tex]
Step 4: Calculate the p-value
The p-value is the probability of obtaining a test statistic at least as extreme as the one observed, given that the null hypothesis is true. Since this is a left-tailed test, we look for the cumulative probability up to our test statistic:
Using the z-table or standard normal distribution calculator, for [tex]\(z \approx -0.5222\)[/tex]:
[tex]\[ \text{P-value} \approx 0.3008 \][/tex]
Conclusion
The test statistic [tex]\(z\)[/tex] is approximately [tex]\(-0.5222\)[/tex], and the p-value is approximately [tex]\(0.3008\)[/tex]. This indicates there isn't convincing evidence to reject the null hypothesis at any common significance level (e.g., 0.05 or 0.01). Hence, we do not have sufficient evidence to conclude that the true proportion of voters who prefer candidate A is less than 55%.
Therefore, the correct choice from the given options is:
[tex]\[ z = \frac{0.52 - 0.55}{\sqrt{\frac{0.55(1-0.55)}{75}}}, \text{P-value} = 0.3015 \][/tex]
(Note: Though the precise p-value may vary slightly depending on tables and rounding, [tex]\(0.3015\)[/tex] is the closest match to our calculated [tex]\(0.3008\)[/tex]).
Step 1: Define the null and alternative hypotheses
The political pollster claims that 55% of voters prefer candidate A. We want to test if the true proportion is less than 55%.
- Null hypothesis ([tex]\(H_0\)[/tex]): [tex]\(p = 0.55\)[/tex]
- Alternative hypothesis ([tex]\(H_1\)[/tex]): [tex]\(p < 0.55\)[/tex]
Step 2: Calculate the sample proportion
The pollster sampled 75 voters, and 39 preferred candidate A.
[tex]\[ \hat{p} = \frac{39}{75} = 0.52 \][/tex]
Step 3: Calculate the test statistic (z-score)
The test statistic for a proportion can be calculated using the formula:
[tex]\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \][/tex]
Where:
- [tex]\(\hat{p}\)[/tex] is the sample proportion (0.52).
- [tex]\(p_0\)[/tex] is the claimed proportion under the null hypothesis (0.55).
- [tex]\(n\)[/tex] is the sample size (75).
Now we substitute these values into the formula:
[tex]\[ z = \frac{0.52 - 0.55}{\sqrt{\frac{0.55(1 - 0.55)}{75}}} \][/tex]
First, calculate the standard error:
[tex]\[ \text{Standard error} = \sqrt{\frac{0.55 \times 0.45}{75}} = \sqrt{\frac{0.2475}{75}} = \sqrt{0.0033} \approx 0.0574 \][/tex]
Then calculate the z-score:
[tex]\[ z = \frac{0.52 - 0.55}{0.0574} \approx \frac{-0.03}{0.0574} \approx -0.5222 \][/tex]
Step 4: Calculate the p-value
The p-value is the probability of obtaining a test statistic at least as extreme as the one observed, given that the null hypothesis is true. Since this is a left-tailed test, we look for the cumulative probability up to our test statistic:
Using the z-table or standard normal distribution calculator, for [tex]\(z \approx -0.5222\)[/tex]:
[tex]\[ \text{P-value} \approx 0.3008 \][/tex]
Conclusion
The test statistic [tex]\(z\)[/tex] is approximately [tex]\(-0.5222\)[/tex], and the p-value is approximately [tex]\(0.3008\)[/tex]. This indicates there isn't convincing evidence to reject the null hypothesis at any common significance level (e.g., 0.05 or 0.01). Hence, we do not have sufficient evidence to conclude that the true proportion of voters who prefer candidate A is less than 55%.
Therefore, the correct choice from the given options is:
[tex]\[ z = \frac{0.52 - 0.55}{\sqrt{\frac{0.55(1-0.55)}{75}}}, \text{P-value} = 0.3015 \][/tex]
(Note: Though the precise p-value may vary slightly depending on tables and rounding, [tex]\(0.3015\)[/tex] is the closest match to our calculated [tex]\(0.3008\)[/tex]).