To find the P-value for the hypothesis test, we can follow these steps:
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State the null and alternative hypotheses:
- Null hypothesis (\(H_0\)): \(p = 0.65\) (the proportion of households with lawn mowers in Omaha is equal to 65%)
- Alternative hypothesis (\(H_a\)): \(p > 0.65\) (the proportion of households with lawn mowers in Omaha is greater than 65%)
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Collect the sample data:
- Sample size (\(n\)) = 497
- Number of successes (households with lawn mowers) = 340
- Sample proportion (\(\hat{p}\)) = \( \frac{340}{497} \approx 0.6847\)
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Calculate the standard error (SE):
- Use the formula for the standard error of the proportion: \[ SE = \sqrt{\frac{p_0(1 - p_0)}{n}} \] where \(p_0 = 0.65\).
\[ SE = \sqrt{\frac{0.65(1 - 0.65)}{497}} = \sqrt{\frac{0.65 \times 0.35}{497}} \approx \sqrt{\frac{0.2275}{497}} \approx \sqrt{0.000457} \approx 0.0214 \]
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Calculate the test statistic (Z): \[ Z = \frac{\hat{p} - p_0}{SE} \] \[ Z = \frac{0.6847 - 0.65}{0.0214} \approx \frac{0.0347}{0.0214} \approx 1.6203 \]
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Find the P-value: Since this is a one-tailed test (because \(H_a: p > 0.65\)), we will find the P-value associated with the calculated Z-value.
Using a Z-table or a calculator, we can find:
- The area to the right of \(Z = 1.6203\).
Using standard normal distribution, we find: \[ P(Z > 1.6203) \approx 0.0526 \]
Thus, the P-value is approximately 0.0526.
Looking at the options:
- A) 0.1118
- B) 0.0505
- C) 0.0559
- D) 0.0252
The closest option we can find that matches our approximate P-value is: C) 0.0559
This indicates that the P-value for this hypothesis test is most appropriately approximated by option C.