To calculate the test statistic and the p-value for the given hypothesis test, we follow these steps:
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Define the null and alternative hypotheses:
- Null hypothesis (\(H_0\)): \(p = 0.87\)
- Alternative hypothesis (\(H_1\)): \(p < 0.87\) (Note: Your provided alternative hypothesis seems to state \(p > 0.87\), but your question indicates that you're checking if the accuracy rate is less than 0.87, which should be \(H_1: p < 0.87\). Please verify if you intended to perform a one-tailed test as described or if it should be two-tailed.)
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Sample proportion: Given:
- Sample size (\(n\)) = 501
- Number of successes (\(x\)) = 421
Calculate the sample proportion (\(\hat{p}\)): \[ \hat{p} = \frac{x}{n} = \frac{421}{501} \approx 0.8403 \]
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Calculate the test statistic: The formula for the test statistic \(z\) for a proportion is: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] where \(p_0\) is the hypothesized population proportion (0.87).
Substituting the values: \[ z = \frac{0.8403 - 0.87}{\sqrt{\frac{0.87 \times (1 - 0.87)}{501}}} \]
First, calculate the standard error: \[ SE = \sqrt{\frac{0.87 \times 0.13}{501}} \approx \sqrt{\frac{0.1131}{501}} \approx \sqrt{0.000225192} \approx 0.015 \]
Now compute \(z\): \[ z = \frac{0.8403 - 0.87}{0.015} \approx \frac{-0.0297}{0.015} \approx -1.98 \]
So, the test statistic is: \[ z \approx -1.980 \]
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Calculate the p-value: To find the p-value for a left-tailed test with the calculated \(z\)-score, we can use standard normal tables or a calculator to find the area to the left of \(z = -1.980\): \[ \text{p-value} = P(Z < -1.980) \]
From standard normal distribution tables or software: \[ \text{p-value} \approx 0.0236 \]
Combining the results, we find:
- Test statistic: \(z \approx -1.980\)
- p-value: \( \approx 0.0236\)
Final answers:
Test statistic = -1.980
p-value = 0.0236
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