To conduct the hypothesis test, we will calculate the test statistic using the formula for the z-test for a single mean:
\[ z = \frac{M - \mu_0}{\sigma / \sqrt{n}} \]
where:
- \( M \) = sample mean
- \( \mu_0 \) = population mean under the null hypothesis
- \( \sigma \) = population standard deviation
- \( n \) = sample size
Given:
- \( M = 67.5 \)
- \( \mu_0 = 74.7 \)
- \( \sigma = 20.1 \)
- \( n = 52 \)
Let's compute the test statistic:
- Calculate the standard error (SE):
\[ SE = \frac{\sigma}{\sqrt{n}} = \frac{20.1}{\sqrt{52}} \approx \frac{20.1}{7.2111} \approx 2.784 \]
- Calculate the z-test statistic:
\[ z = \frac{67.5 - 74.7}{2.784} = \frac{-7.2}{2.784} \approx -2.586 \]
Now we need to round the test statistic to three decimal places:
\[ \text{test statistic} \approx -2.586 \]
Next, we calculate the p-value. Since this is a left-tailed test (since \( H_a: \mu < 74.7 \)), we will look for the probability that \( Z \) is less than the value we just calculated.
To find the p-value, we can use the standard normal distribution table or a calculator:
- Looking up \( z = -2.586 \):
Using a standard normal distribution table or calculator, we find:
\[ \text{p-value} = P(Z < -2.586) \approx 0.0049 \]
We will round the p-value to four decimal places:
\[ \text{p-value} \approx 0.0049 \]
Thus, the final results are:
- Test statistic: \(-2.586\)
- p-value: \(0.0049\)
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