To determine the rejection region for the hypothesis test, we follow these steps:
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Define Hypotheses:
- Null hypothesis, \( H_0: \mu \geq 60 \)
- Alternative hypothesis, \( H_1: \mu < 60 \)
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Determine the significance level (\( \alpha \)):
- Given \( \alpha = 0.05 \)
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Identify the test statistic:
We will use the t-test for the mean. The test statistic \( t \) is given by:\[ t = \frac{\bar{x} - \mu_0}{S / \sqrt{n}} \]
where:
- \( \bar{x} = 72 \) (sample mean),
- \( \mu_0 = 60 \) (hypothesized mean under \( H_0 \)),
- \( S = 20 \) (sample standard deviation),
- \( n = 100 \) (sample size).
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Calculate the test statistic:
\[ t = \frac{72 - 60}{20 / \sqrt{100}} = \frac{12}{2} = 6 \]
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Determine the critical value:
We need the critical value for a one-tailed t-test with \( \alpha = 0.05 \) and \( n - 1 = 99 \) degrees of freedom.Using a t-distribution table or calculator for \( \alpha = 0.05 \) in the left tail and \( 99 \) degrees of freedom (df), the critical value \( t_{\text{critical}} \) is approximately \(-1.660\).
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State the rejection region:
The rejection region for this one-tailed test is:\[ t < -1.660 \]
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Conclusion: If the calculated t-statistic is less than \(-1.660\), we reject the null hypothesis \( H_0 \).
Given the calculated \( t = 6 \) is greater than \(-1.660\), we do not reject the null hypothesis in this case. Thus, the rejection region for \( \alpha = 0.05 \) is \( t < -1.660 \).