To perform the hypothesis test, we will follow these steps:
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Define the hypotheses:
- Null Hypothesis \( H_0 \): The average sleep of smokers is equal to that of non-smokers, \( \mu = 7.7 \) hours.
- Alternative Hypothesis \( H_a \): The average sleep of smokers is less than that of non-smokers, \( \mu < 7.7 \) hours.
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Significance Level: The level of significance given is \( \alpha = 0.25 \).
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Calculate the test statistic: We will use the Z-test for the sample mean since the sample size is large (n = 50). The test statistic is calculated using the formula:
\[ Z = \frac{\bar{X} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \]
Where:
- \( \bar{X} = 7.5 \) hours (sample mean)
- \( \mu_0 = 7.7 \) hours (population mean for non-smokers)
- \( \sigma = 0.4 \) hours (sample standard deviation)
- \( n = 50 \) (sample size)
Plugging in the values:
\[ Z = \frac{7.5 - 7.7}{\frac{0.4}{\sqrt{50}}} \]
First, calculate the standard error:
\[ \text{Standard Error} = \frac{0.4}{\sqrt{50}} \approx \frac{0.4}{7.071} \approx 0.0566 \]
Now, calculate the Z statistic:
\[ Z = \frac{-0.2}{0.0566} \approx -3.53 \]
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Determine the critical value: Since we are conducting a one-tailed test at the 0.25 significance level, we need to find the critical Z value that corresponds to the left tail of the normal distribution.
Using a Z-table or calculator for a one-tailed test with \( \alpha = 0.25 \):
The critical Z value for \( \alpha = 0.25 \) is approximately \( -0.674 \).
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Conclusion of the test:
- Compare the test statistic \( Z \) with the critical value.
- If \( Z \) is less than the critical value, we reject the null hypothesis.
Here, \( Z \approx -3.53 \) is indeed less than \( -0.674 \).
Therefore, we reject the null hypothesis and conclude that smokers tend to need less sleep than non-smokers at the 25% significance level.
Summary:
- Critical value: \( -0.674 \)
- Test statistic: \( -3.53 \)
- Conclusion: Reject \( H_0 \); smokers need less sleep than non-smokers.
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