To test the hypothesis that smokers need more sleep than non-smokers, we can set up our null and alternative hypotheses as follows:
- Null Hypothesis (H0): \( \mu_{\text{smokers}} \leq 7.7 \) (Smokers sleep on average 7.7 hours or less)
- Alternative Hypothesis (H1): \( \mu_{\text{smokers}} > 7.7 \) (Smokers sleep on average more than 7.7 hours)
Given Data:
- Sample Size (\( n \)) = 50
- Sample Mean (\( \bar{x} \)) = 8 hours
- Population Mean under Null Hypothesis (\( \mu_0 \)) = 7.7 hours
- Sample Standard Deviation (\( s \)) = 0.4 hours
- Significance Level (\( \alpha \)) = 0.10
Test Statistic Calculation:
Since we have a sample size of 50, which is greater than 30, we can use the z-test for the hypothesis testing. The test statistic \( z \) is calculated using the formula:
\[ z = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]
Substituting the values we have:
\[ z = \frac{8 - 7.7}{0.4 / \sqrt{50}} \]
Calculating the denominator:
\[ s / \sqrt{n} = 0.4 / \sqrt{50} \approx 0.4 / 7.0711 \approx 0.0566 \]
Now substituting this value back into the test statistic:
\[ z = \frac{8 - 7.7}{0.0566} = \frac{0.3}{0.0566} \approx 5.29 \]
Conclusion:
The test statistic is approximately \( z \approx 5.29 \).
Now, to determine if we can reject the null hypothesis, we would compare this test statistic to the critical value from the z-distribution for a one-tailed test at a 10% level of significance (\( \alpha = 0.10 \)). The critical value for \( z \) at \( \alpha = 0.10 \) is approximately 1.28.
Since \( 5.29 > 1.28 \), we reject the null hypothesis.
This indicates that there is sufficient evidence to conclude that smokers sleep more than non-smokers at the 10% level of significance.
1 answer