To test the hypothesis that smokers need less sleep than non-smokers, we can perform a one-sample \( t \)-test. We will formulate our hypotheses as follows:
- Null Hypothesis (\( H_0 \)): The average hours of sleep for smokers is equal to non-smokers, \( \mu = 8 \) hours.
- Alternative Hypothesis (\( H_a \)): The average hours of sleep for smokers is less than that of non-smokers, \( \mu < 8 \).
Given the information:
- Sample size (\( n \)) = 25
- Sample mean (\( \bar{x} \)) = 7 hours
- Population mean (\( \mu_0 \)) = 8 hours (mean sleep hours for non-smokers)
- Sample standard deviation (\( s \)) = 4 hours
Step 1: Calculate the test statistic
The formula for the \( t \)-statistic is given by:
\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]
Plugging in the values:
- \( \bar{x} = 7 \)
- \( \mu_0 = 8 \)
- \( s = 4 \)
- \( n = 25 \)
We can now calculate:
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Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{4}{\sqrt{25}} = \frac{4}{5} = 0.8 \]
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Calculate the \( t \)-statistic: \[ t = \frac{7 - 8}{0.8} = \frac{-1}{0.8} = -1.25 \]
Step 2: Determine the critical value and make a decision
Since we are performing a one-tailed test at a 1% significance level (\( \alpha = 0.01 \)), we need to find the critical value from the \( t \)-distribution with \( n - 1 = 24 \) degrees of freedom.
Using a \( t \)-table or a calculator, we find that the critical value \( t_{0.01} \) for 24 degrees of freedom is approximately -2.492 (the negative sign is because we are looking for the lower tail).
Step 3: Compare the test statistic to the critical value
- Our calculated test statistic \( t = -1.25 \)
- Critical value \( t_{critical} = -2.492 \)
Since \( -1.25 > -2.492 \), we do not reject the null hypothesis \( H_0 \).
Conclusion
At the 1% level of significance, we do not have enough evidence to support the claim that smokers need less sleep than non-smokers. The calculated test statistic is \( t = -1.25 \).