To test the hypothesis that smokers need more sleep than non-smokers, we will set up our null and alternative hypotheses, calculate the test statistic, and compare it to the critical value.
Hypotheses
- Null Hypothesis (\(H_0\)): \(\mu = 7\) (The mean sleep duration for smokers is equal to that of non-smokers)
- Alternative Hypothesis (\(H_a\)): \(\mu > 7\) (The mean sleep duration for smokers is greater than that of non-smokers)
Given Data
- Sample size (\(n\)) = 16 (smokers)
- Sample mean (\(\bar{x}\)) = 8 hours
- Sample standard deviation (\(s\)) = 4 hours
- Population mean for non-smokers (\(\mu_0\)) = 7 hours
- Significance level (\(\alpha\)) = 0.05
Test Statistic
Since we are using a small sample size (n < 30) and we do not know the population standard deviation, we will use a t-test. The formula for the t-test statistic is:
\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]
Substituting the values into the formula:
- \(\bar{x} = 8\)
- \(\mu_0 = 7\)
- \(s = 4\)
- \(n = 16\)
Calculating the standard error (SE):
\[ SE = \frac{s}{\sqrt{n}} = \frac{4}{\sqrt{16}} = \frac{4}{4} = 1 \]
Now, substituting these values into the t statistic formula:
\[ t = \frac{8 - 7}{1} = \frac{1}{1} = 1 \]
Conclusion
The calculated test statistic is \(t = 1\).
To determine if we will reject or fail to reject our null hypothesis, we need to compare this t-value against the critical t-value for a one-tailed test with \(n - 1 = 16 - 1 = 15\) degrees of freedom at a significance level of \(\alpha = 0.05\).
Using a t-distribution table or calculator, the critical t-value for \(df = 15\) at \(\alpha = 0.05\) (one-tailed) is approximately \(1.753\).
Since \(t = 1\) is less than \(1.753\), we do not reject the null hypothesis.
Final Result
We do not have sufficient evidence at the 5% level of significance to conclude that smokers need more sleep than non-smokers. The calculated test statistic is 1.