A random sample of 16 smokers were questioned about the number of hours they sleep each day. The sample of smokers sleeps on average 8 hours per day with a standard deviation of 4 hours. We want to test the hypothesis that smokers need more sleep than non-smokers, who need an average of 7 hours of sleep. Perform the test at a 5% level of significance. Calculate the test statistic.

1 answer

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.