A random sample of 25 smokers were questioned about the number of hours they sleep each day. The sample of smokers sleeps on average 7 hours per day with a standard deviation of 4 hours. We want to test the hypothesis that smokers need less sleep than non-smokers, who need an average of 8 hours of sleep. Perform the test at a 1% level of significance. Calculate the test statistic.

1 answer

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:

  1. Calculate the standard error (SE): \[ SE = \frac{s}{\sqrt{n}} = \frac{4}{\sqrt{25}} = \frac{4}{5} = 0.8 \]

  2. 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 \( t = -1.25 \).