To test the hypothesis that smokers need less sleep than non-smokers, we can perform a one-sample z-test for the mean. Here are the steps to perform the test:
Step 1: State the Hypotheses
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Null Hypothesis (\(H_0\)): The mean hours of sleep of smokers is equal to that of non-smokers. \[ H_0: \mu = 7.7 \]
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Alternative Hypothesis (\(H_a\)): The mean hours of sleep of smokers is less than that of non-smokers. \[ H_a: \mu < 7.7 \]
Step 2: Calculate the Test Statistic
We will use the formula for the z-test statistic:
\[ z = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}} \]
Where:
- \(\bar{x} = 7.5\) (sample mean)
- \(\mu_0 = 7.7\) (hypothesized population mean)
- \(s = 0.4\) (sample standard deviation)
- \(n = 50\) (sample size)
Now, substituting the values into the formula:
\[ z = \frac{7.5 - 7.7}{\frac{0.4}{\sqrt{50}}} \]
First, let’s calculate the denominator:
\[ \frac{0.4}{\sqrt{50}} \approx \frac{0.4}{7.0711} \approx 0.05657 \]
Now, calculate the z value:
\[ z = \frac{7.5 - 7.7}{0.05657} \approx \frac{-0.2}{0.05657} \approx -3.53 \]
Step 3: Determine the Critical Value
At a 1% significance level (\(\alpha = 0.01\)) for a one-tailed test, we will find the critical value from the standard normal distribution table.
Using a Z-table or a calculator, the critical z value for \(\alpha = 0.01\) (one-tailed) is approximately:
\[ z_{critical} = -2.33 \]
Step 4: Make a Decision
Now, we compare our calculated z value with the critical z value.
- Calculated z: \(-3.53\)
- Critical z: \(-2.33\)
Since \(-3.53 < -2.33\), we reject the null hypothesis.
Conclusion
At the 1% significance level, there is sufficient evidence to conclude that smokers need less sleep than non-smokers. The critical value for this test is \(-2.33\).
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