To solve this problem, we first set up our hypotheses regarding the heights of contemporary males and eighteenth-century males:
- Null hypothesis (H0): \( \mu_1 - \mu_2 \leq 0 \) (where \( \mu_1 \) is the mean height of contemporary males and \( \mu_2 \) is the mean height of eighteenth-century males).
- Alternative hypothesis (H1): \( \mu_1 - \mu_2 > 0 \) (contemporary males are taller).
Given Data:
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Sample size of contemporary males (\( n_1 \)) = 30
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Mean height of contemporary males (\( \bar{x_1} \)) = 70.1 inches
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Standard deviation of contemporary males (\( s_1 \)) = 2.52 inches
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Sample size of eighteenth-century males (\( n_2 \)) = 30
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Mean height of eighteenth-century males (\( \bar{x_2} \)) = 65.2 inches
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Standard deviation of eighteenth-century males (\( s_2 \)) = 3.51 inches
Step 1: Calculate the Test Statistic
We can use a two-sample t-test for this purpose:
\[ \bar{x_1} - \bar{x_2} = 70.1 - 65.2 = 4.9 \]
Calculating the standard error (SE) for the difference in means:
\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{(2.52)^2}{30} + \frac{(3.51)^2}{30}} \]
Calculating the values:
\[ \frac{(2.52)^2}{30} = \frac{6.3504}{30} = 0.21168 \quad \text{and} \quad \frac{(3.51)^2}{30} = \frac{12.3201}{30} = 0.41067 \]
Adding these together and taking the square root:
\[ SE = \sqrt{0.21168 + 0.41067} = \sqrt{0.62235} \approx 0.789 \]
Calculating the t-statistic:
\[ t = \frac{(\bar{x_1} - \bar{x_2}) - 0}{SE} = \frac{4.9}{0.789} \approx 6.21 \]
Step 2: Determine Degrees of Freedom
Degrees of freedom (\( df \)) can be estimated using the formula for two-sample t-tests:
\[ df = \frac{ \left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2 }{ \frac{ \left( \frac{s_1^2}{n_1} \right)^2 }{n_1 - 1} + \frac{ \left( \frac{s_2^2}{n_2} \right)^2 }{n_2 - 1} } \]
However, for simplicity, we will use the simpler method here and roughly assume \( df \approx n_1 + n_2 - 2 = 30 + 30 - 2 = 58 \).
Step 3: Calculate the P-value
Looking up the t-distribution table for \( t \approx 6.21 \) with a one-tailed test at \( df = 58 \):
Since this value is very high, the P-value will be extremely small, effectively approaching 0.
Decision
At a significance level (\( \alpha \)) of 0.01, we find that the P-value is significantly less than 0.01.
Conclusion
According to our calculations and analysis, the right answer is:
D) The P-value is less than 0.00001. There is sufficient evidence to reject the null hypothesis.