Question

You are comparing heights of contemporary males and eighteenth-century males. A random sample of 30 contemporary males has a mean of 70.1 inches and a standard deviation of 2.52 inches. A random sample of 30 eighteenth-century males has a mean of 65.2 inches and a standard deviation of 3.51 inches. Using a confidence level of 0.01, is there sufficient evidence to conclude that contemporary males are taller than eighteenth-century males?

A) The P-value is less than 0.00001. There is insufficient evidence to reject the null hypothesis.
B) The P-value is greater than 0.0001. There is sufficient evidence to reject the null hypothesis.
C) The P-value is greater than 0.00001. There is insufficient evidence to reject the null hypothesis.
D) The P-value is less than 0.00001. There is sufficient evidence to reject the null hypothesis.

Answer 1

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:

- Sample size of contemporary males (\( n_1 \)) = 30
- Mean height of contemporary males (\( \bar{x_1} \)) = 70.1 inches
- Standard deviation of contemporary males (\( s_1 \)) = 2.52 inches

- Sample size of eighteenth-century males (\( n_2 \)) = 30
- Mean height of eighteenth-century males (\( \bar{x_2} \)) = 65.2 inches
- 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.**