To find the 99% confidence interval for the difference in mean mass between the two groups of male students (those who participated in athletics vs. those who did not), we can use the following formula for the confidence interval for the difference between two means:
\[ \text{CI} = (\bar{x}_1 - \bar{x}2) \pm Z{\alpha/2} \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]
Where:
- \(\bar{x}_1\) and \(\bar{x}_2\) are the means of the two groups,
- \(s_1\) and \(s_2\) are the standard deviations of the two groups,
- \(n_1\) and \(n_2\) are the sample sizes of the two groups,
- \(Z_{\alpha/2}\) is the Z-value for the desired confidence level.
Given Data:
-
Group 1 (Athletes):
- \(\bar{x}_1 = 68.2\) kg
- \(s_1 = 3\) kg
- \(n_1 = 50\)
-
Group 2 (Non-athletes):
- \(\bar{x}_2 = 67.5\) kg
- \(s_2 = 2.8\) kg
- \(n_2 = 50\)
Step 1: Calculate the difference in means:
\[ \bar{x}_1 - \bar{x}_2 = 68.2 - 67.5 = 0.7 \text{ kg} \]
Step 2: Find the Z-value for a 99% confidence interval:
For a 99% confidence level, \(\alpha = 0.01\), so \(\alpha/2 = 0.005\). The Z-value corresponding to 99% confidence (using Z-tables or a calculator) is approximately 2.576.
Step 3: Calculate the standard error:
\[ \text{SE} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{3^2}{50} + \frac{2.8^2}{50}} \] \[ = \sqrt{\frac{9}{50} + \frac{7.84}{50}} = \sqrt{\frac{16.84}{50}} = \sqrt{0.3368} \approx 0.58 \]
Step 4: Calculate the confidence interval:
\[ \text{CI} = 0.7 \pm 2.576 \cdot 0.58 \] \[ = 0.7 \pm 1.49408 \]
Calculating the limits:
- Lower limit: \(0.7 - 1.49408 \approx -0.79408\)
- Upper limit: \(0.7 + 1.49408 \approx 2.19408\)
Conclusion:
The 99% confidence interval for the difference in mean mass between the two groups is approximately:
\[ (-0.79 \text{ kg}, 2.19 \text{ kg}) \]
This means we are 99% confident that the difference in mean mass lies within this interval.
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