The mean mass of 50 male students who showed above average participation in athletics was 68.2 kg with a standard deviation of 3 kg, while 50 male students who showed no interest in such participation had mean mass of 67.5 with a standard deviation of 2.8 kg.

Find a
8.1 99%,
8.2 95 %;
8.3 90%;
8.4 80%, confidence interval for the difference between the population means of the two groups.

1 answer

To find the confidence intervals for the difference between the population means of the two groups, we can use the formula for the confidence interval for the difference between two means. The general formula is given by:

\[ CI = (\bar{X_1} - \bar{X_2}) \pm Z \times SE \]

Where:

  • \(\bar{X_1}\) and \(\bar{X_2}\) are the sample means,
  • \(Z\) is the Z-value corresponding to the desired confidence level,
  • \(SE\) is the standard error of the difference between the means.

The standard error (SE) can be calculated using the formula:

\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]

Where:

  • \(s_1\) and \(s_2\) are the standard deviations of the two groups,
  • \(n_1\) and \(n_2\) are the sample sizes.

Given:

  • Group 1 (athletic students):

    • Mean (\(\bar{X_1}\)) = 68.2 kg
    • Standard deviation (\(s_1\)) = 3 kg
    • Sample size \(n_1 = 50\)
  • Group 2 (non-athletic students):

    • Mean (\(\bar{X_2}\)) = 67.5 kg
    • Standard deviation (\(s_2\)) = 2.8 kg
    • Sample size \(n_2 = 50\)

Step 1: Calculate the standard error (SE)

\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{3^2}{50} + \frac{2.8^2}{50}} = \sqrt{\frac{9}{50} + \frac{7.84}{50}} = \sqrt{\frac{16.84}{50}} = \sqrt{0.3368} \approx 0.58 \text{ kg} \]

Step 2: Calculate the difference in means

\[ \Delta \bar{X} = \bar{X_1} - \bar{X_2} = 68.2 - 67.5 = 0.7 \text{ kg} \]

Step 3: Find the Z-values for the different confidence levels

  • For 99% confidence level, \(Z \approx 2.576\)
  • For 95% confidence level, \(Z \approx 1.96\)
  • For 90% confidence level, \(Z \approx 1.645\)
  • For 80% confidence level, \(Z \approx 1.282\)

Step 4: Calculate the confidence intervals

8.1: 99% Confidence Interval

\[ CI = 0.7 \pm 2.576 \times 0.58 \]

\[ = 0.7 \pm 1.4946 \]

\[ = (0.7 - 1.4946, 0.7 + 1.4946) = (-0.7946, 2.1946) \text{ kg} \]

8.2: 95% Confidence Interval

\[ CI = 0.7 \pm 1.96 \times 0.58 \]

\[ = 0.7 \pm 1.1288 \]

\[ = (0.7 - 1.1288, 0.7 + 1.1288) = (-0.4288, 1.8288) \text{ kg} \]

8.3: 90% Confidence Interval

\[ CI = 0.7 \pm 1.645 \times 0.58 \]

\[ = 0.7 \pm 0.9541 \]

\[ = (0.7 - 0.9541, 0.7 + 0.9541) = (-0.2541, 1.6541) \text{ kg} \]

8.4: 80% Confidence Interval

\[ CI = 0.7 \pm 1.282 \times 0.58 \]

\[ = 0.7 \pm 0.7436 \]

\[ = (0.7 - 0.7436, 0.7 + 0.7436) = (-0.0436, 1.4436) \text{ kg} \]

Summary of Confidence Intervals:

  • 99% Confidence Interval: \((-0.7946, 2.1946)\) kg
  • 95% Confidence Interval: \((-0.4288, 1.8288)\) kg
  • 90% Confidence Interval: \((-0.2541, 1.6541)\) kg
  • 80% Confidence Interval: \((-0.0436, 1.4436)\) kg