Questions: Suppose it is desired to compare two physical education training programs for preadolescent girls. A total of 82 girls are randomly selected, with 41 assigned to each program. After a three 6-week period on the program, each girl is given a fitness test that yields a score between 0 to 100. The mean and variances of the scores for the two groups are shown in the table. n X̄ s^2 Program 1 41 78.7 201.64 Program 2 41 75.3 259.21 The 90% confidence interval for the difference of two population means (μ1-μ2) is?

Solution Steps

We have two physical education training programs for preadolescent girls, with the following statistics for their fitness test scores:

• Program 1:

  • Sample size (\(n_1\)) = 41
  • Mean score (\(\bar{x}_1\)) = 78.7
  • Variance (\(s_1^2\)) = 201.64

• Program 2:

  • Sample size (\(n_2\)) = 41
  • Mean score (\(\bar{x}_2\)) = 75.3
  • Variance (\(s_2^2\)) = 259.21

The sample standard deviations for both programs are calculated as follows:

\[ s_1 = \sqrt{s_1^2} = \sqrt{201.64} \approx 14.2 \] \[ s_2 = \sqrt{s_2^2} = \sqrt{259.21} \approx 16.1 \]

To find the 90% confidence interval for the difference between the two population means (\(\mu_1 - \mu_2\)), we use the formula:

\[ (\bar{x}_1 - \bar{x}_2) \pm z \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]

Where \(z\) is the critical value for a 90% confidence level, which is approximately \(1.64\).

Substituting the values:

\[ (78.7 - 75.3) \pm 1.64 \sqrt{\frac{201.64}{41} + \frac{259.21}{41}} \]

Calculating the difference:

\[ 78.7 - 75.3 = 3.4 \]

Calculating the standard error:

\[ \sqrt{\frac{201.64}{41} + \frac{259.21}{41}} = \sqrt{4.91 + 6.32} = \sqrt{11.23} \approx 3.35 \]

Now substituting back into the confidence interval formula:

\[ 3.4 \pm 1.64 \cdot 3.35 \]

Calculating the margin of error:

\[ 1.64 \cdot 3.35 \approx 5.49 \]

Thus, the confidence interval is:

\[ 3.4 - 5.49 \quad \text{to} \quad 3.4 + 5.49 \]

Calculating the bounds:

\[ -2.11 \quad \text{to} \quad 8.91 \]

Final Answer