Answer :
To find a [tex]\(90\% \)[/tex] confidence interval for the difference in the means of students per class between the Portland Public School district ([tex]\( \mu_1 \)[/tex]) and the Beaverton School district ([tex]\( \mu_2 \)[/tex]), we use the following information:
- Sample size for Portland ([tex]\(n_1\)[/tex]): 28
- Sample mean for Portland ([tex]\(\bar{x}_1\)[/tex]): 36
- Sample standard deviation for Portland ([tex]\(s_1\)[/tex]): 4
- Sample size for Beaverton ([tex]\(n_2\)[/tex]): 27
- Sample mean for Beaverton ([tex]\(\bar{x}_2\)[/tex]): 34
- Sample standard deviation for Beaverton ([tex]\(s_2\)[/tex]): 5
Step-by-Step Solution:
1. Calculate the standard error of the difference in means:
[tex]\[ SE = \sqrt{\left(\frac{s_1^2}{n_1}\right) + \left(\frac{s_2^2}{n_2}\right)} \][/tex]
Substituting the values:
[tex]\[ SE = \sqrt{\left(\frac{4^2}{28}\right) + \left(\frac{5^2}{27}\right)} = \sqrt{\left(\frac{16}{28}\right) + \left(\frac{25}{27}\right)} \][/tex]
[tex]\[ SE \approx \sqrt{0.5714 + 0.9259} \approx \sqrt{1.4973} \approx 1.2237 \][/tex]
2. Calculate degrees of freedom using the Welch-Satterthwaite equation:
[tex]\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2} {\left(\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}\right)} \][/tex]
Substituting the values:
[tex]\[ df = \frac{\left(\frac{16}{28} + \frac{25}{27}\right)^2} {\left(\frac{\left(\frac{16}{28}\right)^2}{28-1} + \frac{\left(\frac{25}{27}\right)^2}{27-1}\right)} \][/tex]
[tex]\[ df \approx \frac{(0.5714 + 0.9259)^2} {\left(\frac{0.5714^2}{27} + \frac{0.9259^2}{26}\right)} \approx \frac{1.4973^2} {\left(\frac{0.3266}{27} + \frac{0.8572}{26}\right)} \][/tex]
[tex]\[ df \approx \frac{2.2420} {0.0121 + 0.0330} \approx \frac{2.2420}{0.0451} \approx 49.7061 \approx 50 \; (\text{rounded}) \][/tex]
3. Determine the critical t-value ([tex]\( t_{\text{critical}} \)[/tex]) for a [tex]\(90\% \)[/tex] confidence level and [tex]\(df = 50\)[/tex]:
Using t-distribution tables or statistical software, we find the critical t-value for a 90% confidence interval with [tex]\(df \approx 50\)[/tex] is approximately [tex]\( t_{\text{critical}} \approx 1.676 \)[/tex].
4. Calculate the margin of error:
[tex]\[ \text{Margin of Error} = t_{\text{critical}} \times SE = 1.676 \times 1.2237 \approx 2.0518 \][/tex]
5. Construct the confidence interval:
First, we find the difference in sample means:
[tex]\[ \bar{x}_1 - \bar{x}_2 = 36 - 34 = 2 \][/tex]
Then, apply the margin of error to find the lower and upper bounds:
[tex]\[ \text{Lower bound} = (\bar{x}_1 - \bar{x}_2) - \text{Margin of Error} = 2 - 2.0518 \approx -0.0518 \][/tex]
[tex]\[ \text{Upper bound} = (\bar{x}_1 - \bar{x}_2) + \text{Margin of Error} = 2 + 2.0518 \approx 4.0518 \][/tex]
Thus, the [tex]\(90\%\)[/tex] confidence interval for the difference in the means is:
[tex]\[ -0.0509 < \mu_1 - \mu_2 < 4.0509 \][/tex]
Rounded to 4 decimal places.
- Sample size for Portland ([tex]\(n_1\)[/tex]): 28
- Sample mean for Portland ([tex]\(\bar{x}_1\)[/tex]): 36
- Sample standard deviation for Portland ([tex]\(s_1\)[/tex]): 4
- Sample size for Beaverton ([tex]\(n_2\)[/tex]): 27
- Sample mean for Beaverton ([tex]\(\bar{x}_2\)[/tex]): 34
- Sample standard deviation for Beaverton ([tex]\(s_2\)[/tex]): 5
Step-by-Step Solution:
1. Calculate the standard error of the difference in means:
[tex]\[ SE = \sqrt{\left(\frac{s_1^2}{n_1}\right) + \left(\frac{s_2^2}{n_2}\right)} \][/tex]
Substituting the values:
[tex]\[ SE = \sqrt{\left(\frac{4^2}{28}\right) + \left(\frac{5^2}{27}\right)} = \sqrt{\left(\frac{16}{28}\right) + \left(\frac{25}{27}\right)} \][/tex]
[tex]\[ SE \approx \sqrt{0.5714 + 0.9259} \approx \sqrt{1.4973} \approx 1.2237 \][/tex]
2. Calculate degrees of freedom using the Welch-Satterthwaite equation:
[tex]\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2} {\left(\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}\right)} \][/tex]
Substituting the values:
[tex]\[ df = \frac{\left(\frac{16}{28} + \frac{25}{27}\right)^2} {\left(\frac{\left(\frac{16}{28}\right)^2}{28-1} + \frac{\left(\frac{25}{27}\right)^2}{27-1}\right)} \][/tex]
[tex]\[ df \approx \frac{(0.5714 + 0.9259)^2} {\left(\frac{0.5714^2}{27} + \frac{0.9259^2}{26}\right)} \approx \frac{1.4973^2} {\left(\frac{0.3266}{27} + \frac{0.8572}{26}\right)} \][/tex]
[tex]\[ df \approx \frac{2.2420} {0.0121 + 0.0330} \approx \frac{2.2420}{0.0451} \approx 49.7061 \approx 50 \; (\text{rounded}) \][/tex]
3. Determine the critical t-value ([tex]\( t_{\text{critical}} \)[/tex]) for a [tex]\(90\% \)[/tex] confidence level and [tex]\(df = 50\)[/tex]:
Using t-distribution tables or statistical software, we find the critical t-value for a 90% confidence interval with [tex]\(df \approx 50\)[/tex] is approximately [tex]\( t_{\text{critical}} \approx 1.676 \)[/tex].
4. Calculate the margin of error:
[tex]\[ \text{Margin of Error} = t_{\text{critical}} \times SE = 1.676 \times 1.2237 \approx 2.0518 \][/tex]
5. Construct the confidence interval:
First, we find the difference in sample means:
[tex]\[ \bar{x}_1 - \bar{x}_2 = 36 - 34 = 2 \][/tex]
Then, apply the margin of error to find the lower and upper bounds:
[tex]\[ \text{Lower bound} = (\bar{x}_1 - \bar{x}_2) - \text{Margin of Error} = 2 - 2.0518 \approx -0.0518 \][/tex]
[tex]\[ \text{Upper bound} = (\bar{x}_1 - \bar{x}_2) + \text{Margin of Error} = 2 + 2.0518 \approx 4.0518 \][/tex]
Thus, the [tex]\(90\%\)[/tex] confidence interval for the difference in the means is:
[tex]\[ -0.0509 < \mu_1 - \mu_2 < 4.0509 \][/tex]
Rounded to 4 decimal places.
