To construct the 80% confidence interval for the population mean \( \mu \) when the sample size \( n = 18 \), we use the formula:
\[
\bar{x} \pm t \frac{s}{\sqrt{n}}
\]
Where:
- \( \bar{x} = 106 \) (sample mean)
- \( s = 10 \) (sample standard deviation)
- \( n = 18 \) (sample size)
- \( t \) is the critical value from the t-distribution for \( \alpha = 0.20 \) and \( df = n - 1 = 17 \), which is approximately \( 1.3 \).
Calculating the margin of error:
\[
E = t \frac{s}{\sqrt{n}} = 1.3 \cdot \frac{10}{\sqrt{18}} \approx 3.1
\]
Thus, the confidence interval is:
\[
(106 - 3.1, 106 + 3.1) = (102.9, 109.1)
\]
For the sample size \( n = 26 \), we again use the confidence interval formula:
\[
\bar{x} \pm t \frac{s}{\sqrt{n}}
\]
Where:
- \( n = 26 \)
- The critical value \( t \) for \( \alpha = 0.20 \) and \( df = 25 \) is approximately \( 1.3 \).
Calculating the margin of error:
\[
E = t \frac{s}{\sqrt{n}} = 1.3 \cdot \frac{10}{\sqrt{26}} \approx 2.6
\]
Thus, the confidence interval is:
\[
(106 - 2.6, 106 + 2.6) = (103.4, 108.6)
\]
For the 90% confidence interval with \( n = 18 \), we use the same formula:
\[
\bar{x} \pm t \frac{s}{\sqrt{n}}
\]
Where:
- The critical value \( t \) for \( \alpha = 0.10 \) and \( df = 17 \) is approximately \( 1.7 \).
Calculating the margin of error:
\[
E = t \frac{s}{\sqrt{n}} = 1.7 \cdot \frac{10}{\sqrt{18}} \approx 4.1
\]
Thus, the confidence interval is:
\[
(106 - 4.1, 106 + 4.1) = (101.9, 110.1)
\]
- 80% confidence interval for \( n = 18 \): \( \boxed{(102.9, 109.1)} \)
- 80% confidence interval for \( n = 26 \): \( \boxed{(103.4, 108.6)} \)
- 90% confidence interval for \( n = 18 \): \( \boxed{(101.9, 110.1)} \)