6) HW Score: $68.73 \%, 17.18$ of 25 Points: 0 of 1 Construct the indicated confidence interval for the population mean $\mu$ using the $t$-distribution. Assume the population is normally distributed. \[ c=0.90, \bar{x}=13.2, s=0.85, n=18 \] (Round to one decimal place as needed.)
Solution
Step 1 ：First, find the t-value for the given confidence level and degrees of freedom. The degrees of freedom is \(n-1 = 18-1 = 17\). For a 90% confidence level, the t-value (two-tailed) for 17 degrees of freedom is approximately 1.740.
Step 2 ：Next, calculate the standard error of the mean (SEM), which is the standard deviation divided by the square root of the sample size: \(SEM = \frac{s}{\sqrt{n}} = \frac{0.85}{\sqrt{18}} \approx 0.20\).
Step 3 ：Then, calculate the margin of error (ME), which is the t-value times the standard error of the mean: \(ME = t \times SEM = 1.740 \times 0.20 \approx 0.35\).
Step 4 ：Finally, construct the confidence interval by subtracting and adding the margin of error from/to the sample mean: Lower limit = \(\bar{x} - ME = 13.2 - 0.35 = 12.85\). Upper limit = \(\bar{x} + ME = 13.2 + 0.35 = 13.55\).
Step 5 ：So, the 90% confidence interval for the population mean \(\mu\) is \(\boxed{(12.9, 13.6)}\) when rounded to one decimal place.
