Question

construct the indicated confidence interval for the population mean μ using the t-distribution. assume the population is normally distributed.

c = 0.95, x̄ = 14.8, s = 3.0, n = 7

( , )
(round to one decimal place as needed.)

Explanation:

Step1: Find the degrees of freedom

The degrees of freedom \( df = n - 1 = 7 - 1 = 6 \).

Step2: Find the t - critical value

For a confidence level \( c = 0.95 \), the significance level \( \alpha=1 - c=1 - 0.95 = 0.05 \), and \( \frac{\alpha}{2}=0.025 \). Looking up in the t - distribution table with \( df = 6 \), the critical value \( t_{\alpha/2}=t_{0.025,6}\approx 2.447 \).

Step3: Calculate the standard error

The standard error \( SE=\frac{s}{\sqrt{n}}=\frac{3.0}{\sqrt{7}}\approx\frac{3.0}{2.6458}\approx1.134 \).

Step4: Calculate the margin of error

The margin of error \( E = t_{\alpha/2}\times SE=2.447\times1.134\approx2.775 \).

Step5: Calculate the confidence interval

The confidence interval is \( \bar{x}-E<\mu<\bar{x} + E \). Substituting \( \bar{x} = 14.8 \) and \( E\approx2.775 \), we get \( 14.8-2.775<\mu<14.8 + 2.775 \), which is \( 12.025<\mu<17.575 \). Rounding to one decimal place, we have \( 12.0<\mu<17.6 \).

Answer:

\( 12.0 < \mu < 17.6 \)