Question

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

c = 0.99, x̄ = 12.1, s = 0.78, n = 14

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

Explanation:

Step1: Find the degrees of freedom

The degrees of freedom \( df = n - 1 = 14 - 1 = 13 \).

Step2: Find the t - critical value

For a confidence level \( c = 0.99 \), the significance level \( \alpha=1 - c = 1 - 0.99 = 0.01 \). The two - tailed critical value \( t_{\alpha/2} \) with \( df = 13 \) can be found from the t - distribution table or using a calculator. \( t_{0.005,13}\approx3.012 \)

Step3: Calculate the margin of error \( E \)

The formula for the margin of error when using the t - distribution is \( E=t_{\alpha/2}\times\frac{s}{\sqrt{n}} \). Substituting \( t_{\alpha/2}=3.012 \), \( s = 0.78 \) and \( n = 14 \):
\( E = 3.012\times\frac{0.78}{\sqrt{14}}\approx3.012\times\frac{0.78}{3.7417}\approx3.012\times0.2085\approx0.628 \)

Step4: Calculate the confidence interval

The confidence interval for the population mean \( \mu \) is given by \( \bar{x}-E<\mu<\bar{x} + E \). Substituting \( \bar{x}=12.1 \) and \( E\approx0.628 \):
Lower bound: \( 12.1-0.628 = 11.472\approx11.5 \) (rounded to one decimal place)
Upper bound: \( 12.1 + 0.628=12.728\approx12.7 \) (rounded to one decimal place)

Answer:

\((11.5, 12.7)\)