Question

construct the indicated confidence interval for the population mean \\( \mu \\) using the t-distribution. assume the population is normally distributed.\\( c = 0.89, \bar{x} = 12.1, s = 0.78, n = 14 \\)\\( (\square, \square) \\)\\( \text{(round to one decimal place as needed.)} \\)

Explanation:

Step1: Determine degrees of freedom

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

Step2: Find t - critical value

Confidence level \( c = 0.89 \), so the significance level \( \alpha=1 - c = 1 - 0.89 = 0.11 \). The two - tailed critical value \( t_{\alpha/2,df}=t_{0.055,13}\). Using a t - table or a calculator (for example, in a TI - 84 Plus, we can use the invT function: invT(0.945,13) since \( \alpha/2 = 0.055 \) and the area to the left of the critical value is \( 1 - 0.055=0.945\)), we get \( t_{0.055,13}\approx1.602 \).

Step3: Calculate the standard error

The standard error \( SE=\frac{s}{\sqrt{n}}=\frac{0.78}{\sqrt{14}}\approx\frac{0.78}{3.7417}\approx0.2085 \).

Step4: Calculate the margin of error

Margin of error \( E = t_{\alpha/2,df}\times SE=1.602\times0.2085\approx0.334 \).

Step5: Calculate the confidence interval

The confidence interval for the population mean \( \mu \) is \( \bar{x}-E<\mu<\bar{x} + E \). Substituting \( \bar{x} = 12.1 \) and \( E\approx0.334 \), we get \( 12.1-0.334<\mu<12.1 + 0.334 \), which is \( 11.766<\mu<12.434 \). Rounding to one decimal place, the confidence interval is \( (11.8,12.4) \).

Answer:

\((11.8, 12.4)\)