Question

the state test scores for 12 randomly selected high school seniors are shown on the right. complete parts (a) through (c) below. assume the population is normally distributed. 1427 1228 987 699 725 830 726 748 544 627 1441 950 (a) find the sample mean. x̄ = 911.0 (round to one decimal place as needed.) (b) find the sample standard deviation. s = 304.1 (round to one decimal place as needed.) (c) construct a 95% confidence interval for the population mean μ. a 95% confidence interval for the population mean is (□,□). (round to one decimal place as needed.)

Explanation:

Step1: Determine the degrees of freedom and t - value

The sample size \(n = 12\), so the degrees of freedom \(df=n - 1=12-1 = 11\).
For a 95% confidence interval, the significance level \(\alpha=1 - 0.95=0.05\), and \(\frac{\alpha}{2}=0.025\).
Looking up the t - distribution table, \(t_{\frac{\alpha}{2},df}=t_{0.025,11}=2.201\).

Step2: Calculate the margin of error \(E\)

The formula for the margin of error for a confidence interval when the population standard deviation \(\sigma\) is unknown (using the sample standard deviation \(s\)) is \(E = t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\).
Given \(\bar{x}=911.0\), \(s = 304.1\), \(n = 12\), and \(t_{\frac{\alpha}{2}}=2.201\).
First, calculate \(\frac{s}{\sqrt{n}}=\frac{304.1}{\sqrt{12}}\approx\frac{304.1}{3.464}\approx87.8\).
Then \(E=2.201\times87.8\approx193.2\).

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x} + E\).
Substitute \(\bar{x}=911.0\) and \(E = 193.2\) into the formula.
\(\bar{x}-E=911.0-193.2 = 717.8\).
\(\bar{x}+E=911.0 + 193.2=1104.2\).

Answer:

\((717.8,1104.2)\)