Question

suppose we dont know the population standard deviation. we take a random sample of size 25 and calculate the sample mean \\( \overline { x } = 10 \\) and the sample standard deviation \\( s = 2.5 \\). what is the 99% confidence interval for the population mean based on this sample? hint: round the critical values to three decimal places 8.957, 11.043 9.216, 10.784 8.602,11.399 8.97, 11.03

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\), where \(n = 25\). So \(df=25-1 = 24\).

Step2: Find the critical value \(t_{\alpha/2}\)

For a \(99\%\) confidence interval, \(\alpha=1 - 0.99=0.01\), and \(\alpha/2=0.005\). Using the \(t\) - distribution table or a calculator, \(t_{0.005,24}\approx 2.797\).

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

The formula for the margin of error when the population standard deviation \(\sigma\) is unknown is \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\). Substitute \(t_{\alpha/2}=2.797\), \(s = 2.5\), and \(n = 25\). Then \(E=2.797\times\frac{2.5}{\sqrt{25}}=2.797\times\frac{2.5}{5}=2.797\times0.5 = 1.3985\).

Step4: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{X}-E<\mu<\bar{X} + E\). Substitute \(\bar{X}=10\) and \(E = 1.3985\). So \(10-1.3985<\mu<10 + 1.3985\), which is \(8.6015<\mu<11.3985\approx[8.602,11.399]\) (rounded to three decimal places).

Answer:

\([8.602,11.399]\)