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 95% 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 9.02, 10.98 8.968, 11.032

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 95% confidence interval, \(\alpha=1 - 0.95=0.05\), and \(\alpha/2=0.025\). Looking up the \(t\) - value in the \(t\) - distribution table with \(df = 24\), \(t_{0.025,24}=2.064\) (rounded to three decimal places as per the hint).

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

The formula for the margin of error when the population standard deviation \(\sigma\) is unknown is \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\). Given \(s = 2.5\), \(n = 25\), and \(t_{\alpha/2}=2.064\). Then \(E=2.064\times\frac{2.5}{\sqrt{25}}=2.064\times\frac{2.5}{5}=2.064\times0.5 = 1.032\).

Step4: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{X}\pm E\). Given \(\bar{X}=10\), the lower limit is \(\bar{X}-E=10 - 1.032=8.968\) and the upper limit is \(\bar{X}+E=10 + 1.032=11.032\).

Answer:

[8.968, 11.032]