Question

assume that a sample is used to estimate a population mean μ. find the margin of error m.e. that corresponds to a sample of size 13 with a mean of 25.2 and a standard deviation of 14.7 at a confidence level of 99.9%.
report me accurate to one decimal place because the sample statistics are presented with this accuracy.
m.e. =
answer should be obtained without any preliminary rounding. however, the critical value may be rounded to 3 decimal places.
question help: video post to forum

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\), where \(n = 13\). So \(df=13-1 = 12\).

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

The confidence level is \(C = 99.9\%\), so \(\alpha=1 - C=1-0.999 = 0.001\) and \(\alpha/2=0.0005\).
Using a \(t\)-distribution table or calculator, for \(df = 12\) and \(\alpha/2=0.0005\), \(t_{\alpha/2}\approx4.318\).

Step3: Calculate the margin of error formula

The formula for the margin of error for a population mean (when population standard deviation \(\sigma\) is unknown) is \(M.E.=t_{\alpha/2}\times\frac{s}{\sqrt{n}}\), where \(s = 14.7\) and \(n = 13\).
Substitute the values: \(M.E.=4.318\times\frac{14.7}{\sqrt{13}}\).
First, calculate \(\sqrt{13}\approx3.606\). Then \(\frac{14.7}{3.606}\approx4.077\).
Finally, \(M.E.=4.318\times4.077\approx17.6\).

Answer:

\(17.6\)