For 95% confidence interval for a population mean reported to be 132 to 142 with standard deviation \(\sigma = 17.85\) then the sample size used in this case, is: [Given that: \(Z_{.0125} = 1.96\)]
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 49
The confidence interval is \(\bar{x} \pm E\), where the margin of error is \(E = Z\frac{\sigma}{\sqrt{n}}\) and \(Z = 1.96\) (the value given). E is half the width of the interval:
\( \displaystyle E = \frac{142 - 132}{2} = 5 \)
So
\( \displaystyle 1.96 \times \frac{17.85}{\sqrt{n}} = 5 \)
\( \displaystyle \sqrt{n} = \frac{1.96 \times 17.85}{5} = \frac{34.986}{5} = 6.997 \)
\(n = (6.997)^2 = 48.96\)
So the sample size is about 49.