A 95% confidence interval for a population mean was reported to be 152 to 160, if sample standard deviation \(\sigma = 15\), then sample size used in this study is (Given \(Z_{0.025} = 1.96\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 54
With the z-value given, the interval is the large-sample confidence interval \(\bar{x} \pm Z_{0.025}\frac{\sigma}{\sqrt{n}}\), with the given standard deviation 15 used for \(\sigma\). The mean is the centre of the interval and the margin of error is half its width:
\( \displaystyle \bar{x} = \frac{152 + 160}{2} = 156 \)
\(E = 160 - 156 = 4\)
So
\( \displaystyle 1.96 \times \frac{15}{\sqrt{n}} = 4 \)
\( \displaystyle \sqrt{n} = \frac{1.96 \times 15}{4} = \frac{29.4}{4} = 7.35 \)
\(n = (7.35)^2 = 54.02\)
So the sample size is 54.