For 95% confidence interval for a population mean reported to be 132 to 142 with standard deviation $σ = 17.85$ 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 **
95% confidence interval: $132$ to $142$
Mean estimate = midpoint = $\frac{132+142}{2}=137$
Margin of error (half–width): $E=\frac{142-132}{2}=5$
Formula for confidence interval margin: $E=Z\,\frac{\sigma}{\sqrt{n}}$
Substitute values:
$5=1.96\,\frac{17.85}{\sqrt{n}}$
$\sqrt{n}=1.96\,\frac{17.85}{5}$
$\sqrt{n}=1.96\times 3.57=6.9972$
$n=(6.9972)^{2}\approx 49$
Sample size = $49$