If a 95% confidence interval for a population mean was reported to be 132 to 160 and sample standard deviation \(\sigma = 50\), then the size of the sample in the study is:
(Given \(Z_{0.025} = 1.96\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 49
Note: The given answer is as per the official NTA Answer Key. However, the paper calls 50 the sample standard deviation, and with that reading Option (3) could also be a possible answer, as explained below.
The half-width of a 95% confidence interval is the margin of error:
\( \displaystyle E = \frac{160 - 132}{2} = 14 \)
With standard error \(\frac{\sigma}{\sqrt{n}}\):
\( \displaystyle 14 = 1.96 \times \frac{50}{\sqrt{n}} = \frac{98}{\sqrt{n}},\quad \sqrt{n} = 7,\quad n = 49 \)
With standard error \(\frac{s}{\sqrt{n - 1}}\) (s the sample standard deviation):
\( \displaystyle 14 = \frac{98}{\sqrt{n - 1}},\quad \sqrt{n - 1} = 7,\quad n = 50 \)
Option (1) 90 — Incorrect. Neither form of the standard error gives 90.
Option (2) 95 — Incorrect. 95 is the confidence level, not the sample size.
Option (3) 50 — Should also be correct, though not the answer given by NTA. The question calls 50 the sample standard deviation. When the standard deviation is worked out from the sample itself, many books divide by \(\sqrt{n - 1}\) instead of \(\sqrt{n}\). Then \(\sqrt{n - 1} = 7\), so \(n - 1 = 49\) and n = 50.
Option (4) 49 — Correct. With the usual formula, the margin of error is \(1.96 \times \frac{50}{\sqrt{n}}\). Then \(\sqrt{n} = 7\), so n = 49.