A random sample of 17 values from a normal population has a mean of 105 cm and the sample standard deviation is 8.489. If normal population mean is 110 cm, then the upper limit of confidence at 5% level of significance is : (Given \( t_{16}(0.05)=2.12 \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 109.50
The population standard deviation is not known and the sample is small, so the confidence limits use the t-distribution with n − 1 = 16 degrees of freedom.
Here n = 17, \( \bar{x}=105 \) and the sample standard deviation s = 8.489. With s as the standard deviation of the sample itself, the estimated standard error of the mean is
\( \displaystyle \frac{s}{\sqrt{n-1}}=\frac{8.489}{\sqrt{16}} \)
\( \displaystyle =\frac{8.489}{4} \)
\( \displaystyle =2.12225 \)
The 95% confidence limits are \( \bar{x}\pm t_{16}(0.05)\cdot\frac{s}{\sqrt{n-1}} \). The upper limit is
\( \displaystyle \bar{x}+2.12\times 2.12225=105+4.4992 \)
\( \displaystyle =109.4992 \)
\( \displaystyle \approx 109.50 \)
The population mean 110 cm given in the question is not needed for the confidence limits.
Why the other options are incorrect
Option (2) 100.50 — Incorrect. 100.50 is the lower limit, \( 105-4.50 \).
Option (3) 98.80 — Incorrect. It is below even the lower limit 100.50, so it cannot be the upper limit.
Option (4) 111.20 — Incorrect. The upper limit is 105 + 4.50 = 109.50, not 111.20.