The 95% confidence interval for the mean weight (in kg) of the students of a school, found from a random sample as \(\bar{x} \pm 1.96 \times\) (standard error), is 71.08 kg to 78.92 kg. The sample mean and the standard error of the mean are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 75 kg and 2 kg
The interval is \(\bar{x} \pm E\), where the margin of error E = 1.96 × standard error. So the sample mean is the middle of the interval:
\( \bar{x} = \displaystyle\frac{71.08 + 78.92}{2} = 75 \)
The margin of error is half the width:
\( E = \displaystyle\frac{78.92 - 71.08}{2} = 3.92 \)
So the standard error is
\( \displaystyle\frac{3.92}{1.96} = 2 \)
Why the other options are incorrect
Option (2) 75 kg and 3.92 kg — Incorrect. 3.92 is the margin of error, not the standard error.
Option (3) 75 kg and 4 kg — Incorrect. 4 is the full width 7.84 divided by 1.96; the half-width must be used.
Option (4) 75 kg and 1 kg — Incorrect. A standard error of 1 gives a margin of error of only 1.96, not 3.92.