The monthly expenses of all the households of a district have mean \(\mu\) = Rs. 12,000 and standard deviation \(\sigma\) = Rs. 2,000. A random sample of 400 households from the district has a mean monthly expense \(\bar{x}\) = Rs. 11,880. Match List-I with List-II.
| List-I | List-II (in Rs.) | ||
| (A) | Mean of the sampling distribution of the sample mean for samples of 400 | (I) | 100 |
| (B) | Standard error of the mean for samples of 400 | (II) | −120 |
| (C) | Value of the statistic \(\bar{x}\) for this sample | (III) | 12,000 |
| (D) | Sampling error of this sample mean | (IV) | 11,880 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
The population has the parameters \(\mu\) and \(\sigma\); one sample gives the statistic \(\bar{x}\); and the sampling distribution of \(\bar{x}\) (over all samples of 400) has mean \(\mu\) and standard deviation \(\sigma/\sqrt{n}\).
(A) Mean of the sampling distribution of the sample mean for samples of 400
This matches (III) 12,000: the mean of all the sample means equals the population mean \(\mu\).
(B) Standard error of the mean for samples of 400
\( \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{2000}{\sqrt{400}} = \frac{2000}{20} = 100 \)
This matches (I) 100.
(C) Value of the statistic \(\bar{x}\) for this sample
This matches (IV) 11,880: it is the mean of the 400 households actually surveyed.
(D) Sampling error of this sample mean
\(\bar{x} - \mu = 11880 - 12000 = -120\)
This matches (II) −120.