An SRS of size 30 is taken from a population if size 100.
(a) What is the sampling weight for each unit in the sample?
(b) It is known that the sample mean is 8.23. What is the estimate of population total?
(c) It is known that, the sample variance is 16. Calculate estimated variance V(t). Does
fpe make a difference for this sample?
The Correct Answer and Explanation is :
Let’s address each of the sub-questions one by one:
(a) Sampling Weight for Each Unit in the Sample
In simple random sampling (SRS), the sampling weight of each unit in the sample is inversely related to the probability of being selected. Since the sample size is 30 and the population size is 100, the probability of selecting each unit is:
[
P(\text{selection of a unit}) = \frac{\text{Sample size}}{\text{Population size}} = \frac{30}{100} = 0.3
]
The sampling weight is the reciprocal of this probability:
[
\text{Sampling weight} = \frac{1}{P(\text{selection of a unit})} = \frac{1}{0.3} = 3.33
]
Thus, the sampling weight for each unit in the sample is 3.33.
(b) Estimate of Population Total
To estimate the total population based on the sample mean, we use the formula:
[
\hat{T} = \bar{y} \times N
]
Where:
- (\hat{T}) is the estimated population total,
- (\bar{y}) is the sample mean, and
- (N) is the population size.
Given that the sample mean is 8.23, and the population size is 100, the estimate of the population total is:
[
\hat{T} = 8.23 \times 100 = 823
]
Therefore, the estimated population total is 823.
(c) Estimated Variance of the Total (V(t)) and FPE
The formula for the estimated variance of the population total (V(t)) in simple random sampling is:
[
V(t) = \frac{N^2}{n} \left( 1 – \frac{n}{N} \right) s^2
]
Where:
- (N) is the population size,
- (n) is the sample size,
- (s^2) is the sample variance.
Given:
- (N = 100),
- (n = 30),
- (s^2 = 16).
We can now calculate (V(t)):
[
V(t) = \frac{100^2}{30} \left( 1 – \frac{30}{100} \right) \times 16
]
[
V(t) = \frac{10000}{30} \times 0.7 \times 16 = \frac{10000 \times 0.7 \times 16}{30} = \frac{112000}{30} = 3733.33
]
Thus, the estimated variance of the population total is 3733.33.
FPE (Finite Population Correction)
The Finite Population Correction (FPE) is a factor that adjusts the variance when sampling from a finite population. The FPE is given by:
[
FPE = \sqrt{1 – \frac{n}{N}}
]
For this sample:
[
FPE = \sqrt{1 – \frac{30}{100}} = \sqrt{1 – 0.3} = \sqrt{0.7} \approx 0.8367
]
If we apply the FPE to the estimated variance of the total:
[
V(t)_{\text{adjusted}} = V(t) \times FPE^2 = 3733.33 \times (0.8367)^2 \approx 3733.33 \times 0.699 = 2618.33
]
Thus, the variance with the FPE adjusted is 2618.33.
Does FPE Make a Difference?
Yes, FPE makes a difference in this case. Without applying the FPE, the variance of the total would be higher, and the adjustment reduces the variance when sampling from a finite population. In cases where the sample size is a large proportion of the population, the finite population correction should be applied to avoid overestimating the variance. Since (n = 30) and (N = 100), the correction has a noticeable impact.