A statistician wants to estimate the mean weekly family expenditure on clothes. He believes that the standard deviation of the weekly expenditure is $125. Determine with 99% confidence the number of families that must be sampled to estimate the mean weekly family expenditure on clothes to within $15.
99% = mean ± 2.575 SEm
SEm = SD/√n
2.575 SEm = 2.575 SD/√n = 15
Solve for n.
463
To determine the number of families that must be sampled to estimate the mean weekly family expenditure on clothes with a given level of confidence and margin of error, we can use the formula for sample size calculation:
n = (Z^2 * σ^2) / E^2
Where:
- n is the required sample size
- Z is the Z-score corresponding to the desired level of confidence
- σ is the standard deviation of the population
- E is the desired margin of error
In this case, the statistician wants to estimate the mean weekly family expenditure on clothes with 99% confidence and within a margin of error of $15. The standard deviation of the weekly expenditure is given as $125.
First, we need to find the Z-score corresponding to a 99% confidence level. The Z-score can be obtained from a Z-table or a statistical software. For 99% confidence, the Z-score is approximately 2.576.
Next, we can substitute the values into the formula:
n = (2.576^2 * 125^2) / 15^2
Calculating this expression:
n = (6.642976 * 15625) / 225
n ≈ 458.773
Since we cannot have a fraction of a family, we need to round up the sample size to the nearest whole number. Therefore, the number of families that must be sampled is approximately 459 to estimate the mean weekly family expenditure on clothes to within $15 with 99% confidence.
To determine the number of families that must be sampled to estimate the mean weekly family expenditure on clothes within a certain margin of error, we can use the formula for the sample size of a mean:
n = (Z * σ / E) ^ 2
Where:
n = required sample size
Z = Z-score for the desired confidence level (99% confidence level corresponds to Z = 2.576)
σ = standard deviation of the population (in this case, σ = $125)
E = desired margin of error (in this case, E = $15)
Substituting the given values into the formula, we have:
n = (2.576 * 125 / 15) ^ 2
n = 21.6576 ^ 2
n ≈ 467.912
So, the statistician will need to sample approximately 468 families in order to estimate the mean weekly family expenditure on clothes with a margin of error of $15, with 99% confidence. However, since you can't have a partial family, the sample size should be rounded up to the nearest whole number. Therefore, the statistician should sample at least 468 families.