Suppose a 95% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$36,080, $43,920]. The population standard deviation used for the analysis is known to be $12,000.
a. What is the point estimator of the mean salary of all college graduates in this town?
b. Determine the sample size used for the analysis.
i could not find the answer
a. The point estimator of the mean salary of all college graduates in this town is the midpoint of the confidence interval. Therefore, the point estimator is ($36,080 + $43,920) / 2 = $40,000.
b. To determine the sample size used for the analysis, we need to use the formula for the confidence interval:
n = (Z * σ / E)^2
where:
n = sample size
Z = Z-score corresponding to the desired confidence level
σ = population standard deviation
E = margin of error (half the width of the confidence interval)
In this case, the Z-score corresponding to a 95% confidence level is 1.96 (approximately). The population standard deviation is given as $12,000. The width of the confidence interval is the difference between the upper and lower bounds: $43,920 - $36,080 = $7,840. The margin of error is therefore half of the width: $7,840 / 2 = $3,920.
Plugging in these values, we can solve for the sample size:
n = (1.96 * $12,000 / $3,920)^2
n ≈ (1.96 * 12,000 / 3,920)^2
n ≈ (23520 / 3,920)^2
n ≈ (6)^2
n ≈ 36
Therefore, the sample size used for the analysis is 36.
To answer these questions, we need to understand the concepts of confidence intervals and point estimators.
a. The point estimator of the mean salary is the midpoint of the confidence interval. In this case, the midpoint is the average of the lower and upper bounds:
Point estimator = (Lower bound + Upper bound) / 2
= ($36,080 + $43,920) / 2
= $40,000
Therefore, the point estimator of the mean salary of all college graduates in this town is $40,000.
b. To find the sample size used for the analysis, we need to use the formula for the margin of error:
Margin of error = (Z-score) * (Standard deviation / sqrt(n))
In this case, the Z-score associated with a 95% confidence level is 1.96 (this value can be obtained from a Z-table or using statistical software).
And the known population standard deviation is $12,000.
We can rearrange the formula to solve for the sample size:
n = (Z-score^2) * (Standard deviation^2) / (Margin of error^2)
Substituting in the values:
n = (1.96^2) * ($12,000^2) / [(($43,920 - $36,080) / 2)^2]
Simplifying:
n = (3.8416) * ($144,000,000) / [(37320)^2]
n ≈ 36
Therefore, the sample size used for the analysis is approximately 36.