In developing an interval estimate for a population mean,the population standard deviation $\sigma$ was assumed to be 10.The interval estimate was 50.92 $\pm$2.14.Had $\sigma$ equaled 20,the interval estimate would be
A) 60.92 $\pm$ 2.14
B) 50.92$\pm$12.14
C) 101.84 $\pm$ 4.28
D) 50.92 $\pm$ 4.28

Question

Quizplus · Accepted Answer

The formula for the interval estimate of population mean is: 
interval estimate = sample mean +/- (t-value)*(standard error)
where standard error = population standard deviation/sqrt(sample size) and t-value is obtained from t-table using the desired level of confidence and degrees of freedom (sample size -1).

Given: 
Population standard deviation when assumed to be 20, σ = 20
Interval estimate = 50.92 +/- 2.14

We know that the interval estimate is proportional to the standard error, which itself is proportional to standard deviation (when all other factors remain constant). Therefore, we can expect the interval estimate to increase when standard deviation increases.

To calculate the new interval estimate using a standard deviation of 20:
standard error = 20/sqrt(sample size)
Since standard error was calculated using one sample and that sample did not change, we can assume that sample size and t-value will not change either. Therefore, we can use the old t-value to calculate the new interval estimate.
t-value for 95% confidence level with 9 degrees of freedom (sample size -1) is approximately 2.306.
standard error = 20/sqrt(10) = 6.32
New interval estimate = 50.92 +/- 2.306*6.32 = 50.92 +/- 14.56 = [36.36, 65.48]
The closest option to this new interval estimate is D: 50.92 +/- 4.28 = [46.64, 55.20].

In Developing an Interval Estimate for a Population Mean,the Population σ\sigmaσ

In Developing an Interval Estimate for a Population Mean,the Population $\sigma$