Solve the problem.
-When obtaining a confidence interval for a population mean in the case of a finite population of size N and a sample size n which is greater than 0.05N, the margin of error is multiplied by the following finite population Correction factor: $\sqrt { \frac { N - n } { N - 1 } }$ Find the 95% confidence interval for the mean of 200 weights if a sample of 34 of those weights yields a mean of 155.7 lb and a standard deviation of 24.1 lb.
A) $146.8 \mathrm { lb }

Question

Solve the problem.
-When obtaining a confidence interval for a population mean in the case of a finite population of size N and a sample size n which is greater than 0.05N, the margin of error is multiplied by the following finite population Correction factor: $\sqrt { \frac { N - n } { N - 1 } }$ Find the 95% confidence interval for the mean of 200 weights if a sample of 34 of those weights yields a mean of 155.7 lb and a standard deviation of 24.1 lb. 
A) $146.8 \mathrm { lb } < \mu < 164.6 \mathrm { lb }$
B) $148.3 \mathrm { lb } < \mu < 163.1 \mathrm { lb }$
C) $149.3 \mathrm { lb } < \mu < 162.1 \mathrm { lb }$
D) $147.6 \mathrm { lb } < \mu < 163.8 \mathrm { lb }$

Quizplus · Accepted Answer

The Answer of Solve the problem.
-When obtaining a confidence interval for a population mean in the case of a finite population of size N and a sample size n which is greater than 0.05N, the margin of error is multiplied by the following finite population Correction factor: $\sqrt { \frac { N - n } { N - 1 } }$ Find the 95% confidence interval for the mean of 200 weights if a sample of 34 of those weights yields a mean of 155.7 lb and a standard deviation of 24.1 lb. 
A) $146.8 \mathrm { lb } < \mu < 164.6 \mathrm { lb }$
B) $148.3 \mathrm { lb } < \mu < 163.1 \mathrm { lb }$
C) $149.3 \mathrm { lb } < \mu < 162.1 \mathrm { lb }$
D) $147.6 \mathrm { lb } < \mu < 163.8 \mathrm { lb }$

Solve the Problem N−nN−1\sqrt { \frac { N - n } { N - 1 } }N−1N−n​​

Solve the Problem $\sqrt { \frac { N - n } { N - 1 } }$