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 36 of those weights yields a mean of $151.1 \mathrm { lb }$ and a standard deviation of $23.2 \mathrm { lb }$.
A) $143.5 \mathrm { lb }

Question

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 36 of those weights yields a mean of $151.1 \mathrm { lb }$ and a standard deviation of $23.2 \mathrm { lb }$.
A) $143.5 \mathrm { lb } < \mu < 158.7 \mathrm { lb }$
B) $145.2 \mathrm { lb } < \mu < 157.0 \mathrm { lb }$
C) $144.2 \mathrm { lb } < \mu < 158.0 \mathrm { lb }$
D) $142.8 \mathrm { lb } < \mu < 159.4 \mathrm { lb }$

Quizplus · Accepted Answer

The Answer of 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 36 of those weights yields a mean of $151.1 \mathrm { lb }$ and a standard deviation of $23.2 \mathrm { lb }$.
A) $143.5 \mathrm { lb } < \mu < 158.7 \mathrm { lb }$
B) $145.2 \mathrm { lb } < \mu < 157.0 \mathrm { lb }$
C) $144.2 \mathrm { lb } < \mu < 158.0 \mathrm { lb }$
D) $142.8 \mathrm { lb } < \mu < 159.4 \mathrm { lb }$

When Obtaining a Confidence Interval for a Population Mean in the Case