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:Image for When obtaining a confidence interval for a population mean in the case of a finite population of size N and aFind the 95% confidence interval for the mean of 200 weights if a sample of 32 of those weights yields a mean of 150.6 lb and a standard deviation of 24.4 lb.a. 143.9 lb < µ < 157.3 lbb. 141.4 lb < µ < 159.8 lbc. 142.8 lb < µ < 158.4 lbd. 142.1 lb < µ < 159.1 lb

Answer 1

Answer:

95% Confidence interval for the mean

$142.8 \leq\mu\leq158.4$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the mean of a finite population.

The error is multiplied by the following finite population correction factor:

$cf=\sqrt{\frac{N-n}{N-1} }$

The standard deviation can be estimated as

$\sigma=\frac{s}{\sqrt{n}} \sqrt{\frac{N-n}{N-1} } =\frac{24.4}{\sqrt{32} }* \sqrt{\frac{200-32}{200-1} }=3.963$

The 95% confidence interval has a z value of 1.96, so it becomes:

$M-z*\sigma_c\leq\mu\leq M+z*\sigma_c\\\\150.6-1.96*3.963\leq\mu\leq 150.6+1.96*3.963\\\\ 142.8 \leq\mu\leq 158.4$