Question

Finite Population Correction Factor If a simple random sample of size n is selected without replacement from a finite population of size N, and the sample size is more than 5% of the population size ,better results can be obtained by using the finite population correction factor, which involves multiplying the margin of error E by $\sqrt{\frac{\left(N-n\right)}{\left(N-1\right)}}$For the sample of 100 weights of M&M candies in Data Set 27 “M&M Weights” in Appendix B, we get $\overline{x}=0.8656 g$and$s=0.0518 g$ First construct a 95% confidence interval estimate of , assuming that the population is large; then construct a 95% confidence interval estimate of the mean weight of M&Ms in the full bag from which the sample was taken. The full bag has 465 M&Ms. Compare the results.

Short answer

The 95% confidence interval for the estimate mean, assuming the large population size, is$\text{0.8462 g}<\mu <0.8668 \text{g}$.

The confidence interval for the estimate mean, using the population correction factor, is calculated as $\text{0.8474} \text{g}<\mu <0.8656 \text{g}$.

On comparing the results, the second confidence interval is narrower than the first and gives an accurate estimate when the sample size is large.

Step-by-step solution

Given information

The results of the weights of M&M candies in Data Set 27 ‘M&M Weights’ in Appendix B are

$\text{N}=465$,$\text{n} \text{=} 100$ ,$\overline{x} =0.8565 \text{g}$ and $s =0.0518 \text{g}$.

The confidence level is 95%.

Describe confidence interval

A confidence interval is an estimate of the interval that may contain the true value of a population parameter. It is also known as an interval estimate.

The general formula for the confidence interval estimate of mean is as follows.

$\text{Confidence} \text{interval}=\left(\overline{x}-E,\overline{x}+E\right) ...\left(1\right)$

Here, E is the margin of error, which is calculated as follows.

$\text{E}=t_{\frac{\alpha }{2}}\times \frac{\text{s}}{\sqrt{\text{n}}}$

Find the appropriate distribution

Assume that the sample is randomly selected.

As $\sigma$is unknown and $\text{n}=100$, which is greater than 30, t-distribution is applicable.

Find the critical value

To find the critical value, $t_{\frac{\alpha }{2}}$requires a value for the degrees of freedom.

The degree of freedom is calculated as follows.

$$\begin{gathered} \text{Degree} \text{of} \text{freedom} =n-1 \\ =100-1 \\ =99 \end{gathered}$$

The 95% confidence level corresponds to $\alpha =0.05$. So, there is an area of 0.025 in each of the two tails of the t distribution.

Referring to Table A-3 for critical value of t-distribution.

The critical value role="math" localid="1648036963662" $t_{\frac{\alpha }{2}}=t_{0.025}$is obtained from the intersection of the column with 0.05 for the ‘Area in Two Tails’ and the row value number of degrees of freedom is 99 degrees, which is 1.987.

Find the margin of error

The margin error is calculated as follows.

$$\begin{gathered} \text{E}=t_{\frac{\alpha }{2}}\times \frac{\text{s}}{\sqrt{\text{n}}} \\ =1.987\times \frac{0.0518}{\sqrt{100}} \\ =0.0103 \end{gathered}$$

Find the confidence interval, assuming the large population

The confidence interval is obtained by substituting the value of the margin of error in equation (1) as follows.

$$\begin{gathered} \text{Confidence} \text{interval}=\left(\overline{x}-\text{E},\overline{x}+\text{E}\right) \\ =\left(0.8565-0.0103 , 0.8565+0.0103\right) \\ =\left(0.8462 , 0.8668\right) \end{gathered}$$

Thus, the 95% confidence interval for the estimate mean is $\text{0.8462 g}<\mu <0.8668 \text{g}$.

Find the confidence interval using the finite population correction factor

For a population of 465 marbles, $0.2150\left(\frac{n}{N}=\frac{100}{465}\right)>0.05$.

Thus, the finite population correction factor $\sqrt{\frac{\left(N-n\right)}{N-1}}$will give a better estimate.

The confidence interval, using the finite population correction factor, is calculated as follows.

$\text{C.I}=\left(\overline{x}-E\times \sqrt{\frac{\left(N-n\right)}{\left(N-1\right)}} , \overline{x}+E\times \sqrt{\frac{\left(N-n\right)}{\left(N-1\right)}}\right) ...\left(2\right)$

Substituting the values in equation (2),

$$\begin{gathered} \text{C.I}=\left(\overline{x}-E\times \sqrt{\frac{\left(N-n\right)}{\left(N-1\right)}} , \overline{x}+E\times \sqrt{\frac{\left(N-n\right)}{\left(N-1\right)}}\right) \\ =\left(0.8565-0.0103\times \sqrt{\frac{\left(465-100\right)}{\left(465-1\right)}} , 0.8565+0.0103\times \sqrt{\frac{\left(465-100\right)}{\left(465-1\right)}}\right) \\ =\left(0.8474,0.8656\right) \end{gathered}$$

Thus, the confidence interval for the estimate mean, using the population correction factor, is$\text{0.8474} \text{g}<\mu <0.8656 \text{g}$ .

Compare the results   

The 95% confidence interval for the estimate mean, assuming a large population size, is$\text{0.8462 g}<\mu <0.8668 \text{g}$

The confidence interval for the estimate mean, using the population correction factor, is $\text{0.8474} \text{g}<\mu <0.8656 \text{g}$.

On comparing the above results, the confidence interval obtained from the finite population is narrow when compared to the first, which is indicative of the fact that getting a greater accuracy in estimation can be expected when the relatively large sample is obtained by the method of without replacement from a relatively small finite population.