91影视

The results of the weights of M&M candies in Data Set 27 鈥楳&M Weights鈥� in Appendix B are

$\text{N}=465$,$\text{n}鈥�\text{=}鈥�100$ ,$\stackrel{炉}{x}鈥�=0.8565鈥�\text{g}$ and $s鈥�=0.0518鈥�\text{g}$.

The confidence level is 95%.

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.

$\text{Confidence}鈥�鈥�\text{interval}=\left(\stackrel{炉}{x}-E,\stackrel{炉}{x}+E\right)鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�...\left(1\right)$

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

$\text{E}=t_{\frac{伪}{2}}脳\frac{\text{s}}{\sqrt{\text{n}}}$

Assume that the sample is randomly selected.

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

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

The degree of freedom is calculated as follows.

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

The 95% confidence level corresponds to $伪=0.05$. 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" $t_{\frac{伪}{2}}=t_{0.025}$is obtained from the intersection of the column with 0.05 for the 鈥楢rea in Two Tails鈥� and the row value number of degrees of freedom is 99 degrees, which is 1.987.

The margin error is calculated as follows.

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

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

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

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

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.

$\text{C.I}=\left(\stackrel{炉}{x}-E脳\sqrt{\frac{\left(N-n\right)}{\left(N-1\right)}}鈥�鈥�,鈥�鈥�\stackrel{炉}{x}+E脳\sqrt{\frac{\left(N-n\right)}{\left(N-1\right)}}\right)鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�...\left(2\right)$

Substituting the values in equation (2),

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

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

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

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

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.