91影视

Based on Data set 4 鈥淏irth鈥� in Appendix B, the summary statistics for randomly selected weights of newborn girls as,$\mathbf{n}鈥�\mathbf{=}鈥�\mathbf{205}\mathbf{,}\stackrel{\mathbf{炉}}{\mathbf{x}}鈥�\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{4}鈥�\mathbf{hg}\mathbf{,}\mathbf{s}鈥�\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{1}鈥�\mathbf{hg}$

The confidence level is 95%.

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

The general formula for confidence interval estimate of mean is,

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

Where, E is the margin of error, which is calculated as,

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

If $蟽$ is not known and $\text{n}>30$ then t-distribution is suitable to find the confidence interval.

In this case, $蟽$is unknown and $\text{n}=205$ which means n>30.So, the t-distribution applies here.

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

The degree of freedom is calculated as,

$$\begin{gathered} \text{degree}鈥�鈥�鈥�\text{of}鈥�鈥�鈥�\text{freedom}鈥�鈥�鈥�=n-1 \\ =205-1 \\ =204 \end{gathered}$$

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 critical value of t-distribution, the critical value of${\mathbf{t}}_{\frac{\mathbf{伪}}{\mathbf{2}}}\mathbf{=}{\mathbf{t}}_{\mathbf{0}\mathbf{.}\mathbf{025}}$is obtained from the intersection of column with 0.05 for the 鈥淎rea in Two Tails鈥� (or use the same column with 0.025 for the 鈥淎rea in One Tail鈥�)and the row value number of degrees of freedom is204, which is 1.972.

The margin of error is calculated as,

$$\begin{gathered} \text{E}=t_{\frac{伪}{2}}脳\frac{\text{s}}{\sqrt{\text{n}}} \\ =1.972脳\frac{7.1}{\sqrt{205}} \\ =0.9779 \end{gathered}$$

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

$$\begin{gathered} \text{Confidence}鈥�鈥�\text{Interval}=\left(\stackrel{炉}{x}-\text{E},\stackrel{炉}{x}+\text{E}\right) \\ =\left(30.4-0.9779鈥�鈥�,鈥�鈥�30.4+0.9779\right) \\ =\left(29.4221鈥�鈥�,鈥�鈥�31.3779\right) \end{gathered}$$

Thus, the confidence interval estimate of the mean is$29.4鈥�\text{hg}<渭<31.4鈥�\text{hg}$.

The confidence interval found in the Example-2 with 15 sample values for mean birth weights of girls is$29.2鈥�\text{hg}<渭<32.5鈥�\text{hg}$.

The confidence interval estimate of the mean with 205 sample values is$29.4鈥�\text{hg}<渭<31.4鈥�\text{hg}$

So, the results in these two cases do not differ largely.