Question

Confidence Intervals. In Exercises 9–24, construct the confidence interval estimate of the mean.

Birth Weights of Girls Use these summary statistics given in Exercise 8:$\mathbf{n} \mathbf{=} \mathbf{205}\mathbf{,}\overline{\mathbf{x}} \mathbf{=}\mathbf{30}\mathbf{.}\mathbf{4} \mathbf{hg}\mathbf{,}\mathbf{s} \mathbf{=}\mathbf{7}\mathbf{.}\mathbf{1} \mathbf{hg}$. Use a 95% confidence level. Are the results very different from those found in Example 2 with only 15 sample values?

Short answer

The confidence interval estimate of the mean is$29.4 \text{hg}<\mu <31.4 \text{hg}$.

The results are almost same, when the sample values change.

Step-by-step solution

Given information

Based on Data set 4 “Birth” in Appendix B, the summary statistics for randomly selected weights of newborn girls as,$\mathbf{n} \mathbf{=} \mathbf{205}\mathbf{,}\overline{\mathbf{x}} \mathbf{=}\mathbf{30}\mathbf{.}\mathbf{4} \mathbf{hg}\mathbf{,}\mathbf{s} \mathbf{=}\mathbf{7}\mathbf{.}\mathbf{1} \mathbf{hg}$

The confidence level is 95%.

Describe confidence interval

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,

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

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

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

Find the appropriate distribution

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

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

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,

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

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

Find the margin of error

The margin of error is calculated as,

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

Find the confidence interval

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

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

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

Compare the results.

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

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

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