Using Correct Distribution. In Exercises 5–8, assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) Find the critical value $t_{\frac{\alpha }{2}}$ ,(b) find the critical value $z_{\frac{\alpha }{2}}$,or (c) state that neither the normal distribution nor the t distribution applies.

Birth Weights Here are summary statistics for randomly selected weights of newborn girls:$\mathbf{n} \mathbf{=} \mathbf{205}\mathbf{,}\overline{\mathbf{x}} \mathbf{=}\mathbf{30}\mathbf{.}\mathbf{4} \mathbf{hg}\mathbf{,}\mathbf{s} \mathbf{=}\mathbf{7}\mathbf{.}\mathbf{1} \mathbf{hg}$ (based on Data Set 4 “Births” in Appendix B). The confidence level is 95%.

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

And, the confidence level is 95%.

The critical value is the border value which separates the sample statistics that are significantly high or low from those sample statistics that are not significant.

For randomly selected samples, the conditions for t-distribution and normal distribution are as follow,

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

If $\sigma$is known and $\text{n}>30$then normal distribution is suitable to find the confidence interval.

In this case,$\sigma$ is unknown and $\text{n}=61$, where n>30.So, t-distribution applies here.

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

The degree of freedom is,

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

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 is expressed as,

$$\begin{gathered} t_{\frac{\alpha }{2}}=t_{\frac{0.05}{2}} \\ =t_{0.025} \end{gathered}$$

It 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 is 204, which is 1.972.