Chapter 22, Problem 22E

(a)

To determine

To Explain: that this and experiment or not.

Answer to Problem 22E

Observational study

Explanation of Solution

This study attempted to demonstrate a connection between two groups of mothers or to compare live births. Although it was not a survey and there were no assignments, it was not an experiment, but an observational study.

(b)

To determine

To Test: a suitable hypothesis and give the conclusion.

Explanation of Solution

Formula used:

  $S{E}_{Pooled}\left({\widehat{p}}_{Exposed}-{\widehat{p}}_{NotExpossed}\right)=\sqrt{\frac{{\widehat{p}}_{Pololed}{q}_{Pololed}}{n_{Exposed}}+\frac{{\widehat{p}}_{Pololed}{q}_{Pololed}}{n_{Exposed}}}$

Calculation:

There is required to find out whether there is good evidence that air pollution could be associated with a significantly higher percentage of babies with low birth weight; this is achieved by contrasting data groups: mothers who were exposed to soot and smoke, and mothers who were not. Hence, is there a distinction between the two classes or not? Here is, here is

the hypothesis

This implies that there is no significant difference in the category of two mothers'

  $H_0:{\gamma }_{Exposed}={\gamma }_{NotExpossed}$

This implies that there is a significant difference in the category of two mothers'

  $H_1:{\gamma }_{Exposed}\ne {\gamma }_{NotExpossed}$

This implies that in the category of two mothers, there is a significant difference

To find the p-value

  $\begin{array}{l}{\widehat{p}}_{Exposed}=0.08\\ {\widehat{p}}_{NotExpossed}=0.04\end{array}$

Not find the ${\widehat{p}}_{Pooled}$ :

  $\begin{array}{c}{\widehat{p}}_{Pooled}=\frac{182+15}{2300+92}\\ =0.0824\end{array}$

Next find the $S{E}_{Pooled}$

  $\begin{array}{c}S{E}_{Pooled}\left({\widehat{p}}_{Exposed}-{\widehat{p}}_{NotExpossed}\right)=\sqrt{\frac{{\widehat{p}}_{Pololed}{q}_{Pololed}}{n_{Exposed}}+\frac{{\widehat{p}}_{Pololed}{q}_{Pololed}}{n_{Exposed}}}\\ =\sqrt{\frac{0.0824\times \left(1-\mathrm{0.0.0824}\right)}{2300}+\frac{0.0824\times \left(1-\mathrm{0.0.0824}\right)}{92}}\\ =\sqrt{0.000854724}\\ =0.02924\end{array}$

The value of z is

  $\begin{array}{c}z=\frac{\left({\widehat{p}}_{Exposed}-{\widehat{p}}_{NotExpossed}\right)-0}{S{E}_{Pooled}\left({\widehat{p}}_{Exposed}-{\widehat{p}}_{NotExpossed}\right)}\\ =\frac{\left(0.08-0.04\right)-0}{0.02924}\\ =1.37\end{array}$

Finally, using this value to find p in the Z table.

  $P=2P\left(z\ge 1.37\right)=0.1706$

This p-value is very high then; thus, it does not reject it. There is no evidence of a significant differences between the two groups of the two mothers in this data.

(c)

To determine

To find: the difference with confidence interval and clarify the interval, there is a difference in the conclusion.

Explanation of Solution

There is no need to do an interval because it is failed to reject it. The difference will be calculated by this interval.