91Ó°ÊÓ

a) - The whole number of successful attempts is $x=320$.

- There are $n=400$ drivers within the sample.

- The population mean percent information is $\widehat{p}=\frac{x}{n}=\frac{320}{400}=0.8$

b) Component at unexpected times Average amount of "successes" inside which users claim to carry on tight is $x$.

The proportion of drivers polled who indicate their constantly buckle up is $\widehat{p}$.

c) c) Given $\widehat{p}=0.8$and $n=400$, the dispersion we must follow that compute a little is

$\mathcal{N}\left(0.8,\sqrt{\frac{0.8×0.2}{400}}\right)$

d) If the percent of sightings during a randomized subset $n$ that relate to a category of things is $\widehat{p}$, an estimate $100(1−\mathrm{Î\pm })\mathrm{\%}$ significance level here on proportion $p$ of such populace that belong to the current group is .

$\widehat{p}−z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{\widehat{p}(1−\widehat{p})}{n}}≤p≤\widehat{p}+z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{\widehat{p}(1−\widehat{p})}{n}}$

localid="1649850739294" $\frac{\mathrm{Î\pm }}{2}=\frac{1−0.95}{2}$

localid="1649850746722" $z_{\frac{\mathrm{Î\pm }}{2}}=1.96$

localid="1649850753572" $\begin{array}{r}0.8−1.96\sqrt{\frac{0.8(1−0.8)}{400}}≤p≤0.8+1.96\sqrt{\frac{0.8(1−0.8)}{400}}\\ \end{array}$

localid="1649850761120" $0.8−0.039≤p≤0.8+0.039$

localid="1649850768823" $0.761≤p≤0.839$

This problem's graph is as follows:

- The upper limit of the error limitation is

$EBM=0.039$

Is that if poll were conducted over the phone, the businesses might face three challengesin an exceedingly achieving random results:

- obtain results from people in a same town,

- obtain results from people of comparable ages,

- obtain inaccurate findings.