91Ó°ÊÓ

Insurance companies have an interest interested in knowing the population percent of drivers who always buckle up before riding in a very car.

A survey of $95\%$confident that the population proportion is estimated to be within $0.03$.

so as to search out the critical value we'd like to require in count that we are finding the interval for a proportion, so on this case we'd like to use the distribution. Since our interval is at of confidence, our significance level would incline by $\mathrm{Î\pm }=1−0.95=0.05$and $\mathrm{Î\pm }/2=0.025$.

And therefore the critical value would be lean by:

$z_{\mathrm{Î\pm }/2}=−1.96,z_{1−\mathrm{Î\pm }/2}=1.96$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\mathrm{Î\pm }/2}\sqrt{\frac{\widehat{p}(1−\widehat{p})}{n}}$ $\left(a\right)$

And on this case we've got that $ME=Â\pm 0.03$ and that we have an interest so as to seek out the worth of n, if we solve n from equation (a) we got:

$n=\frac{\widehat{p}(1−\widehat{p})}{{\left(\frac{ME}{z}\right)}^2}$$\left(b\right)$

We do not have a previous estimation for the proportion $\widehat{p}$so we will use $0.5$as an approximation for this case

And replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1−0.5)}{{\left(\frac{0.103}{1.96}\right)}^2}=1067.11$

And rounded up we've got that $n=1068$

Ifit had been later determined thatit absolutely was important to beover$95\%$ confident anda brand new survey was commissioned. The sample sizewould wish to be increased since the critical value increasesbecause the confidence level increases.