91Ó°ÊÓ

Given it costs $ 10 to draw a sample size$n=1$ and measure the attribute of interest.

a.

Here assume that$\mathrm{σ}=14$.

Known that the width of 5 units.

Then, the marginal error (ME) is,

$\begin{array}{c}ME=\frac{5}{2}\\ =2.5\end{array}$

For a 95% confidence interval, the level of significance is $\mathrm{Î\pm }=0.05$.

Then,

$\begin{array}{c}\frac{\mathrm{Î\pm }}{2}=\frac{0.05}{2}\\ \frac{\mathrm{Î\pm }}{2}=0.025\end{array}$

Using the normal distribution tables,

$z_{0.025}=1.960$

The sample size (n) obtained by,

$\begin{array}{c}n={\left(\frac{z_{\frac{\mathrm{Î\pm }}{2}}\mathrm{σ}}{ME}\right)}^2\\ ={\left(\frac{1.960×14}{2.5}\right)}^2\\ ={\left(\frac{27.44}{2.5}\right)}^2\\ ={\left(10.976\right)}^2\\ =120.472576\end{array}$

Hence, there is 121 samples which cost$121\left(\$10\right)=\$1210$,

Yes, $1210 is sufficient funds to estimate the population mean for the attributes interest with 95% confidence interval of 5 in width.

b.

For a 95% confidence interval, the level of significance is $\mathrm{Î\pm }=0.10$.

Then,

$\begin{array}{c}\frac{\mathrm{Î\pm }}{2}=\frac{0.10}{2}\\ \frac{\mathrm{Î\pm }}{2}=0.05\end{array}$

Using the normal distribution tables,

$z_{0.05}=1.645$

The sample size (n) obtained by,

$\begin{array}{c}n={\left(\frac{z_{\frac{\mathrm{Î\pm }}{2}}\mathrm{σ}}{ME}\right)}^2\\ ={\left(\frac{1.645×14}{2.5}\right)}^2\\ ={\left(\frac{23.03}{2.5}\right)}^2\\ ={\left(9.212\right)}^2\\ =84.860944\end{array}$

$n≈85$

Hence, there is 85 samples which cost,$85\left(\$10\right)=\$850$

Yes, still $850 is sufficient funds to estimate the population mean for the attributes interest with 90% confidence level 5 in width.

So, the answer to part (a) is not changed.