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Recently there was traced a poisoning outbreak for a particular ice-cream brand. Then, many consumers refused to buy the other ice-creams even though the poisonous brand is removed from the market.

A survey of customers is taken after 6 months of the event happened by one manufacturer. They took a sample of 244 ice-cream bar consumers and find that 23 consumers still had not purchased any ice-cream bars because of that poisoning incident.

a.

Let鈥檚 consider the sample size$\mathrm{n}=244$ and the number of consumers who have not purchased the ice cream bars, data-custom-editor="chemistry" $\mathrm{x}=23$.

So, the sample proportion, that is the true fraction of the entire market who refuses to purchase bars 6 months after the outbreak is,

$\begin{array}{rcl}\hat{\mathrm{p}}& =& \frac{\mathrm{x}}{\mathrm{n}}\\ & =& \frac{23}{244}\\ & =& 0.1027\end{array}$

Thus, the required point estimate is 0.1027.

b.

Consider,$\hat{\mathrm{q}}$ that is the point estimate of those consumers who purchased bars.

$\begin{array}{rcl}\hat{\mathrm{q}}& =& \left(1-\hat{\mathrm{p}}\right)\\ & =& \left(1-0.1207\right)\\ & =& 0.8973\end{array}$

So, by the normal approximation of the binomial distribution there can be concluded that, if the quantities of $\mathbf{n}\hat{\mathbf{p}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{n}\hat{\mathbf{q}}$are greater than 15 then the normal approximation can be used.

Hence, the sample size is 244, so, $\mathrm{n}\hat{\mathrm{p}}=25.05\mathrm{and}\mathrm{n}\hat{\mathrm{q}}=218.94$.

Therefore, the sample size is large enough to use the normal approximation.

c.

The 95% confidence interval for the binomial proportion is defined as,

$\left[\hat{p}-Z_{伪}\sqrt{\frac{\hat{p}\hat{q}}{n}}鈮�z鈮�\hat{p}+Z_{伪}\sqrt{\frac{\hat{p}\hat{q}}{n}}\right]$

Now, the confidence level is data-custom-editor="chemistry" $\mathrm{CL}=0.95$, then,

$\begin{array}{rcl}\mathrm{伪}& =& 1-\mathrm{CL}\\ & =& 1-0.95\\ & =& 0.05\end{array}$

And $\left(\frac{\mathrm{伪}}{2}\right)=0.025$.

Therefore, by the standard normal probability table,

$\begin{array}{rcl}{\mathrm{Z}}_{\left(\frac{\mathrm{伪}}{2}\right)}& =& {\mathrm{Z}}_{0.025}\\ & =& 1.96\end{array}$.

Thus, the interval is,

$\begin{array}{rcl}\left[\hat{\mathrm{p}}-{\mathrm{Z}}_{0.025}\sqrt{\frac{\hat{\mathrm{p}}\hat{\mathrm{q}}}{\mathrm{n}}},\hat{\mathrm{p}}+{\mathrm{Z}}_{0.025}\sqrt{\frac{\hat{\mathrm{p}}\hat{\mathrm{q}}}{\mathrm{n}}}\right]& =& \left[0.1027-1.96\sqrt{\frac{0.1027脳0.8973}{244}},0.1027+1.96\sqrt{\frac{0.1027脳0.8973}{244}}\right]\\ & =& \left[0.1027-0.039,0.1027+0.039\right]\\ & =& \left[0.06,0.14\right]\end{array}$

Therefore, the 95% confidence that approximately between 6% and 14%.

d.

In terms of this application, the 95% confidence interval constructed in this way that the interval contains the proportion of the point estimator that is the true fraction of the entire market that refuse to purchase bars 6 months after the outbreak.

The 95% confidence interval is between 6% and 14% and the point estimator is approximately 10%.