Salmonella poisoning from eating an ice cream bar.Recently, a case of salmonella (bacterial) poisoning wastraced to a particular brand of ice cream bar, and themanufacturer removed the bars from the market. Despitethis response, many consumers refused to purchase anybrand of ice cream bars for some period of time after the event (McClave, personal consulting). One manufacturerconducted a survey of consumers 6 months after theoutbreak. A sample of 244 ice cream bar consumers wascontacted, and 23 respondents indicated that they wouldnot purchase ice cream bars because of the potential forfood poisoning.

1. What is the point estimate of the true fraction of the entiremarket who refuse to purchase bars 6 months after the out-break?
2. Is the sample size large enough to use the normalapproximation for the sampling distribution of the estimator of the binomial probability? Justify your response.
3. Construct a 95% confidence interval for the true proportionof the market who still refuses to purchase icecream bars 6 months after the event.
4. Interpret both the point estimate and confidence interval in terms of this application.

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’s 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,

$\left[\hat{p}-Z_{\alpha }\sqrt{\frac{\hat{p}\hat{q}}{n}}\le z\le \hat{p}+Z_{\alpha }\sqrt{\frac{\hat{p}\hat{q}}{n}}\right]$

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

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

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

Therefore, by the standard normal probability table,

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

Thus, the interval is,

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

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%.