Salmonella poisoning from eating an ice cream bar(cont’d). Refer to Exercise 6.132. Suppose it is now 1 yearafter the outbreak of food poisoning was traced to icecream bars. The manufacturer wishes to estimate the proportionwho still will not purchase bars to within .02 usinga 95% confidence interval. How many consumers should be sampled?

Referring to exercise 6.132, here the manufacturer wanted to estimate the proportion who still will not purchase bars to within 0.02 using a 95% confidence interval.

Let’s consider that the population size is N. So, by the formula,

$N=\frac{Z_{\alpha }^2\times p\left(1-p\right)}{e^2}$

Where Z is the value from the standard normal probability to the 95% confidence interval. p is the estimated true proportion and e is the desired precision.

And the number of consumers who should be sampled is n. So, by the formula,

$n=\frac{NP}{\left(N+P-1\right)}$

Where P is the original population size. That is 244.

Therefore, the required population number is,

$\begin{array}{rcl}N& =& \frac{{\left(1.96\right)}^2\times 0.1027\times 0.8973}{{\left(0.02\right)}^2}\\ & =& 885.03\\ & \approx & 886\\ & & \end{array}$

Thus, the manufacturer needs almost 886 consumers as the population.

Now, the required sample size is,

$\begin{array}{rcl}n& =& \frac{244\times 886}{\left(244+886-1\right)}\\ & =& 191.48\\ & \approx & 192\\ & & \end{array}$

Therefore, 192 consumers should be sampled.