# Solved: Order in choice: planning a study. How large a sample would be needed to obtain margin of error $\pm$ 0.05 in the study

#### ByDr. Raju Chaudhari

Oct 12, 2020

Order in choice: planning a study. How large a sample would be needed to obtain margin of error $\pm$ 0.05 in the study of choice order for tasting wine? Use the $\hat{p}$ from Exercise 20.40 as your guess for the unknown p.

Exercise 20.40:

Order in choice. Does the order in which wine is presented make a difference? Several choices of wine are presented one at a time, and the subject is then asked to choose his or her preferred wine at the end of the sequence. In one study, subjects were asked to taste two wine samples in sequence. Both samples given to a subject were the same wine, although the subjects were expecting to taste two different samples of a particular variety. Of the 32 subjects in the study, 22 selected the wine presented first when presented with two identical wine samples.
(a) Give a 95% confidence interval for the proportion of subjects who would select the first choice presented.
(b) The subjects were recruited in Ontario, Canada, via advertisements to participate in a study of "attitudes and values towards wine." What assumption are you making about these subjects?

### Solution

The formula to estimate the sample size required to estimate the proportion is

 $$n =p*(1-p)\bigg(\frac{z}{E}\bigg)^2$$
where $p$ is the proportion of success, $z$ is the $Z_{\alpha/2}$ and $E$ is the margin of error.

Given that $\hat{p} =\frac{X}{n}=\frac{22}{32} = 0.6875$.

The margin of error is $E =0.05$. The confidence coefficient is $0.95$.

The critical value of $Z$ is $Z_{\alpha/2} = 1.96$.

The minimum sample size required to estimate the proportion is

 \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.6875(1-0.6875)\bigg(\frac{1.96}{0.05}\bigg)^2\\ &=330.1375\\ &\approx 331. \end{aligned}

Thus, the sample of size $n=331$ will ensure that the $95$\% confidence interval for the proportion will have a margin of error $0.05$.