Question

A random sample of 50 consumers taste-tested a new snack food. Their responses were coded (0: do not like; 1: like; 2: indifferent) and recorded as follows:

a. Use an 80% confidence interval to estimate the proportion of consumers who like the snack food.

b. Provide a statistical interpretation for the confidence interval you constructed in part a.

Short answer

a. The 80% confidence interval is$\left(0.2170,0.3830\right)$

b.One is 80% confident that the true proportion of consumers who like the snack food lies between the interval$\left(0.2170,0.3830\right)$

Step-by-step solution

Given information

A random sample of 50 consumers taste-tested a new snack food. Their responses were coded according to some criteria (0: do not like; 1: like; 2: indifferent). This has been presented in the table above.

(a) Calculation of 80% confidence interval of estimate the proportion of consumers who like the snack food.

The sample size is n= 50

15 of the data value are a 1, which indicates x=15

$p=\frac{x}{n}=\frac{15}{50}=0.3$

Here the confidence coefficient is 80%, hence $\alpha =20\%$.Now from the standard normal distribution table,$z_{\frac{\alpha }{2}}=1.28$

The margin of error

$\begin{array}{rcl}E& =& z_{\frac{\alpha }{2}}\times \sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}\\ & =& 1.28\times \sqrt{\frac{0.3\times \left(1-0.3\right)}{50}}\\ & =& 0.0830\\ & & \end{array}$

Hence 80% lower limit:$\hat{p}-E=0.3-0.0830=0.2170$

80% upper limit:$\hat{p}+E=0.3+0.0830=0.3830$

The 80% confidence interval is$\left(0.2170,0.3830\right)$

(b) Interpretation of confidence interval constructed in part a

One is 80% confident that the true proportion of consumers who like snack food lies between the interval$\left(0.2170,0.3830\right)$