Products “Made in the USA.” Refer to Exercise 2.154 (p. 143) and the Journal of Global Business (Spring 2002) survey to determine what “Made in the USA” means to consumers. Recall that 106 shoppers at a shopping mall in Muncie, Indiana, responded to the question, “Made in the USA” means what percentage of U.S. labor and materials?” Sixty-four shoppers answered,“100%.”

a. Define the population of interest in the survey.

b. What is the characteristic of interest in the population?

c. Estimate the true proportion of consumers who believe “Made in the USA” means 100% U.S. labor and materials using a 90% confidence interval.

d. Give a practical interpretation of the interval, part c.

e. Explain what the phrase “90% confidence” means for this interval.

f. Compute the sample size necessary to estimate the true proportion to within .05 using a 90% confidence interval.

Let the sample size be 106 and number of successes is 64.

a.

All consumers who shop at the shopping mall in Muncie, Indiana constitute the population.

b.

The true population of consumers who believe “Made in the USA” means 100% of U.S. labor and material.

Since, the point estimate $\left(\hat{p}\right)$of the population proportion p is calculated using the following formula:

$\overbrace{p}=\frac{x}{n}$

Therefore,

$\begin{array}{rcl}\hat{p}& =& \frac{64}{106}\\ & =& 0.604\\ & & \end{array}$

Therefore, the point estimate of the population proportion p is 0.604.

The sample size is large enough if the below conditions are satisfied.

$n\hat{p}⩾15,n\left(1-\hat{p}\right)⩾15$

Obtain the value of $n\hat{p}$,

$\begin{array}{rcl}n\hat{p}& =& 106\left(0.604\right)\\ & =& 64.025\left(⩾15\right)\\ & & \end{array}$

Obtain the value of $n\left(1-\hat{p}\right)$

$\begin{array}{rcl}n\left(1-\hat{p}\right)& =& 106\left(1-0.604\right)\\ & =& 106\left(0.396\right)\\ & =& 41.976\left(⩾15\right)\\ & & \end{array}$

Since,

The value of $n\hat{p}$and$n\left(1-\hat{p}\right)$is greater than 15, the sample size is sufficiently large. Thus, it can be concluded that the normal approximation is reasonable.

Let the confidence level be 0.90

$\begin{array}{rcl}1-\alpha & =& 0.90\\ \alpha & =& 1-0.90\\ \alpha & =& 0.10\\ & & \end{array}$

So, $\frac{\alpha }{2}=0.05$

From table, the required value for 90% confidence level is 1.645.

The 90% confidence interval is obtained as

$\begin{array}{rcl}\hat{p}+z_{0.05}\left(\sqrt{\frac{\hat{p}\hat{q}}{n}}\right)& =& 0.604\pm \sqrt{\frac{\left(0.604\right)\left(0.396\right)}{106}}\\ & =& 0.604\pm 1.645\left(\sqrt{0.002256}\right)\\ & =& 0.604\pm 0.0781\\ & & \end{array}$

That is,

$\left(0.604-0.0781,0.604+0.0781\right)=\left(0.5259,0.0621\right)$

Thus, the 90% confidence interval for the population proportion is (0.5259, 0.6821).

The true population of consumers who believe “Made in the USA” means that 100% U.S. labor and materials lies between 0.5259 and 0.6821 at 90% confidence

The “90% confidence means, for each repeated samples size of 106, the 90% of the interval for the true population of consumes who believe “ Made in the USA” means that 100% U.S. labor and materials lie within the upper and lower limit and the confidence interval gives the plausible values for the population proportion. That is, the 90% confidence interval will contain the true population proportion.

The formula for sample size is given below;

$n=\frac{\left(z_{\frac{\alpha }{2}}\right)pq}{{\left(ME\right)}^2}$

Here, the product of pq is unknown, then using the sample fraction of success,$\hat{p}$

Substitute 1.645 for$z_{0.05}$ , 0.604 for $\hat{p}$, 0.396 for $\hat{q}$, and 0.5 for ME in the above equation.

$\begin{array}{rcl}n& =& \frac{{\left(1.645\right)}^2\left(0.604\right)\left(0.396\right)}{{\left(0.05\right)}^2}\\ & =& \frac{0.647}{0.25}\\ & \approx & 259\\ & & \end{array}$

Thus, the required sample size is 259.

Hence, to estimate the true proportion to be within 0.05 by using 90% confidence interval, 259 samples are needed