Question

A local marketing company wants to estimate the proportion of consumers in the Oconee County area who would react favorably to a marketing campaign. Further, the company wants the estimate to have a margin of error of no more than 4 percent with 90 percent confidence. Of the following, which is the closest to the minimum number of consumers needed to obtain the estimate with the desired precision? A. 65 B. 93 C. 423 D. 601

Answer 1

Answer:

C. n=423

Step-by-step explanation:

1) Notation and important concepts

Margin of error for a proportion is defined as "percentage points your results will differ from the real population value"

Confidence=90%=0.9

$\alpha=1-0.9=0.1$ represent the significance level defined as "a measure of the strength of the evidence that must be present in your sample before you will reject the null hypothesis and conclude that the effect is statistically significant".

$\hat p=0.5$ represent the sample proportion of consumers in the Oconee County area who would react favorably to a marketing campaig. For this case since we don't have enough info we use the value of 0.5 since is equiprobable the event analyzed.

$z_{\alpha/2}$ represent a quantile of the normal standard distribution that accumulates ${\alpha/2}$ on each tail of the distribution.

2) Formulas and solution for the problem

For this case the margin of error for a proportion is given by this formula

$ME=z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$   (1)

For this case the confidence level is 90% or 0.9 so then the significance would be

$\alpha=1-0.9=0.1$ and $\alpha/2=0.05$

With $\alpha/2=0.05$, we can find the value for $z_{\alpha/2}$ using the normal standard distribution table or excel.

The calculated value is $z_{\alpha/2}=1.644854$

Now from equation (1) we need to solve for n in order to answer the question.

$\frac{ME}{z_{\alpha/2}}=\sqrt{\frac{\hat p(1-\hat p)}{n}}$  

Squaring both sides:

$(\frac{ME}{z_{\alpha/2}})^2=\frac{\hat p(1-\hat p)}{n}$

And solving for n we got:

$n=\frac{\hat p(1-\hat p)}{(\frac{ME}{z_{\alpha/2}})^2}$

Now we can replpace the values

$n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.644854})^2}=422.739$

And rounded up to the nearest integer we got:

n=423