Question

A marketing company wants to estimate the proportion of consumers in a certain region of the country who would react favorably to a new marketing campaign. Further, the company wants the estimate to have a margin of error of no more than 5 percent with 90 percent confidence. Of the following, which is closest to the minimum number of consumers needed to obtain the estimate with the desired precision?

Answer 1

Answer: 271

Step-by-step explanation:

When the prior estimate of population proportion is not given , then the formula to find the sample size is given by :-

$n=0.25(\dfrac{z^*}{E})^2$

, where E = Margin of error.

z* = Critical z-value.

As per given , we have

E = 5%=0.05

Confidence level = 90%

The critical value of z at 90% is 1.645  (By z-table)

Put all values in the formula , we get

$n=0.25(\dfrac{1.645}{0.05})^2$

$n=0.25\cdot32.9^{2}$

$n=270.6025\approx271$

Thus, the minimum sample size needed = 271

Hence , the correct answer is 271 .

Answer 2

To estimate the proportion of consumers in a certain region of the country who would react favorably to a new marketing campaign, we can use formula of sample size.

The minimum size needed is 271.

Given:

The margin of error is$E=\dfrac{5\%}{100}=0.05$.

Confidence level = 90%

Refer the z-table critical value of z at 90% confidence level is $1.645$.

Calculate the sample size by using following formula.

$n=0.25\left (\dfrac{z^*}{E}\right )^2$

Substitute the value in above equation.

$n=0.25\left(\dfrac{1.645}{0.05}\right)^2\\n=0.25\times 39.9^2\\n=270.6025\\n\approx271$

Thus, the minimum size needed is 271.

Learn more about what  sample size is here:

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