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If a manufacturer conducted a survey among randomly selected target market households and wanted to be 95​% confident that the difference between the sample estimate and the actual market share for its new product was no more than 9​%, what sample size would be​ needed?

User Aristarhys
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1 Answer

7 votes

Answer:

We need a sample size of least 119

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
\pi, and a confidence level of
1-\alpha, we have the following confidence interval of proportions.


\pi \pm z\sqrt{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

The margin of error is:


M = z\sqrt{(\pi(1-\pi))/(n)}

95% confidence level

So
\alpha = 0.05, z is the value of Z that has a pvalue of
1 - (0.05)/(2) = 0.975, so
Z = 1.96.

Sample size needed

At least n, in which n is found when
M = 0.09

We don't know the proportion, so we use
\pi = 0.5, which is when we would need the largest sample size.


M = z\sqrt{(\pi(1-\pi))/(n)}


0.09 = 1.96\sqrt{(0.5*0.5)/(n)}


0.09√(n) = 1.96*0.5


√(n) = (1.96*0.5)/(0.09)


(√(n))^(2) = ((1.96*0.5)/(0.09))^(2)


n = 118.6

Rounding up

We need a sample size of least 119

User Roman Yudin
by
8.1k points

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