233k views
5 votes
A nutritionist is interested in estimating the proportion of consumers who look to purchase organic produce.

How large of a sample would be necessary in order to be 99% confident that the estimate is within 3% of the actual proportion?



(Hint: use a 'worst case scenario' of p=0.50)

User Lonix
by
8.1k points

1 Answer

2 votes

Answer: To determine the sample size needed for estimating the proportion of consumers who look to purchase organic produce with 99% confidence and a margin of error of 3%, we can use the formula for the sample size of a proportion:

Sample size (n) = (Z^2 * p * q) / E^2

where:

Z = Z-score for the desired level of confidence (in this case, for 99% confidence, Z ≈ 2.576)

p = estimated proportion (use a worst-case scenario of p = 0.50 for maximum sample size)

q = 1 - p

E = margin of error (3% = 0.03)

Now, let's plug in the values:

n = (2.576^2 * 0.50 * 0.50) / 0.03^2

n = (6.633776 * 0.25) / 0.0009

n = 1.658444 / 0.0009

n ≈ 1842.71

Since we cannot have a fraction of a participant, we need to round up to the nearest whole number. Therefore, a sample size of 1843 participants would be necessary to be 99% confident that the estimate of the proportion of consumers who look to purchase organic produce is within 3% of the actual proportion, using a worst-case scenario of p = 0.50.

User Pratiked
by
7.3k points