Question

John davis, a manager of a supermarket, wants to estimate the proportion of customers who use food stamps at his store. he has no initial estimate of what the sample proportion will be. how large a sample is required to estimate the true proportion to within 3 percentage points with 98% confidence?

Answer 1

Answer:

1509

Step-by-step explanation:

We know that, formula for number of samples is,

$n= \frac{z^2 \times p \times q}{SE^2}$

When nothing is given, we take $p=0.5$ and we know that $q = 1-p=0.5$.

SE = 0.03 (the true proportion should remain withing 3%)

Putting the values we get,

$n= \frac{(2.33)^2 \times 0.5 \times 0.5}{(0.03)^2}$

$n= \frac{5.4289 \times 0.5 \times 0.5}{0.03 \times 0.03} \approx 1509$

Therefore, the sample required to estimate the true proportion to within 3 percentage points is 1509.