Question

The manager of an orchard expects about 70% of his apples to exceed the weight requirement for ""Grade A"" designation. At least how many apples must he sample to be 90% confident of estimating the true proportion within ± 4%?

Answer 1

Answer:

356

Step-by-step explanation:

We are given that p=0.7, and a 90% confidence interval. Also, we know the margin of error to be 0.04, the sample size is calculated as:

#The sample size is calculated as:

$ME=z_{\alpha/2}\sqrt\frac{p(1-p)}{n}}\\\\0.04=1.645\sqrt\frac{0.7(1-0.7)}{n}\\\\n=\frac{0.7\times 0.3}{(0.04/1.645)^2}\\\\n=355.17\approx 356$

Hence, at least 356 apples must be sampled to get a true proportion with the 4% ME.