Question

2 6. 2pts) Suppose a consumer advocacy group would like to conduct a survey to find the proportion "p" of consumers who bought the newest generation of an MP3 player who were happy with their purchase. How large a sample should they take to estimate "p" with 2% margin oferror and 90% confidence?

Answer 1

$\alpha$ = 0.10

From Table, critical values of Z = $\pm$ 1.645

Since p is not given:
Take p = 0.5

MOE = 0.02

Substituting, we get:

$n=\frac{1.645^{2}\times 0.5\times (1-0.5)}{0.02^{2}}=1692$

So,

Answer is:

1692

Answer 2

How can we just take p=0.5??