Question

What is the minimum sample size needed in order to use normal approximations for an estimate about a population, given a probability of success of 10 percent?

Answer 1

To determine the minimum sample size needed to use normal approximations for a proportion, we typically apply the rule of thumb that requires both \( np \) and \( n(1-p) \) to be at least 5, where \( n \) is the sample size and \( p \) is the proportion of success.

Given a probability of success \( p = 0.10 \), we calculate as follows:

1. **Calculate \( np \)**:
\[
np = n \times 0.10
\]

2. **Calculate \( n(1-p) \)**:
\[
n(1-p) = n \times (1 - 0.10) = n \times 0.90
\]

Now we set up the inequalities:

- For \( np \) to be at least 5:
\[
n \times 0.10 \geq 5 \quad \Rightarrow \quad n \geq \frac{5}{0.10} = 50
\]

- For \( n(1-p) \) to be at least 5:
\[
n \times 0.90 \geq 5 \quad \Rightarrow \quad n \geq \frac{5}{0.90} \approx 5.56
\]

Since the second inequality yields a minimum sample size of approximately 6, the more restrictive condition is the first one. Thus, the minimum sample size required to ensure both conditions are satisfied is:

\[
\text{Minimum sample size } n = 50
\]

In conclusion, **the minimum sample size needed to use normal approximations with a probability of success of 10 percent is 50**.