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 25 percent?(1 point)
Responses

40
40

38
38

39
39

41

Answer 1

To determine the minimum sample size needed to use normal approximations for a proportion when given a probability of success \( p = 0.25 \), we can use the rule of thumb for the normal approximation to the binomial distribution. This rule states that both \( np \) and \( n(1-p) \) should be greater than or equal to 5:

1. The probability of success \( p = 0.25 \).
2. The probability of failure \( q = 1 - p = 0.75 \).

We want to find \( n \) such that:

- \( np \geq 5 \)
- \( n(1 - p) \geq 5 \)

Now we solve these inequalities:

1. From \( np \geq 5 \):
\[
n \cdot 0.25 \geq 5 \implies n \geq \frac{5}{0.25} = 20
\]

2. From \( n(1 - p) \geq 5 \):
\[
n \cdot 0.75 \geq 5 \implies n \geq \frac{5}{0.75} \approx 6.67
\]

Both conditions are satisfied starting from \( n = 20 \). However, to ensure an accurate approximation, it is commonly recommended that both counts be at least 10.

So we check:
1. For \( np \geq 10 \):
\[
n \cdot 0.25 \geq 10 \implies n \geq \frac{10}{0.25} = 40
\]

2. For \( n(1 - p) \geq 10 \):
\[
n \cdot 0.75 \geq 10 \implies n \geq \frac{10}{0.75} \approx 13.33
\]

Thus, the minimum sample size that satisfies both conditions is \( n = 40 \).

Therefore, the correct response is:

**40**