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
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39
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40
40
38
Responses
39
39
41
41
40
40
38
Answers
GPT-4o mini
To determine the minimum sample size needed to use normal approximations for a proportion, we often use the rule of thumb that both \( np \) and \( n(1 - p) \) should be at least 10, where \( p \) is the probability of success and \( n \) is the sample size.
Given \( p = 0.25 \):
1. Calculate \( np \):
\[
np = n \times 0.25
\]
2. Calculate \( n(1-p) \):
\[
n(1 - p) = n \times (1 - 0.25) = n \times 0.75
\]
We need both \( np \geq 10 \) and \( n(1 - p) \geq 10 \).
From \( np \geq 10 \):
\[
n \times 0.25 \geq 10 \implies n \geq \frac{10}{0.25} = 40
\]
From \( n(1 - p) \geq 10 \):
\[
n \times 0.75 \geq 10 \implies n \geq \frac{10}{0.75} \approx 13.33
\]
The more restrictive condition is \( n \geq 40 \).
Thus, the minimum sample size needed is **40**.
The answer is **40**.
Given \( p = 0.25 \):
1. Calculate \( np \):
\[
np = n \times 0.25
\]
2. Calculate \( n(1-p) \):
\[
n(1 - p) = n \times (1 - 0.25) = n \times 0.75
\]
We need both \( np \geq 10 \) and \( n(1 - p) \geq 10 \).
From \( np \geq 10 \):
\[
n \times 0.25 \geq 10 \implies n \geq \frac{10}{0.25} = 40
\]
From \( n(1 - p) \geq 10 \):
\[
n \times 0.75 \geq 10 \implies n \geq \frac{10}{0.75} \approx 13.33
\]
The more restrictive condition is \( n \geq 40 \).
Thus, the minimum sample size needed is **40**.
The answer is **40**.