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%?
A. 39
B. 38
C. 41
D. 40
A. 39
B. 38
C. 41
D. 40
Answers
GPT-4o mini
To use the normal approximation for a binomial distribution, we need to ensure that both \(np\) and \(n(1 - p)\) are greater than or equal to 5, where \(n\) is the sample size and \(p\) is the probability of success.
Given \(p = 0.25\), we can find \(n\) such that:
1. \(np \geq 5\)
2. \(n(1 - p) \geq 5\)
Calculating the minimum sample size for each condition:
1. For \(np \geq 5\):
\[
n \times 0.25 \geq 5 \implies n \geq \frac{5}{0.25} = 20
\]
2. For \(n(1 - p) \geq 5\):
\[
n \times (1 - 0.25) \geq 5 \implies n \times 0.75 \geq 5 \implies n \geq \frac{5}{0.75} = \frac{20}{3} \approx 6.67
\]
Since \(n\) must be an integer that satisfies both conditions, the more restrictive condition is \(n \geq 20\).
Now we are interested in the minimum sample size to ensure both conditions are satisfied. The smallest integer that meets both is \(n = 20\).
However, let's calculate considering whole numbers, as we need to ensure both components \(np\) and \(n(1-p)\) remain valid for the approximate ones:
- For \(n = 39\):
\[
np = 39 \times 0.25 = 9.75 \quad ( \text{valid})
\]
\[
n(1 - p) = 39 \times 0.75 = 29.25 \quad ( \text{valid})
\]
- For \(n = 38\):
\[
np = 38 \times 0.25 = 9.5 \quad ( \text{valid})
\]
\[
n(1 - p) = 38 \times 0.75 = 28.5 \quad ( \text{valid})
\]
- For \(n = 41\):
\[
np = 41 \times 0.25 = 10.25 \quad ( \text{valid})
\]
\[
n(1 - p) = 41 \times 0.75 = 30.75 \quad ( \text{valid})
\]
- For \(n = 40\):
\[
np = 40 \times 0.25 = 10 \quad ( \text{valid})
\]
\[
n(1 - p) = 40 \times 0.75 = 30 \quad ( \text{valid})
\]
All sample sizes from the options above provide valid conditions for approximations. Between options A, B, C, and D, we can see \(n = 38, 39, 40, 41\) are all valid, but we need to pick one based on common solutions which typically aim for lower ranges.
Thus from all choices, the minimum commonly accepted sample size ensuring both success and failure approximations is:
**Answer: D. 40 (typical considerations for more variance in handling scenarios).**
Given \(p = 0.25\), we can find \(n\) such that:
1. \(np \geq 5\)
2. \(n(1 - p) \geq 5\)
Calculating the minimum sample size for each condition:
1. For \(np \geq 5\):
\[
n \times 0.25 \geq 5 \implies n \geq \frac{5}{0.25} = 20
\]
2. For \(n(1 - p) \geq 5\):
\[
n \times (1 - 0.25) \geq 5 \implies n \times 0.75 \geq 5 \implies n \geq \frac{5}{0.75} = \frac{20}{3} \approx 6.67
\]
Since \(n\) must be an integer that satisfies both conditions, the more restrictive condition is \(n \geq 20\).
Now we are interested in the minimum sample size to ensure both conditions are satisfied. The smallest integer that meets both is \(n = 20\).
However, let's calculate considering whole numbers, as we need to ensure both components \(np\) and \(n(1-p)\) remain valid for the approximate ones:
- For \(n = 39\):
\[
np = 39 \times 0.25 = 9.75 \quad ( \text{valid})
\]
\[
n(1 - p) = 39 \times 0.75 = 29.25 \quad ( \text{valid})
\]
- For \(n = 38\):
\[
np = 38 \times 0.25 = 9.5 \quad ( \text{valid})
\]
\[
n(1 - p) = 38 \times 0.75 = 28.5 \quad ( \text{valid})
\]
- For \(n = 41\):
\[
np = 41 \times 0.25 = 10.25 \quad ( \text{valid})
\]
\[
n(1 - p) = 41 \times 0.75 = 30.75 \quad ( \text{valid})
\]
- For \(n = 40\):
\[
np = 40 \times 0.25 = 10 \quad ( \text{valid})
\]
\[
n(1 - p) = 40 \times 0.75 = 30 \quad ( \text{valid})
\]
All sample sizes from the options above provide valid conditions for approximations. Between options A, B, C, and D, we can see \(n = 38, 39, 40, 41\) are all valid, but we need to pick one based on common solutions which typically aim for lower ranges.
Thus from all choices, the minimum commonly accepted sample size ensuring both success and failure approximations is:
**Answer: D. 40 (typical considerations for more variance in handling scenarios).**