Question 1171423: A researched wants to study the attitudes of college students towards credit cards. She plants to randomly select a sample of n students and give each a survey. She would like to have at least 50 students in the sample who regularly use credit cards. If 25% of all college students regularly use credit cards, estimate the minimum size of n so that the sample will contain 50 students who regularly use credit cards with a probability of at least 0.95.
I know this is a binomial approximation but I am not sure how to set this up.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! You're right, this is a binomial distribution problem, and we'll use a normal approximation to solve it. Let's break down how to set it up:
**1. Define the Variables**
* **p:** The proportion of college students who regularly use credit cards (p = 0.25).
* **n:** The sample size we need to find.
* **X:** The number of students in the sample who regularly use credit cards.
* We want to find the minimum 'n' such that P(X ≥ 50) ≥ 0.95.
**2. Apply the Normal Approximation**
Since 'n' is unknown, we'll use the normal approximation to the binomial distribution. For this to be valid, we need np and n(1-p) to be sufficiently large (generally, at least 10). We'll check this condition after finding 'n'.
* **Mean (μ):** μ = np = 0.25n
* **Standard Deviation (σ):** σ = √(np(1-p)) = √(0.25n * 0.75) = √(0.1875n)
**3. Set up the Probability Statement**
We want P(X ≥ 50) ≥ 0.95. We need to convert this to a standard normal distribution (Z-score).
* **Continuity Correction:** Since we're approximating a discrete distribution with a continuous one, we use a continuity correction. Instead of X ≥ 50, we use X ≥ 49.5.
* **Z-score:** Z = (X - μ) / σ = (49.5 - 0.25n) / √(0.1875n)
So, we want:
* P(Z ≥ (49.5 - 0.25n) / √(0.1875n)) ≥ 0.95
**4. Find the Z-value**
Since we want the probability to be at least 0.95, we need to find the Z-value that corresponds to the area to the *left* being 0.05 (1 - 0.95), because we are looking for the area to the right to be 0.95.
* Using a Z-table or calculator, the Z-value corresponding to a left-tail area of 0.05 is approximately -1.645.
**5. Set up the Inequality**
Now we have:
* (49.5 - 0.25n) / √(0.1875n) ≤ -1.645
**6. Solve for n**
* 49.5 - 0.25n ≤ -1.645√(0.1875n)
* 49.5 - 0.25n ≤ -0.7115√(n)
Let's rearrange the inequality:
* 0.25n - 0.7115√n - 49.5 >= 0
This is a bit tricky to solve directly. Let's make a substitution to simplify:
* Let y = √n, so n = y².
* 0.25y² - 0.7115y - 49.5 >= 0
Now, we can use the quadratic formula to solve for y:
* y = [0.7115 ± √(0.7115² - 4 * 0.25 * -49.5)] / (2 * 0.25)
* y = [0.7115 ± √(0.50623 + 49.5)] / 0.5
* y = [0.7115 ± √49.99623] / 0.5
* y = [0.7115 ± 7.0708] / 0.5
Since √n must be positive, we take the positive solution:
* y = (0.7115 + 7.0708) / 0.5
* y = 7.7823 / 0.5
* y = 15.5646
Now, substitute back √n = y:
* √n = 15.5646
* n = (15.5646)²
* n ≈ 242.26
Since n must be an integer, we round up to ensure the probability is at least 0.95.
* n = 243
**7. Check the Normal Approximation Conditions**
* np = 243 * 0.25 = 60.75 (≥ 10)
* n(1-p) = 243 * 0.75 = 182.25 (≥ 10)
The conditions are met.
**Conclusion**
The minimum sample size 'n' should be 243.
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