45
Answer: B) 1201
a) 601
b) 1201
c) 423
d) 13
Answer: B) 1201
n = (Z^2 * p * (1-p)) / E^2
Where:
n = required sample size
Z = Z-score (corresponding to the desired level of confidence)
p = estimated proportion
E = maximum margin of error (expressed as a decimal)
In this case, the maximum margin of error is 4%, which is 0.04 as a decimal. The Z-score for a 95% confidence level is approximately 1.96.
Let's plug in the values:
n = (1.96^2 * p * (1-p)) / (0.04^2)
Since we don't have an estimated proportion, we can be conservative and assume maximum variability, which is p = 0.5 (50%).
n = (1.96^2 * 0.5 * (1-0.5)) / (0.04^2)
Simplifying the equation:
n = (3.8416 * 0.5 * 0.5) / 0.0016
n = 0.9604 / 0.0016
n ≈ 600.25
Since we need a whole number for the sample size, we round up to the nearest whole number.
Therefore, the sample size needed is approximately 601.
The correct answer is (a) 601.
n = (Z^2 * P * (1-P)) / E^2
Where:
n = Sample size
Z = Z-score at the desired level of confidence (for a 95% confidence level, the Z-score is approximately 1.96)
P = Estimated proportion (if no information is available, it is common to use 0.5 as a conservative estimate)
E = Maximum error margin (in this case, 4% which can be expressed as 0.04)
Now, let's calculate the sample size:
n = (1.96^2 * 0.5 * (1-0.5)) / 0.04^2
n ≈ 962.14
Since the sample size must be a whole number, we must round it up to the nearest whole number to ensure that it is large enough. Therefore, the recommended sample size is 963.
Among the given choices, option b) 1201 is the closest to the calculated sample size of 963.