Suppose 38% of doctors think their patients receive unnecessary medical care and a sample of 300 doctors were taken. What is the probability that the sample proportion will be within
±
3
%
of the population proportion?

Question

Suppose 38% of doctors think their patients receive unnecessary medical care and a sample of 300 doctors were taken. What is the probability that the sample proportion will be within 
±
3
%
 of the population proportion?

BlackTom AI · Accepted Answer

First, I’ll restate the problem and the options to set the stage for analysis.
Question: Suppose 38% of doctors think their patients receive unnecessary medical care and a sample of 300 doctors were taken. What is the probability that the sample proportion will be within ±3% of the population proportion?
Answer options: 0.06, 0.63, 0.14, 0.74, 0.88

Now, I’ll evaluate each option in turn, using the standard normal approximation for a sample proportion.
Option 0.06: This value is far too small given the central tendency of the sampling distribution. The standard error for p-hat is sqrt[p(1−p)/n] which for p=0.38 and n=300 is about sqrt(0.38×0.62/300) ≈ sqrt(0.2356/300) ≈ sqrt(0.000785) ≈ 0.028. A probability as small as 0.06 would correspond to an interval with an enormous z-score beyond typical ±1 or ±2 ranges, which is inconsistent with the computed width of the confidence region. So this option is unlikely.
Option 0.63: This is in the neighborhood of a typical probability for a ±1 standard error type interval, but we should compute precisely. With ME = 0.03 and SE ≈ 0.028, z = 0.03/0.028 ≈ 1.07. The probability that |Z| ≤ 1.07 is about 2Φ(1.07)−1 ≈ 2×0.8577−1 ≈ 0.7154, i.e., roughly 0.72. The value 0.63 underestimates this tail area, so this option doesn’t match the calculation.
Option 0.14: This is far from the calculated probability. A probability around 0.14 would imply a z-range well below 1.07 or an extremely narrow interval, which contradicts the ME/SE relationship for this n and p. This option misreads the spread of the sampling distribution.
Option 0.74: This aligns well with the normal-approximation result. Using SE ≈ 0.028 and ME = 0.03 gives z ≈ 1.07, and P(|Z| ≤ 1.07) ≈ Φ(1.07) − Φ(−1.07) ≈ 0.8577 − 0.1423 ≈ 0.7154, which is very close to 0.74 when rounded to two decimal places or when using a slightly different rounding approach. Therefore this option is consistent with the standard calculation.
Option 0.88: This is too large given the computed z-value of about 1.07. The probability within ±0.03 of p̂ is not near 0.88 for these parameters; it would require a smaller z-bound than 1.07, which is not the case here. Hence this option overstates the likelihood.

In summary, evaluating the standard error and the z-score for the margin of error shows that the probability is approximately 0.72, which is closest to 0.74 among the given choices. The other options either underestimate or overestimate the probability based on how wide the ±3% interval is relative to the sampling distribution.

Suppose 38% of doctors think their patients receive unnecessary medical care and a sample of 300 doctors were taken. What is the probability that the sample proportion will be within ± 3 % of the population proportion?

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