Question

Suppose 47G% of the doctors in a hospital are surgeons. If a sample of 460460 doctors is selected, what is the probability that the sample proportion of surgeons will differ from the population proportion by greater than 5%5%

Answer 1

Answer:

3.16% probability that the sample proportion of surgeons will differ from the population proportion by greater than 5%

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

$p = 0.47, n = 460, \mu = 0.47, s = \sqrt{\frac{0.47*0.53}{460}} = 0.0233$

What is the probability that the sample proportion of surgeons will differ from the population proportion by greater than 5%

Sample proportion lower than 0.47 - 0.05 = 0.42 or higher than 0.47 + 0.05 = 0.52.

Since they are equidistant from the mean of 0.47 they are equal. So we find one of them, and multiply by two.

Lower than 0.42:

pvalue of Z when X = 0.42. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.42 - 0.47}{0.0233}$

$Z = -2.15$

$Z = -2.15$ has a pvalue of 0.0158

2*0.0158 = 0.0316

3.16% probability that the sample proportion of surgeons will differ from the population proportion by greater than 5%