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

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

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 = (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 = (\sigma)/(√(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{(p(1-p))/(n)}`

In this question:

`p = 0.47, n = 460, \mu = 0.47, s = \sqrt{(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 = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (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%