Question

A random sample of 356 medical doctors showed that 175 had a solo practice. As a news writer, how would you report the survey results regarding the percentage of medical doctors in solo practice? What is the margin of error based on a 90% confidence interval?

Answer 1

Part a)

Total Number of doctors in sample = n = 356
Number of doctors with solo practice = x = 175

Proportion of doctors with solo practice = p =
`(x)/(n) = (175)/(356)=0.492`

q = 1 - p = 0.508

As both p and q are approximately 0.5, as a new writer the percentage of survey results will be reported as:
About half of the medical doctors opt for the solo practice.

Part b)
Margin of Error for 90% confidence interval.

z score = 1.645
Margin of Error = E

Formula for Margin of Error is:

`E=z \sqrt{ (pq)/(n) }`

Using the values in the formula,we get:


`E=1.645 \sqrt{ (0.492*0.508)/(356) }=0.044`

Therefore, the margin of error based on 90% confidence interval is 0.044.