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 = [tex] \frac{x}{n} = \frac{175}{356}=0.492 [/tex]
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:
[tex]E=z \sqrt{ \frac{pq}{n} } [/tex]
Using the values in the formula,we get:
[tex]E=1.645 \sqrt{ \frac{0.492*0.508}{356} }=0.044 [/tex]
Therefore, the margin of error based on 90% confidence interval is 0.044.
