Homework Answers
Answer: 7. A random sample of 328 medical doctors showed that 171 have a solo practice.
a) Let p represent the proportion of all medical doctors whohave a solo practice. find a point estimate for p.
p = 171/328
p = 0.5213
b) A 95% confidence interval for p:
At 95% confidence interval,
Zα/2 = 1.960
Therefore, CI
= p ± Zα/2*√p(1-p)/n
= 0.5213 ± 1.96 * √(0.5213*0.4787)/328
= 0.5213 ± 1.96 * 0.0276
= 0.5213 ± 0.0541
= (0.4672, 0.5754)
Therefore, a 95% confidence interval for p is between 0.4672 and 0.5754.
c) As a news writer, how would you report the survey results regarding the percentage of medical doctors in solo practice.
At 95% confidence interval,
Margin of error, E = Zα/2 * √(p(1-p)/n)
E = 1.96 * √0.5213(1-0.5213)/328
E = 1.96 * 0.0276
E = 0.0541
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Show work for full credit. Compute confidence intervals and
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