a random sample of 328 medical doctors showed that 171 had a solopractice. a.) l

Question

a random sample of 328 medical doctors showed that 171 had a solopractice.
a.) let p represent the proportion of all medical doctors whohave a solo practice. find a point estimate for p.
b.) find a 95% confidence interval (1.96) for p. give a briefexplanation of the meaning of the interval.
c.) as a news writer, how would you report the survey resultsregarding the percentage of medical doctors in solo practice? whatis the margin of error based on a 95% (1.96) confidenceinterval?
a.) let p represent the proportion of all medical doctors whohave a solo practice. find a point estimate for p.
b.) find a 95% confidence interval (1.96) for p. give a briefexplanation of the meaning of the interval.
c.) as a news writer, how would you report the survey resultsregarding the percentage of medical doctors in solo practice? whatis the margin of error based on a 95% (1.96) confidenceinterval?

Explanation / Answer

a.) let p represent the proportion of all medical doctors whohave a solo practice. find a point estimate for p.
p=171/328=0.52


b.) find a 95% confidence interval (1.96) for p. give a briefexplanation of the meaning of the interval.

p±Z*p*(1-p)/n
--> 0.52±1.96*sqrt(0.52*(1-0.52)/328)
--> ( 0.4659, 0.5741)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.