A market researcher at a telecommunications company is planning to conduct a sur

Question

A market researcher at a telecommunications company is planning to conduct a survey to gather information about the proportion of people in the US between the ages of 18 and 40 who own a smartphone. The researcher plans to construct a 95% confidence interval and would like the maximum error of the estimate (margin of error) to be no more than 0.03 (3%). (a) What sample size would be needed if the researcher has no prior estimate for p? (b) What sample would be needed if prior research indicates that the proportion is about 87%?

Explanation / Answer

a.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Sample Proportion = 0.5
ME = 0.03
n = ( 1.96 / 0.03 )^2 * 0.5*0.5
= 1067.111 ~ 1068
b.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.05 is = 1.96
Sample Proportion = 0.87
ME = 0.03
n = ( 1.96 / 0.03 )^2 * 0.87*0.13
= 482.761 ~ 483

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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