The C.S. Mott Children’s Hospital surveyed a national random sample of parents a

Question

The C.S. Mott Children’s Hospital surveyed a national random sample of parents about their desire for email consultation for their child’s minor illness or injury and their willingness to pay for this convenience.  

The approximate 95% (conservative) confidence interval for the population proportion of all parents who can get email advice from their child’s usual health care practice was reported to be (0.023, 0.097). Approximately how many parents answered this email advice question? (If your answer is not a whole number, round up to the next integer.)

Explanation / Answer

For 95% CI , using the z table we get the Z value as 1.96

sample proportion±zsqrt(sample proportion(1sample proportion)/n)

we are given than

sample proportion + zsqrt(sample proportion(1sample proportion)/n) = 0.097

sample proportion - zsqrt(sample proportion(1sample proportion)/n) = 0.023

where Z = 1.96

sample proportion + 1.96sqrt(sample proportion(1sample proportion)/n) = 0.097

sample proportion - 1.96sqrt(sample proportion(1sample proportion)/n) = 0.023

adding the 2 above equations

2*1.96sqrt(sample proportion(1sample proportion)/n) = 0.097+0.023

putting sample proprotion = 1/2

we calculate n as

2*1.96*sqrt(0.5*0.5/n) = 0.12

solving for n

((0.12/(2*1.96))^2)/(0.5*0.5) = 1/n

n = 266.7 = 267 approx

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

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