4. In a random sample of 800 adults in the U.S.A., it was found that 72 of those

Question

4. In a random sample of 800 adults in the U.S.A., it was found that 72 of those had a pinworm infestation. A) What is the point estimate for the proportion of all adults with pinworms B) What is the 90% confidence interval for the proportion of all adults wi pinworm? c) What is the 99% confidence interval for the proportion of all adults with pinworm? D) The Center for Disease Control (CDC) wants to improve the 99% confidence interval for the proportion of healthy adults with pinworm. They want an estimate that is in error by no more than one percentage point. What sample size is needed to obtain this type of accuracy? Use the prior value of p to calculate this number. Using the equation below Where Z (the critical value for 99% confidence interval is constant- 2.575) P is p calculated from part A) E is margin of error which 0.01 since the interval is 99%.

Explanation / Answer

a.
No. of had a pinworm=72
Sample Size(n)=800
point estimate = Sample proportion = x/n =0.09

b.
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval

Confidence Interval = [ 0.09 ±Z a/2 ( Sqrt ( 0.09*0.91) /800)]
= [ 0.09 - 1.645* Sqrt(0) , 0.09 + 1.65* Sqrt(0) ]
= [ 0.073,0.107]

c.
Confidence Interval For Proportion
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
No. of success(x)=72
Sample Size(n)=800
Sample proportion = x/n =0.09
Confidence Interval = [ 0.09 ±Z a/2 ( Sqrt ( 0.09*0.91) /800)]
= [ 0.09 - 2.576* Sqrt(0) , 0.09 + 2.58* Sqrt(0) ]
= [ 0.064,0.116]

d.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.01 is = 2.576
Sample Proportion = 0.09
ME = 0.01
n = ( 2.576 / 0.01 )^2 * 0.09*0.91
= 5434.701 ~ 5435

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

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