A recent study of 3200 children randomly selected found 20% of them deficient in
ID: 3289502 • Letter: A
Question
A recent study of 3200 children randomly selected found 20% of them deficient in vitamin D. a) Construct the 90% confidence interval for the true proportional of children who are deficient in vitamin D. (Round to three decimal places as needed.) b) Explain carefully what the interval means. A. We are 98% confident that the person of people deficient in vitamin D is 20%. B. We are 98% confident that the interval contains the true proportion of children deficient in vitamin D C. We are 98% confident that the interval of children deficient in vitamin D is 20%. D. We are 98% confident that the interval contains the true proportion of people deficient in vitamin D. a) Explain what confidence About 98% of random samples of size 3200 will produce that contain the of children that are deficient in vitamin D.Explanation / Answer
Answer to part a)
n = 3200
p^ = 0.20
Confidence level = 0.98
Z value = 2.33
.
The formula of standard error is :
SE = sqrt(P^ *Q^/n)
SE = sqrt(0.20*0.80/3200)
[Q^ = 1- P^ = 1 - 0.20 = 0.80]
SE = 0.0071
.
Formula of confidence interval is
P^ - Z*SE , P^ + Z*SE
0.20 - 2.33 *0.0071 , 0.20 + 2.33 *0.0071
2.3135 , 2.3465
.
Answer to part b)
the correct explanation is
we are 98% confident that the interval contains the true proportion of children deficient in vitamin D
The correct answer choice is B
.
Answer to part c)
98% confidence means "about 98% of the samples of size 3200 will produce confidence itnervals that contain the true proportion of children that are deficient of vitamin D
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