Question

In: Statistics and Probability

in a California study 69 of 612/4 graders had blood cholesterol above 200 MG/DL, a level...

in a California study 69 of 612/4 graders had blood cholesterol above 200 MG/DL, a level that is dangerously high for children.

a. Construct a 90% confidence interval estimate for the population proportion with dangerously high blood cholesterol.

b. construct a 96% confidence interval estimate for the population proportion with dangerously high blood cholesterol.

c. which is the accurate (a or b) estimate of population proportion and why?

d. do you have to make any assumptions about normality while answering A and B?

Solutions

Expert Solution

GIVEN THAT :-

According to the question we have that ,

California study 69 of 612/4 graders had blood cholesterol above 200 MG/DL, a level that is dangerously high for children.

now

=> TO FIND :-a)Construct a 90% confidence interval estimate for the population proportion with dangerously high blood cholesterol.

We have that from the question

favorable cases X= 69

samle size N= 612.

now

p ^= X/N

= 0.113

now finding the CI (PROPORTION)

=>TO FIND:-b)construct a 96% confidence interval estimate for the population proportion with dangerously high blood cholesterol.

same as the we ahve the same formula but taking the value of alpha is 0.04 and the Zc = 2.054.

and then solving the above we have the solution as

(0.086,0.139)

=>TO FIND :-c)which is the accurate (a or b) estimate of population proportion and why?

as from the above a,b we have that, option B has higher confidence value .

Answer in B is more accurate.

=>TO FIND :-d)do you have to make any assumptions about normality while answering A and B?

No,

because x and (n-x) > 10 and hence by central limit theorem sampling distribution of proportion follows a normal distribution and hence we don't need to assume anything about normality.

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orchestra answered 1 year ago

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