Mathematical Statistics with Applications
Mathematical Statistics with Applications
7th Edition
ISBN: 9780495110811
Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer
Publisher: Cengage Learning
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Chapter 7.5, Problem 70E

a.

To determine

Prove that p+3pq/n<1 if and only if n>9(p/q).

a.

Expert Solution
Check Mark

Explanation of Solution

Let

p+3pq/n<13pq/n<1p3pq/n<qpq/n<q3

Taking square on both sides

p+3pq/n<1pq/n<q29pn<q9np>9qn>9(pq)

Hence proved

b.

To determine

Prove that 0<p3pq/n if and only if n>9(q/p).

b.

Expert Solution
Check Mark

Explanation of Solution

Let

0<p3pq/n3pq/n<p3pq/n<ppq/n<p3

Taking square on both sides

0<p3pq/npq/n<p29qn<p9nq>9pn>9(qp)

Hence proved

c.

To determine

Prove that the normal approximation to the binomial is adequate if n>9(pq) and n>9(qp) or, equivalently, n>9(largerofpandqsmallerofpandq).

c.

Expert Solution
Check Mark

Explanation of Solution

From the part a, normal approximation to the binomial is adequate if n>9(pq) and from part b, normal approximation to the binomial is adequate if n>9(qp).

By combining the parts a and b, normal approximation to the binomial is adequate if n>9(pq) and n>9(qp).

Or,

n>9max(pq,qp)

If p=q, then

n>9max(pq,qp)=n>9

If p>q, then

n>9max(pq,qp)=n>9(pq)

If p<q, then

n>9max(pq,qp)=n>9(qp)

Then,

n>9max(pq,qp)=n>9(largerofpandqsmallerofpandq)

Hence proved

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Chapter 7 Solutions

Mathematical Statistics with Applications

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