g Suppose that p=0.5 and you ran the experiment above (consisting of 200 coin flips) a total of 1000 times. What is the expected number of experiments such that the estimator n−−√(X¯¯¯¯¯n−0.5)0.5(1−0.5)√ is larger than the value D2 attained in the first experiment? (Round your answer to the nearest integer.)

The expected number of experiment is 198

Step-by-step explanation:

Solution

Given that:

We need to carry out a test

where,

H0 : p =0.5

and

H1 : p ≠ 0.5

n = the number of flip coin which is = 200

x = this is the number of  heads declared = 106

So,

xₙ = x/n = 106/200 = 0.53 = p

Thus,

D₂ =√n (xₙ - 0.5)/√0.5 * (1-0.5)

=√200 * (0.53 - 0.5)/ √0.5 * (1-0.5)

= 0.848528137

D₂ = 0.8485

Now,

p ( z> D₂ ) = p ( z > 0.8485)

=0.198072

Thus,

By applying R,

1 - pnorm (0.8485, 0,1)

That is (1- pnorm (D₂, 0, 1)

Hence,

p ( z> D₂ )≈ 0.198072

So,

We find The expected number of experiment such that the estimator √n (xₙ - 0.5)/√0.5 * (1-0.5)i s larger than the value D₂ when the total is 1000 times or  attained in the first experiment Thus

1000 * p ( z> D₂ )

= 1000 * 0.198072

=198.072

=198

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