816 CHAPTER 39. STATISTICAL TESTS

3. The confidence interval is determined by solving for the ratio σ̂2

σ2 in the inequality

a≤ σ̂2

σ2nS2

nŜ2≤ b

4. Thus the confidence interval is

anŜ2

nS2 ≤σ̂

2

σ2 ≤ bnŜ2

nS2

or in other words,

a∑

nk=1 (Yk− Ȳ )2

∑nk=1 (Xk− X̄)

2 ≤σ̂

2

σ2 ≤ b∑

nk=1 (Yk− Ȳ )2

∑nk=1 (Xk− X̄)

2

You do the same thing if you want a different probability. Just identify a different inter-val corresponding to the different probability and do the above.

39.3 Maximum Likelihood EstimatesThese estimates give a simple way to estimate various parameters. Unlike the above mate-rial on confidence intervals and hypothesis testing, you don’t get from these procedures aconfidence interval associated with a probability that the parameter is in this interval or adirection to reject a hypothesis. You just get the best estimate for the parameter in terms ofmaximizing likelihood. I think it is best to illustrate the technique using specific examples.

Example 39.3.1 You know a random variable is a binomial random variable. Thus

P(X = k) =

(nk

)pkqn−k

where q = 1− p. Find the maximum likelihood estimate for p.

What you do is to write down the “likelihood” obtained by taking a sample X1, · · · ,Xm.Then the likelihood is

L(p)≡m

∏k=1

pXk (1− p)1−Xk

To find an estimate, you seek to pick p in order to maximize this likelihood. Obviously itwould be better to maximize ln(L(p)) which equals

ln(L(p)) =m

∑k=1

[Xk ln(p)+(1−Xk) ln(1− p)]

Then from beginning calculus, we take a derivative with respect to p and set equal to 0 andsolve for p. This is the maximum likelihood estimate for p.

m

∑k=1

Xk

p+(1−Xk)

−11− p

= 0