808 CHAPTER 39. STATISTICAL TESTS

This involves the T distribution. We will assume Xk,k ≤ r+ 1 is an independent samplefrom a normal distribution having mean µ and variance σ2.

Lemma 39.2.1 Let Xk be r + 1 independent random variables normally distributed withmean µ and variance σ2. Then

1√r+1

r+1

∑k=1

Xk−µ

σ

is normally distributed with mean 0 and variance 1.

Proof:

E

(exp

(t

1√r+1

r+1

∑k=1

Xk−µ

σ

))= E

(r+1

∏k=1

t√r+1

(Xk−µ

σ

))

Now recall that Xk−µ

σis normally distributed with mean 0 and variance 1. Therefore, by

independence, this equals

r+1

∏k=1

exp(−1

2t2

r+1

)= exp

(−1

2t2)

which is the moment generating function for a normally distributed random variable withmean 0 and variance 1. ■

Recall that (r+1)S2/σ2 is X 2 (r) . By the above discussion of the T distribution,

1√r+1 ∑

r+1k=1

Xk−µ

σ√∑

r+1k=1(Xk−X̄)2

rσ2

is a T random variable with r degrees of freedom discussed above. However, this expres-sion simplifies quite a bit. It becomes

√r√

r+1 ∑r+1k=1 (Xk−µ)√

∑r+1k=1 (Xk− X̄)

2=

√r√

r+1(r+1)(X̄−µ)√

∑r+1k=1 (Xk− X̄)

2=

√r√

r+1(X̄−µ)√∑

r+1k=1 (Xk− X̄)

2

Notice how the σ disappeared leaving only µ .

Example 39.2.2 Here are 11 numbers from an independent random sample of a normaldistribution having variance σ2 and mean µ . Find a .95 confidence interval for the meanµ . The numbers are

3,4,5,6,2,3.5,5,4,6,2,4.2

After some computations, we find X̄ = 4.0636 and 11S2 = 19.245. Then the statisticabove is of the form √

10√

11(4.0636−µ)√19.245

Using the data cursor in the graph of the function F (x) = P(X ≤ x) for X a T randomvariable with r = 10, we can find an interval corresponding to probability at least .95. A