Lecture 25: Stat 400, Lecture 25 Sampling from N ( , 2 ) and the - - PowerPoint PPT Presentation

lecture 25 stat 400 lecture 25 sampling from n 2
SMART_READER_LITE
LIVE PREVIEW

Lecture 25: Stat 400, Lecture 25 Sampling from N ( , 2 ) and the - - PowerPoint PPT Presentation

Lecture 25: Stat 400, Lecture 25 Sampling from N ( , 2 ) and the CLT 0/ 6 Suppose X 1 , X 2 , . . . , X n is a random sample from a normal population. We have seen that we should use the sample mean X to estimate the population mean and


slide-1
SLIDE 1

Lecture 25: Stat 400, Lecture 25 Sampling from N(µ, σ2) and the CLT

0/ 6

slide-2
SLIDE 2

1/ 6

Suppose X1, X2, . . . , Xn is a random sample from a normal population. We have seen that we should use the sample mean X to estimate the population mean µ and the sample variance S2 to estimate the population variance σ2.

Lecture 25: Stat 400, Lecture 25 Sampling from N(µ, σ2) and the CLT

slide-3
SLIDE 3

2/ 6

X and S2 are random variables.

$64,000 question

How are X and S2 distributed ? The answer is given by the following considerations Any linear combination of Independent normal random variables is again normal so X is normal. Since E(X) = µ and V(X) = σ2 n we have X ∼ N

  • µ, σ2

n

  • Lecture 25: Stat 400, Lecture 25 Sampling from N(µ, σ2) and the CLT
slide-4
SLIDE 4

3/ 6

Suppose Z, Z2, . . . , Zn are independent standard normal random variables. Then Z2

1 + . . . + Z2 b ∼ χ2(n)

(∗) Chi-squared with n degrees of Now Zi = Xi − µ

σ ∼ N(0, 1)

So

n

  • i=1

Xi − µ σ 2 = 1 σ2

n

  • i=1

(Xi − µ)2 ∼ χ2(n)

Now replace µ by its estimation X Rule of thumb - every time you replace a quantity by its estimator you lose one degree of freedom in the chi-squared distribution

Lecture 25: Stat 400, Lecture 25 Sampling from N(µ, σ2) and the CLT

slide-5
SLIDE 5

4/ 6

So by the “rule of thumb” Y = 1

σ2

n

  • i=1

(Xi − X)2 ∼ χ2(n − 1)

Now S2 = 1 n − 1

n

  • i=1

(Xi − X)2

So Y = n − 1

σ2 S2 and we obtain the critical

n − 1

σ2 S2σχ2(n − 1)

(∗∗) Remark This isn’t a proof because we used “the rule of thumb” but the result is true

Lecture 25: Stat 400, Lecture 25 Sampling from N(µ, σ2) and the CLT

slide-6
SLIDE 6

5/ 6

Bottom Line

Theorem Let X1, X2, . . . , Xn be a random Sample from a normal population with mean µ and variance σ2. Then (i) X ∼ N

  • µ, σ2

n

  • (ii) n − 1

σ2 S2 ∼ χ2(n − 1)

(iii) X and S2 are independent. The above statement is on exact statement but if we take a large sample

(n > 30) from any population with mean µ and variance σ2 we may assume

Lecture 25: Stat 400, Lecture 25 Sampling from N(µ, σ2) and the CLT

slide-7
SLIDE 7

6/ 6

Theorem (Cont.) to a good approximation that the population has N(µ, σ2) distribution and we have by CLT Theorem If X1, X2, . . . , Xn is a large (n > 30) random sample from any population with mean µ and variance σ2 then (i) X ≈ N(µ, σ2 n ) (ii) S2 ≈ χ2(n − 1) (iii) X and S2 are approximately independent. (then are not independent unless the population is normal.)

Lecture 25: Stat 400, Lecture 25 Sampling from N(µ, σ2) and the CLT