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Distribution of Sample Statistics - Notes | STA 2023, Study notes of Statistics

University of Florida (UF)Statistics

Material Type: Notes; Class: INTRO TO STATISTICS 1; Subject: STATISTICS; University: University of Florida; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 09/17/2009

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koofers-user-ks5 🇺🇸

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bg1
DISTRIBUTION OF
SOME SAMPLE STATISTICS
Rule 1: Sampling Distribution of
ˆ
p
:
1
X p( p )
ˆ
p ~ N p,
n n
IF
X ~ B(n, p)
and n p ≥ 10
and n (1–p) ≥ 10p) ≥ 10
and N ≥ 10n
Rule 2: Distribution of
X
[Normal Theory]:
X
X
X ~ N , n
IF X ~ N(μX, σX)
Rule 3: Distribution of
X
[CLT]:
X
X
X ~ N , n
IF n is large
Make sure to check that
ALL conditions are satisfied
before you use any of these rules.
pf3

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Download Distribution of Sample Statistics - Notes | STA 2023 and more Study notes Statistics in PDF only on Docsity!

DISTRIBUTION OF

SOME SAMPLE STATISTICS

Rule 1: Sampling Distribution of

p

X p( p )

p ~ N p,

n n

IF

X ~ B(n, p)

and n  p ≥ 10

and n  (1–p) ≥ 10p) ≥ 10

and N ≥ 10n

Rule 2: Distribution of

X

[Normal Theory]:

X

X

X ~ N ,

n

IF X ~ N(μ

X

X

Rule 3: Distribution of

X

[CLT]:

X

X

X ~ N ,

n

 IF n is large

Make sure to check that

ALL conditions are satisfied

before you use any of these rules.

Example on page 156 of notes:

Machines fill bottles of cola. The bottle label says each one contains 295 ml, but there will be

variability in the contents (it is a random variable). Suppose we know that the distribution of the

contents of these bottles (X) is approximately Normal with mean 298 ml, standard deviation 3

ml, i.e., X ~ N(298, 3).

a) Find the probability that one randomly selected bottle contains less than 295 ml.

Since we are given that X ~ N(298, 3),

X

X

X

P( X ) P P( Z. ).

. Thus,

there is a 15.15% chance that a randomly selected bottle will contain less than 295 ml of coke.

b) Find the probability that the average content, of a bottle in a randomly selected six–

pack , is less than 295 ml.

Since X ~ N(298, 3),

X ~ N , , i .e., X ~ N ( ,. )

by rule 2. Hence

X

X

X

P( X ) P P( Z. )..

That is,

0.71% of all samples of size six will have average contents below 295 ml.

c) Between what two values would be the central 95% of the means of samples of size 6?

We are asked to find two numbers, x 1

and x 2

such that

1 2

P( xXx ) 0 9500.

. In part (b)

we have established that

X ~ N ( 298 1 225 ,. )

. Also, from the tables of the standard normal

distribution we find that

P( Z  1 96. ) 0 9750.

,

P( Z   1 96. ) 0 0250.

and hence

P(  1 96.  Z  1 96. ) 0 9500.

. Using all of these we get

1 2

1 2

1 2

X

X

. P( x X x )

x X x

P

x x

P Z

P(. Z. )

So,

1

x

  , which gives x 1

=298 + (–1.96)(1.225)= 295.599=295.6. Similarly,

2

x

yields x 2

= 300.4. That is, 95% of all samples of size 6 from this population

will have sample means between 295.6 and 300.4.