Ch. 18 - Sampling Distributions
Expanded def’n: A parameter is: - a numerical value describing some aspect of a pop’n
- usually regarded as constant
- usually unknown
A statistic is: - a numerical value describing some aspect of a sample
- regarded as random before sample is selected
- observed after sample is selected
The observed value depends on the particular sample selected from the population;
typically, it varies from sample to sample. This variability is called sampling variability.
The distribution of all the values of a statistic is called its sampling distribution.
Def’n: p = proportion of ppl with a specific characteristic in a random sample of size n
p = population proportion of ppl with a specific characteristic
The estimate of the standard deviation of a sampling distribution is called a
General Properties of the Sampling Distribution of p ˆ:
Let pnd p be as above. Also, µ apd σ arepthe mean and standard deviation for the
distribution of p. Then the following rules hold:
Rule 1: µ p = p. (Textusesk µ(p) )
p(1− p pq
Rule 2:σ p= = . (standardor Æ σ )p
Ex18.1) Suppose the population proportion is 0.5.
a) What is the standard deviation of pfor a sample size of 4?
p(1 − p) 0.5(1 0.5)
σ ˆp= = = 0.25
b) How large must n (sample size) be so that the sample proportion has a standard
deviation of at most 0.125?
n = Æ n = p (1p ) = 0.5(1 0.5) =16
σ p σ p 0.125 2
Rule 3: When n is large and p is not too near 0 or 1, the sampling distribution of pis
approximately normal. The farther from p = 0.5, the larger n must be for accurate normal
approximation of pˆ. Thus, if np and n(1 – p) are both sufficiently large (≥ 15), then it is
safe to use a normal approximation.
Further assumptions: the sample should always be random and, if sampling without
replacement, the sample should be less than 10% of the population. Using all 3 rules, the distribution of papproximately normal.
Z N )1,~▯(
Ex18.2) Suppose that the true proportion of people who have heard of Sidney Crosby is
0.87 and that a new sample consists of 158 people.
a) Find the mean and standard deviation of p.
p(1− p) 0.87(1− 0.87)
µp= 0.870 σ p = = = 0.0268
b) What can you say about the distribution of p?
np = 158(0.87) = 137.46 n(1 – p) = 158(1 – 0.87) = 20.54
Since both values are > 15, the distribution ofshould be well approximated by a
c) What is the probability of getting a sample proportion greater than 0.94?
P( p > 0.94) Æ P ⎜ p − p > 0.94 −.87 ⎟ = P(Z > 2.62)
⎜ p(1− p) 0.0268 ⎟
⎝ n ⎠
= 1 – P(Z < 2.62) = 1 – 0.9956 = 0.0044
Sampling Distribution of Mean
How does the sampling distribution of the sample mean compare with the distribution of
a single observation (which comes from a population)?
Ex18.3) An epically gigantic jar contains a large number of balls, each labeled 1, 2, or 3,
with the same proportion for each value.
Let Y be the label on a randomly selected ball. Find µand σ .