7.1 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 standard deviation of a sampling distribution is called a standard error
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 ˆp= p.
p (1− p pq
Rule 2:σ ˆp = . (a.k.a. standard error, see above)
Ex7.1) Suppose the population proportion is 0.5.
a) What is the standard deviation of p for 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?
p(1− p) p(1 p ) 0.5(1 0.5)
n = Æ n = 2 = 2 =16
σ pˆ σ p 0.125
Rule 3: When n is large and p is not too near 0 or 1, the sampling distribution of p
approximately normal. The farther from p = 0.5, the larger n must be for accurate normal
approximation of p . Conservatively speaking, if np and n(1 – p) are both sufficiently
large (e.g. ≥ 15), then it is safe to use a normal approximation. Using all 3 rules, the distribution of
p − p
Z = p(1− p)
is well approximated by the standard normal distribution.
Ex7.2) 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)
µ p 0.870 σ pˆ= = = 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 op should be well approximated by a
c) Find P( p> 0.94). Note: not a z-score!
⎜ p − p 0.94 −.87 ⎟
P( p > 0.94) ÆP ⎜ > ⎟ = P(Z > 2.64)
⎜ p(1− p) 0.0268 ⎟
⎝ n ⎠
= 1 – P(Z < 2.64) = 1 – 0.9959 = 0.0041
7.2 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)?
Ex7.3) An epically gigantic jar contains a large number of balls, each labeled 1, 2, or 3,
with the same proportion for each value.
Let X be the label on a randomly selected ball. Find µ and σ.
µ = ∑ xiP(X = x i = 1(1/3) + 2(1/3) + 3(1/3) = 2
2 2 2 2 2
σ = ∑ (x − µ