Lecture 10:
Depicting Sampling Distributions
of a Sample Proportion
Chapter 5: Probability and
Sampling Distributions
2/10/12
Lecture 10
1
Sample Proportion
•  “1” is assigned to population members having
a specified characteristic and “0” is assigned
to those who don’t. The parameter of interest
in this situation is p (or called π), the
proportion of the population that has the
characteristic of interest.
•  Denote p̂ as the proportion of members
having such characteristic (or say, successes),
also called the sample proportion, in a
random sample of size n.
2/10/12
Lecture 10
2
Sampling Distribution of Sample Proportion
•  If X ~ B(n, p), the sample proportion is defined as
X count of successes in sample
pˆ = =
.
n
size of sample
•  Mean & variance of a sample proportion:
µ pˆ = p,
2/10/12
σ pˆ = p(1 − p) / n .
Lecture 10
3
2/10/12
Lecture 10
4
Example: Clinton's vote
•  43% of the population voted for Clinton in
1992.
•  Suppose we survey a sample of size 2300 and
see if they voted for Clinton or not in 1992.
•  We are interested in the sampling
distribution of the sample proportion p̂ , for
samples of size 2300.
•  What's the mean and variance of p̂ ?
2/10/12
Lecture 10
5
Count & Proportion of “Success”
•  A BoilerMaker basketball player is a 95% freethrow shooter.
•  Suppose he will shoot 5 free-throws during each
practice.
•  X: number of free-throws he makes during
practice.
•  p̂ : proportion of made-free-throws during
practice.
•  P( p̂ =0.6) = ?
•  P(X=3) = ?
2/10/12
Lecture 10
6
Normal Approximation for Counts and
Proportions
•  Let X ~ B(n, p) and pˆ = X / n is the sample
proportion.
•  If n is large*, then
X is approx. N (np , np (1-p) )
pˆ is approx. N ( p,
p (1-p) / n ).
•  *Rule of Thumb: np ≥ 5, n(1 - p) ≥ 5.
2/10/12
Lecture 10
7
Switches Inspection
•  A quality engineer selects an SRS of 100
switches from a large shipment for detailed
inspection.
•  Unknown to the engineer, 10% of the
switches in the shipment fail to meet the
specifications.
•  What is the probability that at most 9
switches fail the standard test in the sample?
•  Use Normal Approximation here. Try to use
continuity correction as well. (Similar to ex. 43 in
Hw4)
2/10/12
Lecture 10
8
Switches Inspection
•  Express the probability in terms of X: P(X ≤ 9)
•  The normal approximation to the probability of no
more than 9 bad switches is the area to the left of X
= 9 under the normal curve, which has
µ X = np = (100)(.1) = 10, σ X = np(1 − p) = 100(.1)(.9) = 3.
•  In this case np = 10, and n(1-p) = 90, both
satisfying the condition for “rule of thumb”.
•  Also, don’t forget the CONTINUITY CORRECTION
when applying normal approximation to the
Binomial distribution.
2/10/12
Lecture 10
9
Continuity Correction
•  The normal approximation is more accurate if
we consider X = 9 to extend from 8.5 to 9.5
•  Example (Cont.):
X − 10 9.5 − 10
P ( X ≤ 9) = P ( X ≤ 9.5) = P (
≤
)
3
3
≈ P( Z ≤ −.17) = .4325.
2/10/12
Lecture 10
10
Continuity Correction
P(X ≤ 8) replaced by P(X < 8.5)
P(X ≥ 14) replaced by P(X > 13.5)
P(X < 8) = P(X<=7), then replaced by P(X
< 7.5)
For large n the effects of the
continuity correction factor is
very small and will be omitted.
2/10/12
Lecture 10
11
Coin Tossing Example
•  Toss a fair coin 200 times, what is the probability that
the total number of heads is between 90 and 110? Use
the normal approximation and continuity correction
for Binomial distribution to solve the problem.
•  X= the total number of heads
•  X ~ Bin(200, 0.5). Want: P(90 ≤ X ≤ 110).
•  µX =200 × .5 = 100, σX = (200× .5× .5)1/2 = 7.07.
•  With continuity correction:
P(90 ≤ X ≤ 110) = P(89.5 ≤ X ≤ 110.5)
≈ P( Z ≤ (110.5 - 100)/7.07) - P (Z ≤ (89 .5 - 100)/7.07)
= P (Z ≤ 1.48) - P (Z ≤ -1.48)= .8611.
2/10/12
Lecture 10
12
Example
Note: If not required to use Binomial knowledge, we can use
the sampling distribution for sample proportion as well.
•  The Laurier company’s brand has a market share of 30%.
In a survey 1000 consumers were asked which brand they
prefer. What is the probability that more than 32% of the
respondents say they prefer the Laurier brand?
•  Solution: The number of respondents who prefer Laurier is
binomial with n = 1000 and p = .30. Also, np = 1000(.3)
>5, n(1-p) = 1000(1-.3) > 5. Therefore we can use normal
approximation to sample proportion here:
⎛
⎞
ˆ
p
−
p
.
32
−
.
30
⎟ = P( Z > 1.38) = .0838.
P( pˆ > .32) = P⎜
>
⎜ p(1 − p) n .01449 ⎟
⎝
⎠
2/10/12
Lecture 10
13
Summary
•  Sampling Distribution
•  Sampling distribution for a sample mean
–  Mean and Variance/S.D. for a sample mean
–  Central Limit Theorem
–  Normal calculation
•  Sampling distribution for a sample
proportion
–  Continuity correction for Binomial Distribution
–  Rule of thumb, when n is large
2/10/12
Lecture 10
14
After Class…
•  Review Sec 5.4 through 5.6 .
•  Start Hw 4 – due by next Monday, Feb 14th
•  Review all the notes up to today, go over the practice test.
Come to Office Hours for Hw#1-#4 grading issues.
•  About Exam 1.
–  Start editing your own cheat-sheet
–  Practice Test is posted. Try to do it in an hour
2/10/12
Lecture 10
15