Download Week 7 - Massey University

Week 7
Sample Means & Proportions
Variability of Summary Statistics

Variability in shape of distn of sample

Variability in summary statistics


Mean, median, st devn, upper quartile, …
Summary statistics have distributions
Parameters and statistics

Parameter describes underlying population




Summary statistic



Constant
Greek letter (e.g. , , , …)
Unknown value in practice
Random
Roman letter (e.g. m, s, p, …)
We hope statistic will tell us about corresponding
parameter
Distn of sample vs
Sampling distn of statistic

Values in a single random sample have a
distribution

Single sample --> single value for statistic

Sample-to-sample variability of statistic is its
sampling distribution.
Means




Unknown population mean, 
Sample mean, X, has a distribution — its
sampling distribution.
Usually x ≠ 
A single sample mean, x, gives us information
about 
Sampling distribution of mean
If sample size, n, increases:

Spread of distn of sample is (approx) same.

Spread of sampling distn of mean gets
smaller.


x is likely to be closer to 
x becomes a better estimate of 
Sampling distribution of mean
Population with mean , st devn 
Random sample (n independent values)
Sample mean, X, has sampling distn with:

 Mean, 
X


St devn,

 
X
n
(We will deal
later with the problem that  and 

are unknown in practice.)
Weight loss
Estimate mean weight loss for those attending
clinic for 10 weeks


Random sample of n = 25 people
Sample mean, x
How accurate?
Let’s see, if the population distn of weight loss is:
X ~ normal  8lb,  5lb
Some samples
Four random samples of n = 25 people:
1. Mean = 8.32 pounds, st devn = 4.74 pounds
2. Mean = 8.32 pounds, st devn = 4.74 pounds
3. Mean = 8.48 pounds, st devn = 5.27 pounds
4. Mean = 7.16 pounds, st devn = 5.93 pounds
N.B. In all samples, x ≠ 
Sampling distribution
Means from simulation
of 400 samples
Theory:
mean =  = 8 lb,
s.d.( x ) =

n

5
 1 lb
25
(How does this compare to simulation? To popn distn?)
Errors in estimation
Population
X ~ normal  8lb,  5lb
Sampling
distribution of mean

mean =  = 8 lb,


n

5
 1 lb
25
From 70-95-100 rule



s.d.( x ) =
x will be almost certainly within 8 ± 3 lb
x is unlikely to be more than 3 lb in error
Even if we didn’t know 

x is unlikely to be more than 3 lb in error
Increasing sample size, n
If we sample n = 100 people instead of 25:
s.d.( x ) =

5

 0.5 lb.
n
100
Larger samples 
more accurate estimates
Central Limit Theorem

If population is normal (, )
  
X ~ normal,


n 

If popn is non-normal with (, ) but n is large

  
X approx ~ normal,


n 
Guideline: n > 30 even if very non-normal

Other summary statistics
E.g. Lower quartile, proportion, correlation

Usually not normal distns

Formula for standard devn of samling distn
sometimes

Sampling distn usually close to normal if n is
large
Lottery problem
Pennsylvania Cash 5 lottery
 5 numbers selected from 1-39
 Pick birthdays of family members (none 32-39)
 P(highest selected is 32 or over)?
Statistic:
H = highest of 5 random numbers (without
replacement)
Lottery simulation
Theory? Fairly hard.
Simulation: Generated 5 numbers (without
replacement) 1560 times
Highest number > 31 in about 72% of repetitions
Normal distributions



Family of distributions (populations)
Shape depends only on parameters
 (mean) &  (st devn)
All have same symmetric ‘bell shape’
= 65 inches,
 = 2.7 inches
Importance of normal distn

A reasonable model for many data sets

Transformed data often approx normal

Sample means (and many other statistics)
are approx normal.
Standard normal distribution

Z ~ Normal ( = 0,  = 1)
-3

Prob ( Z < z* )
-2
-1
0
1
2
3
Probabilities for normal (0, 1)
Check from tables:
P(Z  -3.00)
P(Z  −2.59)
P(Z  1.31)
P(Z  2.00)
P(Z  -4.75)
=
=
=
=
=
0.0013
0 .0048
0 .9049
0 .9772
0 .000001
Probability Z > 1.31
P(Z > 1.31) = 1 – P(Z  1.31)
= 1 – .9049 = .0951
Prob ( Z between –2.59 and 1.31)
P(-2.59  Z  1.31)
= P(Z  1.31) – P(Z  -2.59)
= .9049 – .0048 = .9001
Standard devns from mean

Normal (, )





Heights
of students
 = 65 inches,
 = 2.7 inches
Probability and area
X ~ normal ( = 65 ,  = 2.7 )
P (X ≤ 67.7) = area
Probability and area (cont.)

Normal (, )




Exactly
 P(X within  of ) = 0.683
 P(X within 2 of ) = 0.954
 P(X within 3 of ) = 0.997
70-95-100 rule
approx 70%
approx 95%
approx 100%
Finding approx probabilities
Ht of college woman, X ~ normal ( = 65 ,  = 2.7 )
Prob (X ≤ 62 )?
1. Sketch normal density
2. Estimate area
P (X ≤ 62) = area
About 1/8
Translate question from X to Z


X ~ Normal (, )
Find P(X ≤ x*)



x*

Translate to z-score:
X 

Z 

Z ~ Normal ( = 0,  = 1)

-3
-2
z*-1
0
1
2
3
Finding probabilities
Prob (height of randomly selected college woman ≤ 62 )?

62  65 
P X  62  P Z 

2.7


 P Z  1.11  .1335
About 13%.
Prob (X > value)
Ht of college woman, X ~ normal ( = 65 ,  = 2.7 )
Prob (X > 68 inches)?
68  65 

P X  68  P Z 
  PZ  1.11  1  PZ  1.11
2.7 

 1  .8665  .1335
Finding upper quartile
Blood Pressures are normal with mean 120 and
standard deviation 10. What is the 75th percentile?
Step 1: Solve for z-score
Closest z* with area of 0.7500 (tables)
z = 0.67
Step 2: Calculate x = z* + 
x = (0.67)(10) + 120 = 126.7 or about 127.
Probabilities about means

Blood pressure ~ normal ( = 120,  = 10)

8 people given drug

If drug does not affect blood pressure,
 Find P(average blood pressure > 130)
P ( X > 130) ?







X ~ normal ( = 120,  = 10)
n=8


10
X ~ normal  X  120,  X 
 3.54


8
130 120
z 
 2.83
3.54
prob = 0.0023
Very little chance!
Distribution of sum
X ~ distn with (, )

e.g. miles
to kilometers
aX ~ distn with (a, a)
  
X ~ distn with ,


n 
X


 nX ~ distn with n,
n

 Central Limit Theorem implies approx normal
Probabilities about sum

Profit in 1 day ~ normal (= $300, = $200)

Prob(total profit in week < $1,000)?

Total =
X

~ normal 7  2,100,
1000  2100
z 
  2.08
529

 Prob = 0.0188


7  529
Assumes
independence
Categorical data

Most important parameter is
  = Prob (success)

Corresponding summary statistic is
 p = Proportion (success)
N.B. Textbook uses p and p^
Number of successes


Easiest to deal with count of successes
before proportion.
If…
1.
2.
3.
4.

n “trials” (fixed beforehand).
Only “success” or “failure” possible for each trial.
Outcomes are independent.
Prob (success), remains same for all trials, .
• Prob (failure) is 1 – .
X = number of successes ~ binomial (n, )
Examples
Binomial Probabilities
nk
n!
k
P X  k  
 1  
for k = 0, 1, 2, …, n
k! n  k !
You won’t need to use this!!
Prob (win game) = 0.2
Plays of game are independent.
What is Prob (wins 2 out of 3 games)?
What is P(X = 2)?
32
3!
2
P X  2 
.2 1 .2
2! 3  2!
 3(.2)2 (.8)1  0.096
Mean & st devn of Binomial
For a binomial (n, )
Mean
  n
Standard deviation   n 1  

Extraterrestrial Life?
50% of large population would say “yes” if asked,
“Do you believe there is extraterrestrial life?”
Sample of n = 100
X = # “yes” ~ binomial (n = 100,  = 0.5)
Mean
  E X   100(.5)  50
Standard deviation   100(.5).5  5
Extraterrestrial Life?
Sample of n = 100
X = # “yes” ~ binomial (n = 100,  = 0.5)
  E X   100(.5)  50
  100(.5).5  5
70-95-100 rule of thumb for # “yes”

 About 95% chance of between 40 & 60
 Almost certainly between 35 & 65
Normal approx to binomial
If X is binomial (n , ), and n is large, then X is also
approximately normal, with
Mean
  E X   n
Standard deviation   n 1  
Conditions: Both n and n(1 – ) are at least 10.

(Justified by Central Limit Theorem)
Number of H in 30 Flips
X = # heads in n = 30 flips of fair coin
X ~ binomial ( n = 30, = 0.5)
Bell-shaped & approx normal.
  E X   30(.5)  15
  30(.5).5  2.74
Opinion poll
n = 500 adults; 240 agreed with statement
If  = 0.5 of all adults agree, what P(X ≤ 240) ?
X is approx normal with
  E X   500(.5)  250
  100(.5).5  11.2

240  250
P X  240  P Z 
 P Z  .89  .1867
11.2 


Not unlikely to see 48% or less, even if 50% in population
agree.
Sample Proportion

Suppose (unknown to us) 40% of a population carry
the gene for a disease, ( = 0.40).

Random sample of 25 people; X = # with gene.

X ~ binomial (n = 25 ,  = 0.4)
p = proportion with gene
X
p 
n
Distn of sample proportion


X ~ binomial (n , )
 X  n
 X  n 1  
X
p 
n
p  

p 

Large n:
p is approx normal
 1  
n
(n ≥ 10 & n (1 – ) ≥ 10)
Examples

Election Polls: to estimate proportion who favor a
candidate; units = all voters.

Television Ratings: to estimate proportion of households
watching TV program; units = all households with TV.

Consumer Preferences: to estimate proportion of
consumers who prefer new recipe compared with old; units =
all consumers.

Testing ESP: to estimate probability a person can
successfully guess which of 5 symbols on a hidden card;
repeatable situation = a guess.
Public opinion poll
Suppose 40% of all voters favor Candidate A.
Pollsters sample n = 2400 voters.
Propn voting for A is approx normal
 p    0.4
p 
 1  
n
Simulation 400
times & theory.

0.4  0.6
 0.01
2400
Probability from normal approx
If 40% of voters favor Candidate A, and n = 2400 sampled
 p  0.4
 p  0.01
Sample proportion, p, is almost certain to be between 0.37
and 0.43

Prob 0.95 of p being between 0.38 and 0.42