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Week 7 Sample Means & Proportions
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 Constant Greek letter (e.g. , , , …) Unknown value in practice
Summary statistic Random Roman letter (e.g. m, s, p, …)
We hope statistic will tell us about corresponding parameter
Distn of sample vsSampling 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
Sample mean, X, has sampling distn with: Mean,
St devn,
Population with mean , st devn
(We will deal later with the problem that and are unknown in practice.)
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X
= μ
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X
= σn
Random sample (n independent values)
Weight loss
Random sample of n = 25 people Sample mean, x
Estimate mean weight loss for those attending clinic for 10 weeks
How accurate?
Let’s see, if the population distn of weight loss is:
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X ~ normal μ =8lb, σ =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:
(How does this compare to simulation? To popn distn?)
mean = = 8 lb, s.d.( ) = lbx 125
5==
n
Errors in estimation
From 70-95-100 rule 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
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X ~ normal μ =8lb, σ =5lb ⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
mean = = 8 lb, s.d.( ) = lbx 125
5==
n
Population
Sampling distribution of mean
Increasing sample size, n
If we sample n = 100 people instead of 25:
s.d.( ) = lb.x 5.0100
5==
n
Larger samples more accurate estimates
Central Limit Theorem
If population is normal (, )
If popn is non-normal with (, ) but n is large
Guideline: n > 30 even if very non-normal
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X ~ normal, n
⎛
⎝ ⎜
⎞
⎠ ⎟
€
X approx ~ normal, n
⎛
⎝ ⎜
⎞
⎠ ⎟
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
Simulation: Generated 5 numbers (without replacement) 1560 times
Theory? Fairly hard.
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)
0 1 2 3-1-2-3
Prob ( Z < z* )
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 (, )
= 65 inches, = 2.7 inches
Heightsof students
Probability and area
X ~ normal ( = 65 , = 2.7 )
P (X ≤ 67.7) = area
Probability and area (cont.)
Exactly 70-95-100 rule
P(X within of ) = 0.683 approx 70% P(X within 2 of ) = 0.954 approx 95% P(X within 3 of ) = 0.997 approx 100%
Normal (, )
Finding approx probabilities
Prob (X ≤ 62 )?
About 1/8
Ht of college woman, X ~ normal ( = 65 , = 2.7 )
P (X ≤ 62) = area
1. Sketch normal density
2. Estimate area
Translate question from X to Z
Translate to z-score:
Z ~ Normal ( = 0, = 1)0 1 2 3-1-2-3
X ~ Normal (, ) Find P(X ≤ x*)
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Z = X −
x*
z*
Finding probabilities
Prob (height of randomly selected college woman ≤ 62 )?
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P X ≤6( ) =P Z≤6 −65.7
⎛
⎝ ⎜
⎞
⎠ ⎟
=P Z≤−1.11( ) =.15About 13%.
Prob (X > value)
( ) ( ) ( )
1335.8665.1
11.1111.17.2
658668
=−=
≤−=>=⎟⎠
⎞⎜⎝
⎛ −>=> ZPZPZPXP
Prob (X > 68 inches)?
Ht of college woman, X ~ normal ( = 65 , = 2.7 )
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
Step 2: Calculate x = z* + x = (0.67)(10) + 120 = 126.7 or about 127.
Closest z* with area of 0.7500 (tables)
z = 0.67
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) ?
prob = 0.0023
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X ~ normal X =10 , X =108 = .54
⎛
⎝ ⎜
⎞
⎠ ⎟
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z = 10 −10.54
= .8
Very little chance!
X ~ normal ( = 120, = 10) n = 8
Distribution of sum
X ~ distn with (, )
aX ~ distn with (a, a)
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X ~ distn with , n
⎛
⎝ ⎜
⎞
⎠ ⎟
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X = n∑ X ~ distn with n, n( )
Central Limit Theorem implies approx normal
e.g. milesto kilometers
Probabilities about sum
Profit in 1 day ~ normal ( = $300, = $200)
Prob(total profit in week < $1,000)?
Total =
Prob = 0.0188
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X∑ ~ normal 7 =,100, 7 =59( )
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z = 1000 −10059
= −.08 Assumesindependence
Categorical data
Most important parameter is = Prob (success)
Corresponding summary statistic is p = Proportion (success)
N.B. Textbook uses p and p̂
Number of successes
Easiest to deal with count of successes before proportion.
If…
1. n “trials” (fixed beforehand).
2. Only “success” or “failure” possible for each trial.
3. Outcomes are independent.
4. Prob (success), remains same for all trials, .• Prob (failure) is 1 – .
X = number of successes ~ binomial (n, )
Examples
Binomial Probabilities
Prob (win game) = 0.2
Plays of game are independent.
What is Prob (wins 2 out of 3 games)?
What is P(X = 2)?
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P X =k( ) =n!
k! n−k( )! k 1−( )
n−kfor k = 0, 1, 2, …, n
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P X =( ) =!
! −( )!. 1−.( )
−
=(.) (.8)1 =0.096
You won’t need to use this!!
Mean & st devn of Binomial
For a binomial (n, )
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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)
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Mean =E X( ) =100 (.5) =50
Standard deviation = 100 (.5) .5( ) =5
Extraterrestrial Life?
70-95-100 rule of thumb for # “yes” About 95% chance of between 40 & 60 Almost certainly between 35 & 65
Sample of n = 100
X = # “yes” ~ binomial (n = 100, = 0.5)
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=E X( ) =100(.5) =50
= 100(.5) .5( ) =5
Normal approx to binomial
If X is binomial (n , ), and n is large, then X is also approximately normal, with
Conditions: Both n and n(1 – ) are at least 10.
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Mean =E X( ) =n
Standard deviation = n 1−( )
(Justified by Central Limit Theorem)
Number of H in 30 Flips
X = # heads in n = 30 flips of fair coinX ~ binomial ( n = 30, = 0.5)
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=E X( ) =0(.5) =15
= 0(.5) .5( ) =.74
Bell-shaped & approx normal.
Opinion poll
n = 500 adults; 240 agreed with statement
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=E X( ) =500(.5) =50
= 100(.5) .5( ) =11.
X is approx normal with
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P X ≤40( ) ≈P Z ≤40 −5011.
⎛
⎝ ⎜
⎞
⎠ ⎟=P Z ≤−.89( ) =.1867
Not unlikely to see 48% or less, even if 50% in population agree.
If = 0.5 of all adults agree, what P(X ≤ 240) ?
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
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p = Xn
Distn of sample proportion
X ~ binomial (n , )
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X =n
X = n 1−( )
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p = Xn
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p =
p = 1−( )
n
Large n:p is approx normal
(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 pollSuppose 40% of all voters favor Candidate A.
Pollsters sample n = 2400 voters.
Simulation 400 times & theory.
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p = = 0.4
p = 1−( )
n = 0.4 ×0.6
400 = 0.01
Propn voting for A is approx normal
Probability from normal approx
If 40% of voters favor Candidate A, and n = 2400 sampled
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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
