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.)

€

X

= μ

€

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:

€

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

€

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

€

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*)

€

Z = X −

x*

z*

Finding probabilities

Prob (height of randomly selected college woman ≤ 62 )?

€

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

€

X ~ normal X =10 , X =108 = .54

⎛

⎝ ⎜

⎞

⎠ ⎟

€

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)

€

X ~ distn with , n

⎛

⎝ ⎜

⎞

⎠ ⎟

€

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

€

X∑ ~ normal 7 =,100, 7 =59( )

€

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)?

€

P X =k( ) =n!

k! n−k( )! k 1−( )

n−kfor k = 0, 1, 2, …, n

€

P X =( ) =!

! −( )!. 1−.( )

−

=(.) (.8)1 =0.096

You won’t need to use this!!

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?

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)

€

=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.

€

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)

€

=E X( ) =0(.5) =15

= 0(.5) .5( ) =.74

Bell-shaped & approx normal.

Opinion poll

n = 500 adults; 240 agreed with statement

€

=E X( ) =500(.5) =50

= 100(.5) .5( ) =11.

X is approx normal with

€

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

€

p = Xn

Distn of sample proportion

X ~ binomial (n , )

€

X =n

X = n 1−( )

€

p = Xn

€

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.

€

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

€

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