Week 10

Comparing Two Means or Proportions

Generalising from sample

Individuals Measurement Groups Question

Children aged 10

Mark in maths test

Boys & girls

Are male marks higher on average?

Plots in field Yield of wheat Varieties A & B

Which gives higher yields?

Cars leaving production line

CO emissions from exhaust

Production lines 1 & 2

Are both lines same?

Generalising from sample

Individuals Measurement Groups Question

Children aged 10

Pass/fail in maths test

Boys & girls Are males more likely to pass?

Cabbages in field

Infected by cabbage butterfly

Varieties A & B

Which is less likely to be infected?

Cars leaving production line

Rattle in exhaust

Production lines 1 & 2

Do both lines have same chance of rattle?

Numerical measurements: means

Difference in average weight loss for those who diet compared to those who exercise to lose weight?

Difference is there between the mean foot lengths of men and women?

Population parameter 2 – 1 = difference between population means

Sample estimate x2 – x1 = difference between sample means

Categorical measurements: propns

Difference between the proportions that would quit smoking if taking the antidepressant buproprion (Zyban) versus wearing a nicotine patch?

Difference between proportion who have heart disease of men who snore and men who don’t snore?

Population parameter 2 – 1 = difference between population

proportions

Sample estimate p2 – p1 = difference between sample proportions

Requirement: independent samples

Random samples taken separately from 2 populations

Randomised experiment with 2 treatments

One random sample, but a categorical variable splits individuals into 2 groups.

Two samples are called independent samples when the measurements in one sample are not related to the measurements in the other sample.

Model for numerical data

Sample 1 ~ population (mean 1, s.d. 1)Sample 2 ~ population (mean 2, s.d. 2)

Estimation: estimate (2 – 1) with Standard error? Confidence interval?

Testing: is (2 – 1) zero? p-value

€

x2 − x1( )

Model for categorical data

Sample 1 ~ population (proportion 1)Sample 2 ~ population (proportion 2)

Estimation: estimate (2 – 1) with (p2 – p1) Standard error? Confidence interval?

Testing: is (2 – 1) zero? p-value

Distribution of difference

In both cases, we need to find distribution of difference (p2 – p1) or

Independent samples >> difference of independent random variables.

We already know distns of the two parts — what is distn of their difference?

€

x2 − x1( )

Sum of 2 variables

Sample mean:

Sample total:

€

X1 + X22

~ distn, 2

⎛

⎝ ⎜

⎞

⎠ ⎟

€

X1 + X2 ~ distn2, 2( )

Same distns

Different distns

€

X1 + X2 ~ distn1 +2 , 12 +2

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

Difference between 2 variables

Same standard devn as sum

€

X2 − X1 ~ distn2 −1, 12 +2

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

If X1 and X2 are normal

€

X2 − X1 ~ normal2 −1, 12 +2

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

Remember that X1 and X2 must be independent

Example

Husband height ~ normal(1.85, 0.1)Wife height ~ normal(1.7, 0.08)

Assume independent. (Probably not!!)

Prob that wife is taller than husband?

(Husband - Wife) ~

€

normal0.15, 0.12 + 0.082 ⎛ ⎝ ⎜

⎞ ⎠ ⎟ = normal0.15, 0.1281( )

Example

Husband height ~ normal(1.85, 0.1)Wife height ~ normal(1.7, 0.08)

Husband - Wife ~ normal(0.15, 0.1281)

P (diff ≤ 0) = area

0.15 0.28 0.410.02-0.11

€

z = −0.150.1281

= −0.534 Prob = 0.297

Difference between proportions

If X1 and X2 are independent,

€

X2 − X1 ~ distn2 −1, 12 +2

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

If p1 and p2 are independent,

€

p2 − p1 ~ distn2 −1, 1 1−1( )

n1+2 1−2( )

n2

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

For large samples, p1 and p2 are approx normal, so their difference is too.

n1 = n2 = 244 randomly assigned to each treatment

Std error for difference in propns

Nicotine patches vs Antidepressant (Zyban)?

Zyban: 85 out of 244 quit smokingPatch: 52 out of 244 quit smoking

€

s.e.(p1−p2 ) =p1 1−p1( )

n1+p2 1−p2( )

n2

So,

€

p1−p2 =.348 −.213 =.135

€

and s.e.(p1−p2 ) =.348 1−.348( )

244+.213 1−.213( )

244=.040

Approximate 95% C.I.

Best you can do for difference between proportions

For means, CI can be improved by replacing ‘2’ by a different value.

For sufficiently large samples, the interval

Estimate 2 Standard error

is an approximate 95% C.I.

Patch vs Antidepressant

Approx 95% C.I. .135 2(.040) => .135 .080 => .055 to .215

Study: n1 = n2 = 244 randomly assigned to each group

Zyban:85 of the 244 Zyban users quit smoking = .348

Patch: 52 of the 244 patch users quit smoking = .2131p̂2p̂

So, 135.213.348.ˆˆ 21 =−=−pp 040.)ˆˆ.(. and 21 =−ppes

We are 95% confident that Zyban gives an improvement of between 5.5% and 21.5% of the probability of quitting smoking.

Difference between means

If X1 and X2 are independent,

€

X2 − X1 ~ distn2 −1, 12 +2

2 ⎛ ⎝ ⎜

⎞ ⎠ ⎟

If X1 and X2 are independent,

€

X2 − X1 ~ distn2 −1, 12

n1+22

n2

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

If both populations are normal, so is the difference.

n1 = 42 men on diet, n2 = 27 men on exercise routine

Std error for difference in means

Lose More Weight by Diet or Exercise?

2

22

1

21

21 ).(.n

s

n

sxxes +=−

Diet: Lost an average of 7.2 kg with std dev of 3.7 kgExercise: Lost an average of 4.0 kg with std dev of 3.9 kg

So, kg 2.30.42.721 =−=−xx

( ) ( )81.0

47

9.3

42

7.3).(. and

22

21 =+=− xxes

We are 95% confident that those who diet lose on average 1.58 to 4.82 kg more than those who exercised.

Approximate 95% Confidence Interval: 3.2 2(.81) => 3.2 1.62 => 1.58 to 4.82 kg

Study: n1 = 42 men on diet, n2 = 27 men exercise

Diet: Lost an average of 7.2 kg with std dev of 3.7 kgExercise: Lost an average of 4.0 kg with std dev of 3.9 kg

So, kg 2.30.42.721 =−=−xx kg 81.0).(. and 21 =−xxes

Diet vs Exercise

A CI for the Difference Between Two Means(Independent Samples):

where t* is a value from t-tables.

2

22

1

21*

21 n

s

n

stxx +±−

Better C.I. for mean

d.f. = min(n1–1, n2–1) Welch’s approx gives a different d.f. (higher)

but is a complicated formula t* is approx 1.96 if d.f. is high

Randomized experiment: Researchers either stared or did not stare at drivers stopped at a campus stop sign; Timed how long (sec) it took driver to proceed from sign to a mark on other side of the intersection.

Estimate difference between the mean crossing times.

No Stare Group (n = 14): 8.3, 5.5, 6.0, 8.1, 8.8, 7.5, 7.8, 7.1, 5.7, 6.5, 4.7, 6.9, 5.2, 4.7

Stare Group (n = 13): 5.6, 5.0, 5.7, 6.3, 6.5, 5.8, 4.5, 6.1, 4.8, 4.9, 4.5, 7.2, 5.8

Effect of a stare on driving

No outliers; no strong skewness.

Crossing times in stare group seem faster & less variable.

Checking data

A 95% CI for 2–1 is

Effect of stare on driving

Using df = min(n1–1, n2–1) = 12, gives t* = 2.179

€

x2 − x1 = 6.63 −5.59 = 1.04 sec

€

s.e.(x 2 − x 1) =1.36( )

2

14+

0.822( )2

13= 0.43

€

1.04 ± 2.179 × 0.43 = 0.10 to 1.98 sec

Slightly narrower C.I. that we got with d.f. = 12.

N.B. C.I. is based on df = 21 (Welch’s approx)

Effect of stare on driving

Minitab

Interpretation

We are 95% confident that it takes drivers between 0.17 and 1.91 seconds less on average to cross intersection if someone stares at them.

A 95% CI for 2–1 is 0.17 to 1.91 sec

Testing two proportions

Hypotheses

H0: 1 – 2 = 0 HA: 1 – 2 ≠ 0

or 1 – 2 < 0or 1 – 2 > 0

Watch how Population 1 and 2 are defined. Data requirements

Independent samples n1 p1, n1(1-p1), n2 p2, n2(1-p2) all at least 5, preferably ≥10

Test statistic

Based on p1 – p2

Standardise:

€

z = p1 − p2( ) − 0

se p1 − p2( )

€

se p1 − p2( ) = π1 1− π1( )

n1

+π 2 1− π 2( )

n2

=1

n1

+1

n2

⎛

⎝ ⎜

⎞

⎠ ⎟π 1− π( ) if H0 is true

Test statistic

If H0 is true, best estimate of is

€

z = p1 − p2( ) − 0

1

n1

+1

n2

⎛

⎝ ⎜

⎞

⎠ ⎟p 1− p( )

€

p = x1 + x2( )n1 + n2( )

So we use test statistic

If H0 is true, this has standard normal distn p-value from normal distn

Prevention of Ear Infections

Does the use of sweetener xylitol reduce the incidence of ear infections?

Randomized Experiment:Of 165 children on placebo, 68 got ear infection.Of 159 children on xylitol, 46 got ear infection.

Hypotheses: H0: 1 – 2 = 0 Ha: 1 – 2 > 0 Data check: At least 5 success & failure in each group

123.ˆˆ and ,289.159

46ˆ ,412.

165

68ˆ 2121 =−==== pppp

Prevention of Ear Infections

Overall propn getting infection

35.324

114

159165

4668ˆˆˆ

21

2211 ==++

=++

=nnpnpn

p

( ) ( )32.2

1591

1651

35.135.

0123.

11ˆ1ˆ

0ˆˆ

21

21 =

⎟⎠⎞

⎜⎝⎛ +−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−−=

nnpp

ppz

Test statistic

p-value = 0.01

Conclusion: Strong evidence xylitol reduces

chance of ear infection

Testing two means

Hypotheses

H0: 1 – 2 = 0 HA: 1 – 2 ≠ 0

or 1 – 2 < 0or 1 – 2 > 0

Watch how Population 1 and 2 are defined. Data requirements

Fairly large n1 and n2 (say 30 or more), or Not much skewness & no outliers (normal model reasonable)

Test statistic

Based on

Standardise:

€

t = x 1 − x 2s1

2

n1

+s2

2

n2€

se x 1 − x 2( ) = σ 1

2

n1

+σ 2

2

n2

≈s1

2

n1

+s2

2

n2

€

x 1 − x 2

€

t = x 1 − x 2( ) − 0

se x 1 − x 2( )

Test

Test statistic:

If H0 is true, this has approx t-distn with

d.f. = min(n1–1, n2–1) Same d.f. as CI for 1 – 2

p-value from t distn Minitab or Excel

€

t = x 1 − x 2s1

2

n1

+s2

2

n2

n1 and n2 ≥ 30 Use normal tables

Randomized experiment: Researchers either stared or did not stare at drivers stopped at a campus stop sign; Timed how long (sec) it took driver to proceed from sign to a mark on other side of the intersection.

Test whether stare speeds up crossing times.

No Stare Group (n = 14): 8.3, 5.5, 6.0, 8.1, 8.8, 7.5, 7.8, 7.1, 5.7, 6.5, 4.7, 6.9, 5.2, 4.7

Stare Group (n = 13): 5.6, 5.0, 5.7, 6.3, 6.5, 5.8, 4.5, 6.1, 4.8, 4.9, 4.5, 7.2, 5.8

Effect of a stare on driving

Small sample sizes, but

No outliers; no strong skewness.

Checking data

Effect of stare on driving

€

x 1 − x 2 = 6.63 − 5.59 = 1.04 sec

HypothesesH0: 1 – 2 = 0 HA: 1 – 2 > 0

where 1 = no-stare, 2 = stare

€

t = x 1 − x 2

se x 1 − x 2( ) =

1.04

0.429 = 2.42

€

s.e.(x 2 − x 1) =1.36( )

2

14+

0.822( )2

13= 0.429

Effect of stare on driving

Test statistic

df = min(n1–1, n2–1) = 12

Upper tail area of t-distn (12 d.f.)

p = 0.016

P-value

€

t = x 1 − x 2

se x 1 − x 2( ) =

1.04

0.429 = 2.42

Strong evidence that stare speeds up crossing

Very similar p-value and same conclusion

N.B. Test is based on df = 21 (Welch’s approx)

Effect of stare on driving

Minitab

Strong evidence that stare speeds up crossing

Paired data and 2-sample data

Make sure you distinguish between: 2 measurements on each individual (e.g. before &

after)

Measurements from 2 independent groups

Different cars assessed for insurance claims in garages A and B

Same cars assessed by both garages

2 independent samples

Paired data