Confidence Intervals and Hypothesis Tests for the Difference

between Two Population Means µ1 - µ2: Independent Samples

Confidence Intervals and Hypothesis Tests for the Difference

between Two Population Means µ1 - µ2: Independent Samples

Inference for m1 - m2

1

Confidence Intervals for the Difference between Two Population Means µ1 - µ2: Independent Samples

• Two random samples are drawn from the two populations of interest.

• Because we compare two population means, we use the statistic .

2

21 xx

3

Population 1 Population 2

Parameters: µ1 and 12 Parameters: µ2 and 2

2 (values are unknown) (values are unknown)

Sample size: n1 Sample size: n2

Statistics: x1 and s12 Statistics: x2 and s2

2

Estimate µ1 µ2 with x1 x2

2 21 2

1 21 2

( )s s

SE x xn n

s12

n1

s2

2

n2

df

m1-m2

x 1 x 2

Sampling distribution model for ? 1 2x x

22 21 2

1 22 22 2

1 2

1 1 2 2

1 11 1

s sn n

dfs s

n n n n

An estimate of the degrees of freedom is

min(n1 − 1, n2 − 1).

1 2 1 2

2 21 2

1 21 2

( )

( )

E x x

SD x xn n

Shape?

Estimate using

Two sample t-confidence interval

C

t*−t*

Practical use of t: t*

C is the area between −t* and

t*.

We find the value of t* in the

line of the t-table for the correct

df and the column for

confidence level C.

Confidence Interval for m1 – m2

6

*

*

2 21 2( )

1 21 2

where is the value from the t-table

that corresponds to the confidence level

df

df

Confidence interval

s sx x t

n n

t

22 21 2

1 22 22 2

1 2

1 1 2 2

1 11 1

s sn n

dfs s

n n n n

An estimate of the degrees of freedom is

min(n1 − 1, n2 − 1).

Hypothesis test for m1 – m2

7

H0: 1 – 2 = 0 ; Ha: 1 – 2 >0 (or <0, or ≠0)

Test statistic: 1 2

2 21 2

1 2

( ) 0x xt

s sn n

22 21 2

1 22 22 2

1 2

1 1 2 2

1 11 1

s sn n

dfs s

n n n n

An estimate of the degrees of freedom is

min(n1 − 1, n2 − 1).

Example: confidence interval for 1 – 2 using min(n1 –1, n2 -1) to approximate the df

• Example– Do people who eat high-fiber cereal for

breakfast consume, on average, fewer calories for lunch than people who do not eat high-fiber cereal for breakfast?

– A sample of 150 people was randomly drawn. Each person was identified as a consumer or a non-consumer of high-fiber cereal.

– For each person the number of calories consumed at lunch was recorded. 8

Example: confidence interval for m1 – m2

9

Consmers Non-cmrs568 705498 819589 706681 509540 613646 582636 601739 608539 787596 573607 428529 754637 741617 628633 537555 748

. .

. .

. .

. .

Solution:• The parameter to be tested is the difference between two means. • The claim to be tested is: The mean caloric intake of consumers (m1) is less than that of non-consumers (m2).22 2

1 2

1 22 22 2

1 2

1 1 2 2

122.61 1

1 1

s sn n

dfs s

n n n n

Let’s use df = min(43-1, 107-1) = min(42, 106) = 42;t42* = 2.0181

1 2

1 2

2 2

1 2

43 107

604.02 633.239

4103 10670

n n

x x

s s

Example: confidence interval for m1 – m2

• df = 42; t42* = 2.0181• The confidence interval estimator for the difference between

two means using the formula is

10

*42

2 21 2( )

1 21 2

4103 10670(604.02 633.239) 2.0181

43 107

29.21 28.19 57.40, 1.02

s sx x t

n n

Interpretation

• The 95% CI is (-57.40, -1.02).• Since the interval is entirely negative (that is,

does not contain 0), there is evidence from the data that µ1 is less than µ2. We estimate that non-consumers of high-fiber breakfast consume on average between 1.02 and 57.40 more calories for lunch.

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Beware!! Common Mistake !!!

A common mistake is to calculate a one-sample

confidence interval for m1, a one-sample confidence interval for

m2, and to then conclude that m1 and m2 are equal if the

confidence intervals overlap.

This is WRONG because the variability in the sampling

distribution for from two independent samples is more

complex and must take into account variability coming from both

samples. Hence the more complex formula for the standard error.

2

22

1

21

n

s

n

sSE

21 xx

INCORRECT Two single-sample 95% confidence intervals: The confidence interval for the male mean and the confidence interval for the female mean overlap,

suggesting no significant difference between the true mean for males and the true mean for females.

Male interval: (18.68, 20.12)Male Female

mean 19.4 17.9

st. dev. s 2.52 3.39

n 50 50

Female interval: (16.94, 18.86)

2 2* 1 2

1 2 .025,1 2

The 2-sample 95% confidence interval of the form

( ) for the difference between the means

is . Interval is entirely positive,

dfs sy y t n n

CORRECT

(.313, 2.69) suggesting signi

male female

between the true mean for males and the true mean for females

(evidence that true male mean is larger than true female mean).

ficant difference

0 1.5.313 2.69

Reason for Contradictory Result

14

2 21 2 1 2

1 2 1 2

1 2 1 2

It's always true that

. Specifically,

( ) ( ) ( )

a b a b

s s s s

n n n n

SE x x SE x SE x

Example: hypothesis test for m1 – m2

15

Consmers Non-cmrs568 705498 819589 706681 509540 613646 582636 601739 608539 787596 573607 428529 754637 741617 628633 537555 748

. .

. .

. .

. .

Solution:• The parameter to be tested is the difference between two means. • The claim to be tested is: The mean caloric intake of consumers (m1) is less than that of non-consumers (m2).

1 2

1 2

2 2

1 2

43 107

604.02 633.239

4103 10670

n n

x x

s s

Example: hypothesis test for m1 – m2

(cont.)

16

H0: 1 – 2 = 0 ; Ha: 1 – 2 < 0

Test statistic:

1 2

2 2

1 2

1 2

( ) 0 604.02 633.2392.09

4103 10670

43 107

x xt

s s

n n

Let’s use df = min(n1 − 1, n2 − 1) = min(43-1, 107-1) = min(42, 106) = 42

From t-table: for df=42,-2.4185 <t=-2.09 <-2.0181 .01 < P-value < .025

Conclusion: reject H0 and conclude high-fiber breakfast eaters consume fewer calories at lunch

Does smoking damage the lungs of children exposed

to parental smoking?

Forced vital capacity (FVC) is the volume (in milliliters) of

air that an individual can exhale in 6 seconds.

FVC was obtained for a sample of children not exposed to

parental smoking and a group of children exposed to

parental smoking.

We want to know whether parental smoking decreases

children’s lung capacity as measured by the FVC test.

Is the mean FVC lower in the population of children

exposed to parental smoking?

Parental smoking FVC s n

Yes 75.5 9.3 30

No 88.2 15.1 30

x

Parental smoking FVC s n

Yes 75.5 9.3 30

No 88.2 15.1 30

We are 95% confident that lung capacity is between 19.33 and 6.07 milliliters LESS in children of smoking parents.

x

95% confidence interval for (µ1 − µ2), with

df = min(30-1, 30-1) = 29 t* = 2.0452:2 21 2

1 21 2

2 2

( ) *

9.3 15.1(75.5 88.2) 2.0452

30 3012.7 2.0452*3.24

12.7 6.63 ( 19.33, 6.07)

s sx x t

n n

m1 = mean FVC of children with a smoking parent;

m2 = mean FVC of children without a smoking parent

Do left-handed people have a shorter life-expectancy than

right-handed people? Some psychologists believe that the stress of being left-

handed in a right-handed world leads to earlier deaths

among left-handers. Several studies have compared the life expectancies of

left-handers and right-handers. One such study resulted in the data shown in the table.

We will use the data to construct a confidence interval

for the difference in mean life expectancies for left-

handers and right-handers.

Is the mean life expectancy of left-handers less

than the mean life expectancy of right-handers?

Handedness Mean age at death s n

Left 66.8 25.3 99

Right 75.2 15.1 888

x

left-handed presidents

star left-handed quarterback Steve Young

We are 95% confident that the mean life expectancy for left-handers is between 3.26 and 13.54 years LESS than the mean life expectancy for right-handers.

95% confidence interval for (µ1 − µ2), with

df = min(99-1, 888-1) = 98 t* = 1.9845:2 21 2

1 21 2

2 2

( ) *

(25.3) (15.1)(66.8 75.2) 1.9845

99 8888.4 1.9845*2.59

8.4 5.14 ( 13.54, 3.26)

s sx x t

n n

m1 = mean life expectancy of left-handers;

m2 = mean life expectancy of right-handers

Handedness Mean age at death s n

Left 66.8 25.3 99

Right 75.2 15.1 888

The “Bambino”,left-handed Babe Ruth, baseball’s all-time best

player.

Matched pairs t proceduresSometimes we want to compare treatments or conditions at the

individual level. These situations produce two samples that are not

independent — they are related to each other. The members of one

sample are identical to, or matched (paired) with, the members of the

other sample.

– Example: Pre-test and post-test studies look at data collected on the

same sample elements before and after some experiment is performed.

– Example: Twin studies often try to sort out the influence of genetic

factors by comparing a variable between sets of twins.

– Example: Using people matched for age, sex, and education in social

studies allows canceling out the effect of these potential lurking

variables.

Matched pairs t procedures• The data:

– “before”: x11 x12 x13 … x1n

– “after”: x21 x22 x23 … x2n

• The data we deal with are the differences di of the paired values:

d1 = x11 – x21 d2 = x12 – x22 d3 = x13 – x23 … dn = x1n – x2n

• A confidence interval for matched pairs data is calculated just like a confidence interval for 1 sample data:

• A matched pairs hypothesis test is just like a one-sample test:H0: µdifference= 0 ; Ha: µdifference>0 (or <0, or ≠0) 22

*1

dn

sd t

n

Sweetening loss in colasThe sweetness loss due to storage was evaluated by 10 professional

tasters (comparing the sweetness before and after storage):

Taster

• 1 2.0 95% Confidence interval:• 2 0.4 1.02 2.2622(1.196/sqrt(10)) = 1.02 2.2622(.3782)• 3 0.7 = 1.02 .8556 =(.1644, 1.8756)• 4 2.0• 5 −0.4• 6 2.2• 7 −1.3• 8 1.2• 9 1.1• 10 2.3Summary stats: = 1.02, s = 1.196

We want to test if storage results in a

loss of sweetness, thus:

H0: mdifference = 0

versus Ha: mdifference > 0

Before sweetness – after sweetness

This is a pre-/post-test design and the variable is the cola sweetness

before storage minus cola sweetness after storage.

A matched pairs test of significance is indeed just like a one-sample

test.

d

Sweetening loss in colas hypothesis test

• H0: mdifference = 0 vs Ha: mdifference > 0

• Test statistic

• From t-table: for df=9,2.2622 <t=2.6970<2.8214 .01 < P-value < .025

• ti83 gives P-value = .012263…

• Conclusion: reject H0 and conclude colas do lose sweetness in storage (note that CI was entirely positive.

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1.02 0 1.022.6970

1.196 .378210

t

Does lack of caffeine increase depression?

Individuals diagnosed as caffeine-dependent are

deprived of caffeine-rich foods and assigned

to receive daily pills. Sometimes, the pills

contain caffeine and other times they contain

a placebo. Depression was assessed (larger number means more depression).

– There are 2 data points for each subject, but we’ll only look at the difference.

– The sample distribution appears appropriate for a t-test.

SubjectDepression

with CaffeineDepression

with PlaceboPlacebo - Cafeine

1 5 16 112 5 23 183 4 5 14 3 7 45 8 14 66 5 24 197 0 6 68 0 3 39 2 15 1310 11 12 111 1 0 -1

11 “difference” data points.

-5

0

5

10

15

20

DIF

FER

ENC

E

-2 -1 0 1 2Normal quantiles

Hypothesis Test: Does lack of caffeine increase depression?For each individual in the sample, we have calculated a difference in depression score

(placebo minus caffeine).

There were 11 “difference” points, thus df = n − 1 = 10.

We calculate that = 7.36; s = 6.92

H0 :mdifference = 0 ; Ha: mdifference > 0

53.311/92.6

36.70

ns

xt

SubjectDepression

with CaffeineDepression

with PlaceboPlacebo - Cafeine

1 5 16 112 5 23 183 4 5 14 3 7 45 8 14 66 5 24 197 0 6 68 0 3 39 2 15 1310 11 12 111 1 0 -1

For df = 10, 3.169 < t = 3.53 < 3.581 0.005 > p > 0.0025ti83 gives P-value = .0027

Caffeine deprivation causes a significant increase in depression.

x