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Chapter 11

- Comparing Two Populations or Treatments

Two Means Using Independent Samples

- An example to compare two populations A

university financial aid director wants to

determine whether the mean cost of textbooks is

different for students in the engineering college

than for students in the liberal arts college. - Population 1 all students enrolled in the

engineering college. - Population 2 all students enrolled in the

liberal arts college. - Goal To compare their respective mean textbook

costs - Two random samples one from each population.
- An example to compare two treatments An

agriculture scientist wish to compare weight

gains for animals placed on two different diets.

11.1 Inferences Concerning the Difference Between

Two Populations

Two Means Using Independent Samples

- Two samples are said to be independent samples if

the selection of the individuals or objects that

make up one sample does not influence the

selection of individuals or objects in the other

sample. - A comparison of means focuses on the difference

µ1-µ2 - µ1 - µ2 0 is equivalent to µ1 µ2
- µ1 - µ2 gt 0 is equivalent to µ1 gt µ2
- µ1 - µ2 lt 0 is equivalent to µ1lt µ2
- is the point estimate of µ1 - µ2 .
- is calculated from sample values,

and therefore, it has a sampling distribution.

Properties of the Sampling Distribution of

- If the random samples on which and are

based are selected independently of one another,

then

- If n1 and n2 are both large or if the population

distributions are (at least approximately)

normal, then the sampling distribution of

are also normal or approximately normal.

The Two-Sample t Test for Comparing Two

Population Means

- Null hypothesis H0 µ1 µ2 hypothesized value
- Alternative Hypothesis
- (1) Ha µ1 µ2 gt hypothesized value
- P-Value Area to the right of the computed t
- (2) Ha µ1 µ2 lt hypothesized value
- P-Value Area to the left of the computed t
- (3) Ha µ1 µ2 ? hypothesized value
- P-Value

- 2(area to the right of the computed t

) if t gt 0, OR 2(area to the left of the

computed t ) if t lt 0.

The Two-Sample t Test for Comparing Two

Population Means

Test Statistics

Degrees of Freedom for the Two-Sample t Test

where

Truncate (round down) df to obtain an integer

value.

Comments for Two-Sample t Test

- Assumption 1 The two samples are independently

selected random samples. - Assumption 2 The sample sizes are large (in

general, 30 or larger), or the population

distributions are (at least approximately)

normal. - If s is known, two-sample z test can also be

used.

- Example Brain Size Do children diagnosed with

ADHD (attention deficit/hyperactivity disorder)

have smaller brains than children without this

condition? Brain scans were completed for 152

children with ADHD and 139 children without ADHD.

Summary values are given below.

Is there evidence that the mean brain

volume of children with ADHD is smaller than the

mean for children with ADHD? Use a

.05? Solution The two samples are

independently selected and n1 and n2 are both

large, so it is reasonable to proceed with the

two sample t test for comparing the two

population means. Let µ1 true mean brain

volume for children with ADHD, and µ2

true mean brain volume for children without ADHD

Then µ1 -µ2 difference in mean brain volume.

Continued on next slide

Solution to the Example of Brain Size

H0 µ1 - µ2 0 Ha µ1 - µ2 lt 0 Significance

level a .05.

Appendix Table 4 shows that the area under the t

curve with 288 df to the left of -3.36 is 0.

Therefore, P-value 0 lt .05. Conclusion We

reject H0. The is convincing evidence that the

mean brain volume for children with ADHD is

smaller than the mean for children without ADHD.

- Example (Oral Contraceptive Use and Bone Mineral

Density) - To assess the impact of oral contraceptive use

on bone mineral density (BMD), researchers

carried out a study comparing BMD (in grams per

centimeter) for women who had used oral

contraceptives for at least 3 months to BMD for

women who have never used oral contraceptives - Never used oral contraceptives
- 0.82, 0.94, 0.96, 1.31, 0.94, 1.21, 1.26, 1.09,

1.13, 1.14 - Used oral contraceptives
- 0.94, 1.09, 0.97, 0.98, 1.14, 0.85, 1.30, 0.89,

0.87, 1.01 - Is there evidence that women who use oral

contraceptives have a lower mean BMD? Use a

.05. - Solution The two samples are independently

selected, so it is reasonable to proceed with the

two sample t test for comparing the two

population means. - Assumption The sample size are small . We must

be willing to assume that the two populations can

be viewed as at least approximately normally

distributed

Continued on next slide

Solution to the Example Oral Contraceptive and

Bone Mineral Density

The boxplots based on the two samples are

reasonably symmetric and because there are no

outliers, it is plausible to assume the

populations are approximately normal.

- Let
- µ1 true mean bone mineral density for women

who never used oral contraceptives, and - µ2 true mean bone mineral density for women

who used oral contraceptives - Then µ1 -µ2 difference in mean bone mineral

density.

Continued on next slide

Solution to the Example of Oral Contraceptive and

Bone Mineral Density

H0 µ1 - µ2 0 Ha µ1 - µ2 gt 0 Significance

level a .05.

Appendix Table 4 shows that the area under the t

curve with 17 df to the right of 1.1 is 0.143.

Therefore, P-value .143 gt .05. Conclusion We

fail to reject H0. The is no convincing evidence

to support the claim that the mean bone mineral

density is lower for women who use oral

contraceptives.

- Exercise Comparing Two Treatments
- First year nursing students enrolled in a

science course were randomly assigned to one of

the two groups the control group in which

students were provided with conventional

study-skill guidelines, and the treatment group,

in which students were provided homework

prescription based on their identified learning

style. Does the treatment improve science grades?

Use a significance level of .01.

Answer P-value 0. There is evidence that the

mean science score for the treatment group is

higher.

The Two-Sample t Confidence Interval for the

Difference Between Two Population or Treatment

Means

The confidence interval for µ1 - µ2 is

The t critical value is based on the degree of

freedom

where

Truncate (round down) df to obtain an integer

value.

Assumptions for the Two Sample t Confidence

Interval for µ1 µ2

- The two samples are independently chosen random

samples, and - The sample sizes are both large (in general, n1 ?

30 and n2 ? 30) or the population distributions

are approximately normal,

- Example Effect of Talking on Blood Pressure
- Does talking elevate blood pressure?

Patients with high blood pressure were randomly

assigned to one of two groups. Those in the first

group (the talking group) were asked questions

about their medical history and sources of stress

before their blood pressure was measured. Those

in the second group (the counting group) were

asked to count from 1 to 100 four times before

their blood pressure was measured. The following

data values are the diastolic blood pressure (in

mm Hg) for the two groups. - Talking 104 110 107 112 108 103 108 118
- n1 8, mean 108.75, standard deviation

4.74 - Counting 110 96 103 98 100 109 97 105
- n2 8, mean 102.25, standard deviation

5.39 - Construct a 95 confidence interval to estimate

µ1 µ2, the difference in mean diastolic blood

pressure for the two treatments.

Solution to the Example Effect of Talking on

Blood Pressure To estimate µ1 µ2, the

difference in mean diastolic pressure for the two

treatments, we calculate a 95 confidence

interval

In the 13-df row of Appendix Table 3, the t

critical value for a 95 confidence level is

2.16. The 95 interval is then

The 95 confidence for µ1 µ2 is (1.02,

11.98). We estimate that the mean diastolic blood

pressure when talking is higher than the mean

when counting by somewhere between 1.02 and 11.98

mm Hg.

11.2 Inference Concerning the Difference Between

Two Population or Treatment Means Using Paired

Samples

- Independent samples Two samples for which the

individuals or objects in the first sample are

selected independently from those in the second

sample. - Paired samples Two samples for which each

observation in one sample is paired in a

meaningful way with a particular observation in

the second sample.

- An Example of Paired Samples Alcohol Content

of Wine - Are wine drinkers being misled by bottle

labels? The London Times reported the results of

a study that compared the actual alcohol content

on wine to the alcohol content printed on the

label. A representative subset (six bottles) of

the resulting data is given in the following

table

The two samples are paired rather than

independent because both samples are composed of

observations on the same six bottles. Let µ1

the mean actual alcohol content for the

population of all wines and µ2 the mean labeled

alcohol content for the population of all wines.

Hypotheses of interest might be

H0 µ1 - µ2 0 versus Ha µ1 - µ2 gt 0.

Paired t Test

µd the mean value of the difference

population µd µ1 µ2 sd the standard

deviation of the difference population Let µ0 be

hypothesized value.

- Equivalent Hypothesis when samples are paired
- H0 µd µ0
- Ha µd gt µ0
- Ha µd lt µ0
- Ha µd ? µ0

- Hypothesis
- H0 µ1 µ2 µ0
- Ha µ1 µ2 gt µ0
- Ha µ1 µ2 lt µ0
- Ha µ1 µ2 ? µ0

Paired t Test

- Assumptions
- The samples are paired.
- The n sample differences can be viewed as a

random sample from a population of differences.

(Because inferences about µd can be based on the

n observed sample differences, the original

two-sample problem becomes a familiar one-sample

problem.) - The number of sample differences is large (in

general, at least 30) or the population

distribution of differences is approximately

normal.

Paired t Test for Computing Two Population Means

- H0 µ0hypothesized value
- Test statistic
- where and sd are the sample mean and

standard deviation of the differences,

respectively. - Alternative Hypothesis P-Value
- Ha µd gt hypothesized value area in upper tail

- Ha µd lt hypothesized value area in lower tail
- Ha µd ? hypothesized value sum of area in two

tails

- Example Improve Memory by Playing Chess?
- Can taking chess lessons and playing chess

daily improve memory? Twelve sixth-grade students

who have never previously played chess

participated in a program where they took chess

lessons and played chess daily for 9 months.

Their memory test (the Test of Cognitive Skills)

scores before and after the program are given in

the table below. - Do the students who participated in the

chess program tend to achieve higher memory

scores after completion of the program? - Use a significance level a .05.

Solution to the Example Improve memory by

playing chess

Let µ1 mean memory score for six-graders with

no chess training µ2 mean memory score for

six-graders after chess training and µd µ1 -

µ2 mean memory score difference between

students with no chess training and students who

have completed chess training

- The researchers wants to know if students

who have completed chess training have higher

memory scores ( µ1 lt µ2, or µd lt 0 ), so a

lower-tailed paired t-test with a .05 is used. - Null and Alternative hypotheses H0 µd 0

Ha µd lt 0 - Using the 12 differences in the table

we compute

P-value From the df 12 - 1 11 column of

Appendix Table 4, we find P-value area

to the left of - 4.56 area to the right of 4.56

0 Conclusion Reject H0 because P-value lt a.

The data support the theory that mean memory

score is higher for six-graders after completion

of the chess training.

Using Excel for Paired t-Test Input the

data of two samples in two columns, and then go

to Data Analysis like we did mny times before. In

the Data Analysis dialog box, choose t-test

Paired Two Sample for Means.

In the dialog box, enter Variable 1 range

(A1A13) and Variable 2 range (B1B13). Enter

also the hypothesized value (0 in this problem),

and significance level (0.05) for Alpha (a)

The results are shown in the following outcome

box. The P-value for one-tailed test is

0.000405682, which is almost 0.

- Exercise Spatial Ability
- A study discussed the effects of prenatal

exposure to DES (a drug banned by FDA in 1971) on

spatial ability in males born to mothers treated

with DES. In the study 10 males exposed to DES

and their unexposed brothers underwent a spatial

ability test, and the results are

Is the mean spatial ability score higher for

those unexposed to DES? Using significance level

a.05.

Answer P-value .020. There is evidence that

the mean spatial ability score is lower for those

who were exposed to DES than those who were

unexposed.

Paired t Confidence Interval for µd

- When
- The samples are paired,
- The n sample differences can be viewed as a

random sample from a population of differences,

and - The number of sample differences is large (in

general, at least 30) or the population

distribution of differences is approximately

normal, - The paired t confidence interval for µd is
- For a specified confidence level, the (n-1) df

row of the table inside the back cover gives the

appropriate t critical value.

Example Lactic Acid in the Blood After Exercise

- To exam the effect of exercise on the amount of

lactic acid in blood, eight males were selected

at random from those attending a week-long

training camp. Blood lactate levels were measured

before and after playing three racquetball games

and the results were shown in the table. - Construct a 95 confidence interval.

- Solution to Example Lactic Acid in the Blood

After Exercise - Solution Because n 8, df 8 - 1 7, and

the t critical value for 95 confidence level is

2.37. Use the 8 differences we compute

The 95 confidence interval is (-20.568, -6.692).

We are 95 confident that the mean increase in

blood lactate level is somewhere between -20.568,

-6.692.

Analysis The 8 men were selected at random

from a training camp. The box plot of the 8

sample differences is consistent with a

difference population that is approximately

normal, so the paired t confidence interval is

appropriate.

In Chapter 9, we learned how to use Excel to find

confidence intervals. This slide and next will

help us to review the process. Just need to

remember that for paired t confidence intervals

we should use the differences.

The 95 confidence interval for the difference in

mean blood lactate level is -13.63 (mean)

6.922 (-20.5, -6.7).

- Exercise Benefits of Ultrasound Revisited.
- Ultrasound is often used in the treatment of

soft tissue injuries. In an experiment of

investigate the effect of an ultrasound and

stretch therapy on knee extension, range of

motion was measured both before and after

treatment for a sample of physical therapy

patients. A subset of the data is given in the

accompanying table. - Construct a 95 confidence interval.

Answer 95 confidence interval (-6.60, -0.18)