Chapter 10Comparing Two Means Target Goal:

I can use two-sample t procedures to compare two means.10.2a

h.w: pg. 626: 29 – 32, pg. 652: 35, 37, 57

a bag of M&M's for day 1 chapter 11.Bring

Two-Sample Problems Comparative studies are more convincing

than single single-sample investigations, so one-sample inference (for matched pairs) is not as common as comparative (two-sample) inference.

In a comparative study: the units or sample sizes can be different.1) We may want to compare two treatments, or we may

want to compare two populations. 2) The samples must be chosen randomly and

independently in order to perform statistical inference.

Because match pairs are NOT chosen independently, we will NOT use two-sample inference for a matched pairs design.

Recall: for a matched pairs design, apply the one-sample t procedures to the observed differences.

Otherwise, we may use two-sample inference to compare two treatments or two populations.

Significance Tests for µ1 – µ2

The null hypothesis is that there is no difference between the two parameters.

The alternative hypothesis could be that

Before you begin, check your assumptions! Conditions for Comparing Two Means

Both samples must be a Random sample or randomized experiment. must be chosen independently both populations must be normally

distributed.

(Check the data for outliers or skewness.)

Confidence Intervals for µ 1 – µ 2. On page 634C

omparing T

wo M

eans

Two-Sample t Interval for a Difference Between Means

1 2

2 21 2

1 21 2

When the Random, Normal, and Independent conditions are met, an

approximate level C confidence interval for ( ) is

( ) *

where * is the critical v

x x

s sx x t

n n

t

1 2degrees of freed

alue for confidence level C for the distribution with

from either technology or theom smaller of 1 . and 1n

t

n

Confidence Intervals for µ 1 – µ 2. On page 634C

omparing T

wo M

eans

Two-Sample t Interval for a Difference Between Means

1

2

1 2

of size from

Population 1 and a random sample of size from Population 2 by

two groups

The data are produc

of size

ed by a random sample

or

in a randomized experime and nt.

n

n

n n

Random

Normal Both population distributions are Normal OR both sample group sizes are large (n1 30 and n2 30).

Both the samples or groups themselves and the individual

observations in each sample or gr

When sampling

without replacement, check that the

o

two populations are at

up are independ

least

en .

t

Independent

10 times

as large as the corresponding samples (the 10% condition).

If these assumptions hold, then the difference in sample means is an unbiased estimator of the difference in population means,

so is equal to .

1 2The Sampling Distribution of :x x

1 2x x 1 2

The variance of is the sum of the variances of and ,

which is

Note: variances add, standard deviations don’t.

Furthermore, if both populations are normally distributed, then is also normally distributed.

1 2x x

1x2x

1 2

2 21 2

1

2

2x x n n

1 2x x

Standardizing Subtract the mean and divide by the standard

deviation:

1 2x x

1 2 1 2

2 21 2

1 2

x xz

n n

If we do not know and , we will substitute with the standard error SE using and for the standard deviation.

This gives the standardized t value:

1 2

1 2 1 2

2 21 2

1 2

x xt

s sn n

1s

n2s

n

Ex: Calcium and blood pressure Does increasing the amount of calcium in our diet

reduce blood pressure? The subjects were 21 healthy black men. A randomly chosen group of 10 received calcium supplement for 12 weeks. A control group of 11 men received a placebo (double blinded experiment).

The response variable is the decrease in systolic (heart contracted) blood pressure. An increase appears as a negative response (drop in blood pressure).

Group 1: 7, -4, 18, 17, -3, -5,1,10,11,-2 Group 2: -1, 12, -1,-3, 3, -5, 5, 2, -11, -1, -3 Calculate summary statistics: (Enter data into

lists.) Group Treatment n s1 Calcium2 Placebo

x10 5.000 8.743 11 -0.273 5.901

Is this outcome good evidence that calcium decreases blood pressure? Perform a significance test.

Step 1. State - Identify the population of interest and the parameter you want to draw a conclusion about. State the null and alternative hypothesis in words and symbols.

: The mean decrease in blood pressure for those taking calcium is the same as the mean decrease in blood pressure for those taking a placebo.

: The mean decrease in blood pressure for those taking calcium is greater than the mean decrease in blood pressure for those taking a placebo.

0 1 2:H

1 2:aH

Step 2. Plan –If the conditions are met, we will construct a two-sample t test for

Verify the conditions (random, indep, normal). Random: Because of randomization, we must

assume the calcium and placebo groups as SRS.

Normal: Examine the stem plot in notes. Are there outliers?

1 2.

Step 2. Plan –If the conditions are met, we will construct a two-sample t test for

Normal: Examine the stem plot in notes. Are there outliers? no

1 2.

Are there any departures from normality that would prevent us from using a t-test?

Calcium is a little irregular but this is not unusual for a small sample.

Independent: Yes, due to random assignment, the populations can be viewed as independent.

Individual observations in each group should also be independent.

Proceed with t procedures.

Step 3. Carry out the inference procedure. Compute the

test statistic and P-value To test compute:

= 1.604

0 1 2:H

1 2 1 2

2 21 2

1 2

( )x xt

s s

n n

22

0

5.9018.7431

(5.00 ( 0.273

0 11

))

1 2 Note: x -x as a statistic, measures the advantage of calcium over a placebo.

0 1 2: 0H

For df, use the smaller of n1 – 1, n2 – 1.

There are 9 df,

Find P tcdf(

P = .072

Step 4: Interpret your results: The experiment found evidence that calcium

reduces blood pressure but not at the traditional 5% and 1% levels as the p value is .071.

We fail to reject Ho at these levels.

We will use graphing calculators to perform two-sample T tests!

2-SampTTest for a hypothesis test 2-SampTInt for a confidence interval

Confirm the P-value and construct a 90% confidence interval

STAT:TESTS: Pooled: no Use data if you have it or Stats if you are

given 1 2 1 2, , , .x x s s

CI = (-.4767, 11.022). The 90 % C.I. covers 0 (both neg. and pos.

changes). We can not reject Ho against the two sided

alternative at the α = 0.10 level of significance. Read pg. 625 - 637