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CHAPTER 9

Homework: 1,6,15,24,47,49,53,57

Sec 9.1: Sampling Distribution for the difference of two sample means:

1. Sampling distribution for the difference of two independent sample means:(a). The central limit theory ensures that the sampling distribution for the difference of two independent sample means is approximately normal for sufficiently large samples.

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(b). The difference of the two sample means is an unbiased estimator for the difference of the two population means.

(c). The standard deviation for the difference of two independent sample means is

where and 2 are the population standard deviations and n1 and n2 are the respective sample sizes.

dn n

= 12

1

22

2

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2. Problem of two paired samples

Sometimes the two samples are paired. For example, suppose x1, x2, .., xn are weights of n people before they are enrolled onto a dietary course, and y1,y2,..,yn are the corresponding weights of these n people after they complete the dietary course. Here the two samples are not independent, because xi and yi are the weights of the same person -- a paired measurements. We call these two samples paired samples. In order to

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assess the usefulness of the dietary course, we would be interested in making inference on the mean weight reduction after completing the course, d. For this, we can base our inference on y1-x1, y2-x2,..,yn-xn, which can be regarded as a random sample taken from the population of weight reduction after completing the course. Therefore the problem is reduced to a one-sample problem which has already been dealt with in chapters 7 and 8. We will not discuss two-paired-sample problem further in this chapter.

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Sec 9.2: Large sample inference for the difference of two independent population means

(1) The confidence interval

Under the assumption that two random samples taken independently from two populations have large enough sample sizes, the 1- confidence interval for difference of the two population means is given by

where sd s

n

s

n= 1

2

1

22

2

x x z sd1 2 2 /

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(2) Hypotheses testing:

(a). Alternative Hypothesis: (i). Two-Tailed Test: Ha: ()d0

(ii). Right-Tailed Test: Ha: ()d0

(iii). Left-Tailed Test: Ha: ()d0

(b). Null Hypothesis: (i). Two-Tailed Test: H0: ()d0

(ii). Right-Tailed Test: H0: () d0

(iii). Left-Tailed Test: H0: ()d0

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(c). Test Statistic

(d). Rejection RegionA size test has the rejection region

(i).Two-Tailed Test: Z > Zor Z < -Z ,(ii). Right-Tailed Test: Z > Z ,(iii). Left-Tailed Test: Z < -Z.

Zx x d

sd

1 2 0

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(e). p-values of this test:

(i). Two-tailed test: p-value=2*P(z > |Z|)(ii). Right-tailed test:p-value=P(z > Z)(iii). Left-tailed test:p-value=P(z < Z)

A size test rejects the null hypothesis if and only if the p-value is less than , and this test always reaches the same rejection/acceptance decision as the test in (d).

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(EX 9.1) (Basic -- Large Sample Inference)Two independent random samples were

selected from two independent populations. The sample sizes, sample means, and sample variances are as follows:

I II

Sample Sizes 35 49

Sample Mean 12.7 7.4

Sample Variance 1.38 4.14

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(a). Can we use the large sample inference procedure? Explain.(b). Find a 95% confidence interval for the mean difference.(c). Test that the difference of these two means is larger than 5 at. Report the p-value and draw your conclusion.

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(EX 9.2) (Basic)An experiment was conducted to compare

two diets A and B designed for weight reduction. Two groups of 30 overweight dieters each was randomly selected. One group was placed for diet A and another on diet B and their weight loss was recorded over a thirty-day period. The result is presented in table 1.

Table I Weight Loss

Diet A Diet B

sample mean 21.3 13.4

sample s.d. 2.6 1.9

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(a). Can we used the large sample inference procedure? Explain.(b). Find a 90% C.I. for the difference of mean weight loss for these two diets.(c). Does the difference larger than 9 at 0.01. Report the p-value and draw your conclusion.

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Sec 9.3: Small sample inference for the difference of two independent population means

Assumptions:

two independent populations are normally distributed;

the two populations have a common variance.

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The pooled estimator, Sp2 , of the common

variance is given by

where n1 and n2 are the sample sizes.

The standard deviation of the difference of the two sample mean is given by

S Sn nx x p( ) ( ) .

1 2

2

1 2

1 1

Sn S n S

n np2 1 1

22 2

2

1 2

1 1

2

( ) ( ).

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1. Confidence Interval

The (1100% confidence interval for the population mean difference is given by

( ) ,/ , ( ) .x x t sn n x x1 2 2 21 2 1 2

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2 Hypotheses Testing

(a). Alternative Hypothesis: (i). Two-Tailed Test: Ha: ()d0

(ii). Right-Tailed Test: Ha: ()d0

(iii). Left-Tailed Test: Ha: ()d0

(b). Null Hypothesis: (i). Two-Tailed Test: H0: ()d0

(ii). Right-Tailed Test: H0: () d0

(iii). Left-Tailed Test: H0: ()d0

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(c) Test statistics is given by

tx x d

scx x

( ).

( )

1 2 0

1 2

(d) Rejection region

A size has the the rejection region

(i).Two-Tailed Test: > tn1+n2-2 or < -t,n1+n2-2 ,(ii). Right-Tailed Test: > tn1+n2-2 ,(iii). Left-Tailed Test: < -tn1+n2-2.

tc tc

tc

tc

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(e) p-values of this test

(i). Two-tailed test: p-value=2*P(tn1+n2-2 > |tc|)(ii). Right-tailed test:p-value=P(tn1+n2-2 > tc)(iii). Left-tailed test:p-value=P(tn1+n2-2 < tc)

A size test rejects the null hypothesis if and

only if the p-value is smaller than .

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(EX 9.3) In a study of iron deficiency among infants,

samples of infants following different feeding regimens were compared. One group contained breast-fed infants, while the children in another group were fed a standard baby formula without any iron supplements. Here are summary results on blood hemoglobin levels at 12 months of age.

Group n mean s.d.Breast-fed 23 13.3 1.7

Formula 19 12.41.8

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(a). Set up a test to test that the mean hemoglobin level is higher among breast-fed babies?

(i). State the null and alternative hypothesis.(ii). Carry out an appropriate test.(iii). What kind assumptions do you need to

carry out your test?(iv). Give the P-value and state your

conlusions at = 0.05.

(b). Give a 95% C.I. for the mean difference in hemoglobin level between these two populations of infants.

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(EX 9.4) The following table presented the

summary statistics from an experiment for decreasing the blood pressure.

meanGroupTreatment n decrease s.d.1 Calcium 10 5.08.7432 Placebo 11 -.2735.901

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(a). Set up a test to test that the mean blood pressure decreased is larger in Calcuim group?

(i). State the null and alternative hypotheses.

(ii). Carry out an appropriate test.(iii). What kind assumptions do you need?(iv). Give the P-value and state your

conlusions at = 0.05.

(b). Give a 95% C.I. for the mean difference in blood presure decreased between the two populations.

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Sec 9.4: Large sample inference for the difference of two independent population proportions

We need to assume that the two samples are selected from two independent binomial populations and the both sample sizes are sufficiently large.

The difference of the two sample proportions is approximately normally distributed with mean (p1-p2) and variance

Sp p

n

p p

np ( ) ( )

.p1

2

2 1 1

1

2 2

2

1 1

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1. Confidence interval

A confidence interval for the difference of two population proportions is given by

where

( ) )./ ( p p z S p p1 2 2 1 2

S(p1 ) ( ) ( )

.

pp p

n

p p

n2

1 1

1

2 2

2

1 1

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2 Hypotheses Testing

(a). Alternative Hypothesis: (i). Two-Tailed Test: Ha: ()0(ii). Right-Tailed Test: Ha: ()0(iii). Left-Tailed Test: Ha: ()0

(b). Null Hypothesis: (i). Two-Tailed Test: H0: ()0(ii). Right-Tailed Test: H0: () 0(iii). Left-Tailed Test: H0: ()0

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(c) Test Statistics:

(d) Rejection Region of this test:A size test has the rejection region

(i).Two-Tailed Test: Z > Zor Z < -Z ,(ii). Right-Tailed Test: Z > Z ,(iii). Left-Tailed Test: Z < -Z.

Zp p

sp p

1 2

1 2

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(e) p-values of this test:

(i). Two-tailed test: p-value=2*P(z > |Z|)(ii). Right-tailed test:p-value=P(z > Z)(iii). Left-tailed test:p-value=P(z < Z)

A size test rejects the null hypothesis if and only if the p-value is smaller than .

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(EX 9.5) (Basic)Samples of n1 = n2 = 500 observations were

selected from two independent binomial populations, and x1 = 120 and x2 = 147 were observed. (a). Find the 95% C.I. for the difference of the two proportions.(b). Are these two proportions different at 0.05 level?(c). What assumptions do we need to use a large sample inference procedure?

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(EX 9.6). (Applications)A sampling of political candidates -- 200

randomly chosen from the west and 200 from the east – was classified according to whether the candidate received backing by a national labor union and whether the candidate won. A summary of the data is shown below.

West EastWinners (union) 120 142

(a). Find a 95% C.I. interval for the difference of the proportions of union-backed winners in the west verse the east.

(b). Is this difference significant at 0.05?