Lecture 9

• Inference about the ratio of two variances (Chapter 13.5)

• Inference about the difference between population proportions (Chapter 13.6)

13.5 Inference about the ratio 13.5 Inference about the ratio of two variancesof two variances

13.5 Inference about the ratio 13.5 Inference about the ratio of two variancesof two variances

• In this section we draw inference about the ratio of two population variances.

• This question is interesting because:– Variances can be used to evaluate the consistency of

production processes.

– It may help us decide which of the equal- or unequal-variances t-test of the difference between means to use.

– Because variance can measure risk, it allows us to compare risk, for example, between two portfolios.

• Parameter to be tested is 12/2

2

• Statistic used is 22

22

21

21

ss

F

Parameter and Statistic Parameter and Statistic

• Sampling distribution of 12/2

2

– The statistic [s12/1

2] / [s22/2

2] follows the F distribution with 1 = n1 – 1, and 2 = n2 – 1.

F Distribution

• Tables 6(a)-(c) only give the right tail quantiles (critical values) of the F distribution

• To find quantiles for the left tail of the F distribution, use the fact

12

21

,,,,1

1

AA F

F

– Our null hypothesis is always

H0: 12 / 2

2 = 1

– Under this null hypothesis the F statistic becomes

F =S1

2/12

S22/2

2

22

21

ss

F 22

21

ss

F

Parameter and Statistic Parameter and Statistic

(see Xm13-01)In order to perform a test regarding average consumption of calories at people’s lunch in relation to the inclusion of high-fiber cereal in their breakfast, the variance ratio of two samples has to be tested first.

Example 13.6 (revisiting Example 13.1)

The hypotheses are:

H0:

H1: 1

1

Testing the ratio of two population variances Testing the ratio of two population variances

Estimating the Ratio of Two Population Variances

• From the statistic F = [s12/1

2] / [s22/2

2] we can isolate 1

2/22 and build the following

confidence interval:

1nand1nwhere

Fs

sF

1

s

s

221

1,2,2/22

21

22

21

2,1,2/22

21

1nand1nwhere

Fs

sF

1

s

s

221

1,2,2/22

21

22

21

2,1,2/22

21

• Example 13.7– Determine the 95% confidence interval

estimate of the ratio of the two population variances in Example 13.1

Estimating the Ratio of Two Population Variances Estimating the Ratio of Two Population Variances

Inference about the difference between two population proportions

• In this section we deal with two populations whose data are nominal.

• For nominal data we compare the population proportions of the occurrence of a certain event.

• Examples– Comparing the effectiveness of new drug versus older

one– Comparing market share before and after advertising

campaign– Comparing defective rates between two machines

Parameter and Statistic

• Parameter– When the data are nominal, we can only count

the occurrences of a certain event in the two populations, and calculate proportions.

– The parameter is therefore p1 – p2.

• Statistic– An unbiased estimator of p1 – p2 is

(the difference between the sample proportions).

21 p̂p̂

Sample 1 Sample size n1

Number of successes x1

Sample proportion

Sample 1 Sample size n1

Number of successes x1

Sample proportion

• Two random samples are drawn from two populations.• The number of successes in each sample is recorded.• The sample proportions are computed.

Sample 2 Sample size n2

Number of successes x2

Sample proportion

Sample 2 Sample size n2

Number of successes x2

Sample proportionx

n1

1

ˆ p1

2

22 n

xp̂

Sampling Distribution ofSampling Distribution of 21 p̂p̂

• The statistic is approximately normally distributed if n1p1, n1(1 - p1), n2p2, n2(1 - p2) are all greater than or equal to 5.

• The mean of is p1 - p2.

• The variance of is (p1(1-p1) /n1)+ (p2(1-p2)/n2)

21 p̂p̂

21 p̂p̂

21 p̂p̂

Sampling distribution ofSampling distribution of 21 p̂p̂

2

22

1

11

2121

)1()1(

)()ˆˆ(

n

pp

n

pp

ppppZ

2

22

1

11

2121

)1()1(

)()ˆˆ(

n

pp

n

pp

ppppZ

The z-statisticThe z-statistic

Because and are unknown the standard errorBecause and are unknown the standard errormust be estimated using the sample proportions. must be estimated using the sample proportions. The method depends on the null hypothesis The method depends on the null hypothesis

1p 2p

Testing the p1 – p2

• There are two cases to consider:

Case 1: H0: p1-p2 =0

Calculate the pooled proportion

21

21

nn

xxp̂

Then Then

Case 2: H0: p1-p2 =D (D is not equal to 0)Do not pool the data

2

22 n

xp̂

1

11 n

xp̂

)n1

n1

)(p̂1(p̂

)pp()p̂p̂(Z

21

2121

)n1

n1

)(p̂1(p̂

)pp()p̂p̂(Z

21

2121

2

22

1

11

21

n)p̂1(p̂

n)p̂1(p̂

D)p̂p̂(Z

2

22

1

11

21

n)p̂1(p̂

n)p̂1(p̂

D)p̂p̂(Z

• Example 13.8– The marketing manager needs to decide which of two

new packaging designs to adopt, to help improve sales of his company’s soap.

– A study is performed in two supermarkets:• Brightly-colored packaging is distributed in supermarket 1.• Simple packaging is distributed in supermarket 2.

– First design is more expensive, therefore,to be financially viable it has to outsell the second design.

Testing p1 – p2 Testing p1 – p2

• Summary of the experiment results– Supermarket 1 - 180 purchasers of Johnson Brothers

soap out of a total of 904

– Supermarket 2 - 155 purchasers of Johnson Brothers soap out of a total of

1,038

– Use 5% significance level and perform a test to find which type of packaging to use.

Testing p1 – p2 Testing p1 – p2

• Example 13.9 (Revisit Example 13.8)– Management needs to decide which of two new

packaging designs to adopt, to help improve sales of a certain soap.

– A study is performed in two supermarkets:– For the brightly-colored design to be financially

viable it has to outsell the simple design by at least 3%.

Testing p1 – p2 Testing p1 – p2

Estimating p1 – p2 Estimating p1 – p2

• Estimating the cost of life saved– Two drugs are used to treat heart attack victims:

• Streptokinase (available since 1959, costs $460)

• t-PA (genetically engineered, costs $2900).

– The maker of t-PA claims that its drug outperforms Streptokinase.

– An experiment was conducted in 15 countries. • 20,500 patients were given t-PA

• 20,500 patients were given Streptokinase

• The number of deaths by heart attacks was recorded.

• Experiment results– A total of 1497 patients treated with

Streptokinase died.– A total of 1292 patients treated with t-PA died.

• Estimate the cost per life saved by using t-PA instead of Streptokinase.

Estimating p1 – p2 Estimating p1 – p2

Practice Problems

• 13.88, 13.92, 13.102,13.104, 13.106

• Next Class: Chapters 15.1-15.3

15.1 Introduction to ANOVA

• Analysis of variance compares two or more populations of interval data.

• Specifically, we are interested in determining whether differences exist between the population means.

• The procedure works by analyzing the sample variance.

• The analysis of variance is a procedure that tests to determine whether differences exits between two or more population means.

• To do this, the technique analyzes the sample variances

One Way Analysis of Variance

• Example 15.1 - continued

– An experiment was conducted as follows:

• In three cities an advertisement campaign was launched .

• In each city only one of the three characteristics (convenience,

quality, and price) was emphasized.

• The weekly sales were recorded for twenty weeks following

the beginning of the campaigns.

One Way Analysis of Variance

One Way Analysis of Variance

Convnce Quality Price529 804 672658 630 531793 774 443514 717 596663 679 602719 604 502711 620 659606 697 689461 706 675529 615 512498 492 691663 719 733604 787 698495 699 776485 572 561557 523 572353 584 469557 634 581542 580 679614 624 532

Convnce Quality Price529 804 672658 630 531793 774 443514 717 596663 679 602719 604 502711 620 659606 697 689461 706 675529 615 512498 492 691663 719 733604 787 698495 699 776485 572 561557 523 572353 584 469557 634 581542 580 679614 624 532

See file Xm15 -01

Weekly sales

Weekly sales

Weekly sales

• Solution– The data are interval

– The problem objective is to compare sales in three cities.

– We hypothesize that the three population means are equal

One Way Analysis of Variance