Statistics- Tests of Hypotheses-1

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Tests of Hypotheses:

Small SamplesChapterTheimage cannotbedisplayed. Your computer may nothave enough memory toopen theimage,or theimagemay havebeen corrupted.Restartyour computer,and then open thefileagain. Ifthe red x stillappears, you may havetod eletetheimage and then insertit again.

Rejection

region


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CHAPTER GOALS

TO DESCRIBE THE MAJOR CHARACTERISTICS OF

STUDENTS t-DISTRIBUTION.

TO UNDERSTAND THE DIFFERENCE BETWEEN THE t -

DISTRIBUTION AND THE z -DISTRIBUTION.


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CHAPTER GOALS

TO TEST A HYPOTHESIS INVOLVING ONE POPULATION

MEAN.

TO TEST A HYPOTHESIS INVOLVING THE DIFFERENCE

BETWEEN TWO POPULATION MEANS.

TO CONDUCT A TEST OF HYPOTHESIS FOR THEDIFFERENCE BETWEEN A SET OF PAIRED OBSERVATIONS.


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CHARACTERISTICS OF STUDENTS

t-DISTRIBUTION

The t-distribution hasthe following properties: It is continuous, bell shaped and symmetrical about zero like

the z-distribution.

There is a family oft-distributions with mean of zero but one

for each sample size. The t-distribution is more spread out and flatter at the center

than the z-distribution, but approaches the z-distribution as

the sample size gets larger.


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z-distribution

t-distribution

The degrees of

freedom for

the t-distribution

is df = n - 1.


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A TEST FOR A POPULATION MEAN:

SMALL SAMPLE, POPULATION

STANDARD DEVIATION UNKNOWN

The test statistic for a one sample case is given by equation

(9-1)below

(9-1)

tX

nS!

Q

/


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EXAMPLE 1

The current rate for producing 5 amp fuses at Monarch

Electric Company is 250 per hour. A new machine has beenpurchased and installed that, according to the supplier, will

increase the production rate. A sample of 10 randomly

selected hours from last month revealed the mean hourly

production on the new machine was 256, with a samplestandard deviation of 6 per hour. At the 0.05 significance

level can Monarch conclude that the new machine is faster?


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EXAMPLE 1 (continued)

Step 1:State the null and the alternative hypotheses.

H0:Q e 250 H1:Q" 250

Step 2:State the decision rule.

H0isrejectedif t> 1.833, df = 9(Appendix F).

Step 3:Compute the value of the test statistic.

t = [256 - 250]/[6/10]= 3.16.

Step 4:What is the decision on H0?

H0

is rejected. The new machine is faster.


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STUDENT tDISTRIBUTION

Level of significance for one-tailed test

df .10 .05 .025 .01 .005 .0005

Level of significance for two-tailed test

.20 .10 .05 .02 .01 .0011 3.078 6.314 12.706 31.821 63.657 636.6192 1 .886 2.920 4.303 6.965 9.925 31 .5993 1.638 2.353 3.182 4.541 5.841 12.924

4 1.533 2.132 2.776 3.747 4.604 8.610

5 1.476 2.015 2.571 3.365 4.032 6.869

6 1.440 1.943 2.447 3.143 3.707 5.959

7 1.415 1.895 2.365 2.998 3.499 5.408

8 1.397 1.860 2.306 2.896 3.355 5.041

9 1.383 1.833 2.262 2.821 3.250 4.781

10 1.372 1.812 2.228 2.764 3.169 4.587

11 1.363 1.796 2.201 2.718 3.106 4.437

12 1.356 1.782 2.179 2.681 3.055 4.318

13 1.350 1.771 2.160 2.650 3.012 4.221

14 1.345 1.761 2.145 2.624 2.977 4.140

15 1.341 1.753 2.131 2.602 2.947 4.073

16 1.337 1.746 2.120 2.583 2.921 4.015

17 1.333 1.740 2.110 2.567 2.898 3.965

18 1.330 1.734 2.101 2.552 2.878 3.922

19 1.328 1.729 2.093 2.539 2.861 3.883

20 1.325 1.725 2.086 2.528 2.845 3.850

21 1.323 1.721 2.080 2.518 2.831 3.819

22 1.321 1.717 2.074 2.508 2.819 3.792

23 1.319 1.714 2.069 2.500 2.807 3.768

24 1.318 1.711 2.064 2.492 2.797 3.745

25 1.316 1.708 2.060 2.485 2.787 3.725

26 1.315 1.706 2.056 2.479 2.779 3.707

27 1.314 1.703 2.052 2.473 2.771 3.690

28 1.313 1.701 2.048 2.467 2.763 3.674

29 1.311 1.699 2.045 2.462 2.756 3.659

30 1.310 1.697 2.042 2.457 2.750 3.646

40 1.303 1.684 2.021 2.423 2.704 3.551

60 1 .296 1 .671 2.000 2.390 2.660 3.460

120 1.289 1.658 1.980 2.358 2.617 3.373

1 .282 1 .645 1 .960 2.326 2.576 3.291


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Display of the Rejection Region, Critical

Value, and the computed Test Statistic

Region of

rejection

t

Critical

value

1.833

0.05

df= 9


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COMPARING TWO POPULATIONS MEANS

To conduct this test, three assumptions are required:

1.The populations must be normally or approximately

normally distributed.

2.The populations must be independent.

3.The population variances must be equal. Let subscript 1 and 2 be associated with population 1 and 2

respectively.


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POOLED SAMPLE VARIANCE AND TEST

STATISTIC

Pooled Sample Variance

sn s n s

n n

Test Statistic

t X X

s n n

p

p

=

1

2 1 12

2 22

1 2

2

2

1 2

1 1

2

1 1

!

( ) ( )

(9-3)

(9-2)


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Step 1:State the null and the alternative hypotheses.

H0:Q e Q H1:Q " Q

Step 2:State the decision rule.

H0isrejectedif t> 1.708, df= 25.

Step 3:Compute the value of the test statistic.

t = 1.64 (Verify).

Step 4:What is the decision on H0?

H0

is not rejected. Insufficient sample evidence to claim a

higher mpg on the imported cars.



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df .10 .05 .025 .01 .005 .0005


.20 .10 .05 .02 .01 .0011 3.078 6.314 12.706 31.821 63.657 636.6192 1 .886 2.920 4.303 6.965 9.925 31 .5993 1.638 2.353 3.182 4.541 5.841 12.924

4 1.533 2.132 2.776 3.747 4.604 8.610

5 1.476 2.015 2.571 3.365 4.032 6.869

6 1.440 1.943 2.447 3.143 3.707 5.959

7 1.415 1.895 2.365 2.998 3.499 5.408

8 1.397 1.860 2.306 2.896 3.355 5.041

9 1.383 1.833 2.262 2.821 3.250 4.781

10 1.372 1.812 2.228 2.764 3.169 4.587

11 1.363 1.796 2.201 2.718 3.106 4.437

12 1.356 1.782 2.179 2.681 3.055 4.318

13 1.350 1.771 2.160 2.650 3.012 4.221

14 1.345 1.761 2.145 2.624 2.977 4.140

15 1.341 1.753 2.131 2.602 2.947 4.073

16 1.337 1.746 2.120 2.583 2.921 4.015

17 1.333 1.740 2.110 2.567 2.898 3.965

18 1.330 1.734 2.101 2.552 2.878 3.922

19 1.328 1.729 2.093 2.539 2.861 3.883

20 1.325 1.725 2.086 2.528 2.845 3.850

21 1.323 1.721 2.080 2.518 2.831 3.819

22 1.321 1.717 2.074 2.508 2.819 3.792

23 1.319 1.714 2.069 2.500 2.807 3.768

24 1.318 1.711 2.064 2.492 2.797 3.745

25 1.316 1.708 2.060 2.485 2.787 3.725

26 1.315 1.706 2.056 2.479 2.779 3.707

27 1.314 1.703 2.052 2.473 2.771 3.690

28 1.313 1.701 2.048 2.467 2.763 3.674

29 1.311 1.699 2.045 2.462 2.756 3.659

30 1.310 1.697 2.042 2.457 2.750 3.646

40 1.303 1.684 2.021 2.423 2.704 3.551

60 1 .296 1 .671 2.000 2.390 2.660 3.460

120 1.289 1.658 1.980 2.358 2.617 3.373

1 .282 1 .645 1 .960 2.326 2.576 3.291


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Sampling Distribution for the Statistic t for a

Two-Tailed Test, 0.05 Level of Significance

Critical

value

2.06

Critical

value

-2.06

0.95

Do notreject H0

Region of

rejection

Region of

rejection

0.025 0.025

t-2.06 2.06

df= 25

0


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HYPOTHESIS TESTING INVOLVING PAIRED

OBSERVATIONS

Use the following test when the samples are dependent.

For example, suppose you were collecting data on the price

charged by two different body shops because you suspect

that one is charging more than the other.

In this case, the same wrecked vehicle will be assessed by thetwo shops.

Because of this, the samples will be dependent.

Here we will take the difference of the two estimates and

perform a test on the differences.


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TEST STATISTIC

(9- 4)

d-baris the average of the differences

sd is the standarddeviation of the differences

n is the number of pairs (differences)

td

s nd

!/


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EXAMPLE 3

An independent testing agency is comparing the daily rental

cost for renting a compact car from Hertz and Avis. Arandom sample of eight cities is obtained and the following

rental information obtained. At the 0.05 significance level

can the testing agency conclude that there is a difference in

the rental charged?NOTE: These samples are dependent since the same type of car

(compact) is being rented from the two companies in the

same cities.


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State the null and the alternative hypotheses:

H0:Qd = H1:Qd {

State the decision rule.

H0 is rejected ift 2.365.

Compute the value of the test statistic.

t = (1.00)/[3.162/]= 0.89, (verify).

What is the decision on the null hypothesis?

H0

is not rejected. There is no difference in the charge.


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df .10 .05 .025 .01 .005 .0005


.20 .10 .05 .02 .01 .001

1 3.078 6.314 12.706 31.821 63.657 636.6192 1 .886 2.920 4.303 6.965 9.925 31 .5993 1.638 2.353 3.182 4.541 5.841 12.924

4 1.533 2.132 2.776 3.747 4.604 8.610

5 1.476 2.015 2.571 3.365 4.032 6.869

6 1.440 1.943 2.447 3.143 3.707 5.959

7 1.415 1.895 2.365 2.998 3.499 5.408

8 1.397 1.860 2.306 2.896 3.355 5.041

9 1.383 1.833 2.262 2.821 3.250 4.781

10 1.372 1.812 2.228 2.764 3.169 4.587

11 1.363 1.796 2.201 2.718 3.106 4.437

12 1.356 1.782 2.179 2.681 3.055 4.318

13 1.350 1.771 2.160 2.650 3.012 4.221

14 1.345 1.761 2.145 2.624 2.977 4.140

15 1.341 1.753 2.131 2.602 2.947 4.073

16 1.337 1.746 2.120 2.583 2.921 4.015

17 1.333 1.740 2.110 2.567 2.898 3.965

18 1.330 1.734 2.101 2.552 2.878 3.922

19 1.328 1.729 2.093 2.539 2.861 3.883

20 1.325 1.725 2.086 2.528 2.845 3.850

21 1.323 1.721 2.080 2.518 2.831 3.819

22 1.321 1.717 2.074 2.508 2.819 3.792

23 1.319 1.714 2.069 2.500 2.807 3.768

24 1.318 1.711 2.064 2.492 2.797 3.745

25 1.316 1.708 2.060 2.485 2.787 3.725

26 1.315 1.706 2.056 2.479 2.779 3.707

27 1.314 1.703 2.052 2.473 2.771 3.690

28 1.313 1.701 2.048 2.467 2.763 3.674

29 1.311 1.699 2.045 2.462 2.756 3.659

30 1.310 1.697 2.042 2.457 2.750 3.646

40 1.303 1.684 2.021 2.423 2.704 3.551

60 1 .296 1 .671 2.000 2.390 2.660 3.460

120 1.289 1.658 1.980 2.358 2.617 3.373

1 .282 1 .645 1 .960 2.326 2.576 3.291


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2.3650

Computed

t= 0.89

Rejection

region

-2.365

Rejection

region

0.89

df= 7

t


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Homework for Chapter 11,12,13

Chapter 11 :CD-ROM

Chapter 12 :CD-ROM

Chapter 13 :CD-ROM

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Statistics- Tests of Hypotheses-1