1

Tests of Hypotheses:

Small Samples

Tests of Hypotheses:

Small SamplesChapter

Rejectionregion

2CHAPTER GOALSCHAPTER GOALSTO DESCRIBE THE MAJOR

CHARACTERISTICS OF STUDENT’S t-DISTRIBUTION.

TO UNDERSTAND THE DIFFERENCE BETWEEN THE t -DISTRIBUTION AND THE z -DISTRIBUTION.

TO DESCRIBE THE MAJOR CHARACTERISTICS OF STUDENT’S t-DISTRIBUTION.

TO UNDERSTAND THE DIFFERENCE BETWEEN THE t -DISTRIBUTION AND THE z -DISTRIBUTION.

3CHAPTER GOALSCHAPTER GOALSTO 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 THE DIFFERENCE BETWEEN A SET OF PAIRED OBSERVATIONS.

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 THE DIFFERENCE BETWEEN A SET OF PAIRED OBSERVATIONS.

4The t-distribution has the following properties:It is continuous, bell shaped and symmetrical

about zero like the z-distribution.There is a family of t-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.

The t-distribution has the following properties:It is continuous, bell shaped and symmetrical

about zero like the z-distribution.There is a family of t-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.

CHARACTERISTICS OF STUDENT’S t-DISTRIBUTION

CHARACTERISTICS OF STUDENT’S t-DISTRIBUTION

5z-distribution

t-distribution

The degrees of freedom forthe t-distribution is df = n - 1.

6A TEST FOR A POPULATION MEAN: SMALL SAMPLE, POPULATION

STANDARD DEVIATION UNKNOWN

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)

The test statistic for a one sample case is given by equation (9-1) below

(9-1) t X

nS

/

7The current rate for producing 5 amp fuses at

Monarch Electric Company is 250 per hour. A new machine has been purchased 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 sample standard deviation of 6 per hour. At the 0.05 significance level can Monarch conclude that the new machine is faster?

The current rate for producing 5 amp fuses at Monarch Electric Company is 250 per hour. A new machine has been purchased 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 sample standard deviation of 6 per hour. At the 0.05 significance level can Monarch conclude that the new machine is faster?

EXAMPLE 1EXAMPLE 1

8EXAMPLE 1 (continued)EXAMPLE 1 (continued)

Step 1: State the null and the alternative hypotheses.

H0: 250 H1: 250

Step 2: State the decision rule.H0 is rejected if 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.

Step 1: State the null and the alternative hypotheses.

H0: 250 H1: 250

Step 2: State the decision rule.H0 is rejected if 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.

9STUDENT t DISTRIBUTION

Level of significance for one-tailed testdf .10 .05 .025 .01 .005 .0005

Level of significance for two-tailed test.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 12.9244 1.533 2.132 2.776 3.747 4.604 8.6105 1.476 2.015 2.571 3.365 4.032 6.8696 1.440 1.943 2.447 3.143 3.707 5.9597 1.415 1.895 2.365 2.998 3.499 5.4088 1.397 1.860 2.306 2.896 3.355 5.0419 1.383 1.833 2.262 2.821 3.250 4.781

10 1.372 1.812 2.228 2.764 3.169 4.58711 1.363 1.796 2.201 2.718 3.106 4.43712 1.356 1.782 2.179 2.681 3.055 4.31813 1.350 1.771 2.160 2.650 3.012 4.22114 1.345 1.761 2.145 2.624 2.977 4.14015 1.341 1.753 2.131 2.602 2.947 4.07316 1.337 1.746 2.120 2.583 2.921 4.01517 1.333 1.740 2.110 2.567 2.898 3.96518 1.330 1.734 2.101 2.552 2.878 3.92219 1.328 1.729 2.093 2.539 2.861 3.88320 1.325 1.725 2.086 2.528 2.845 3.85021 1.323 1.721 2.080 2.518 2.831 3.81922 1.321 1.717 2.074 2.508 2.819 3.79223 1.319 1.714 2.069 2.500 2.807 3.76824 1.318 1.711 2.064 2.492 2.797 3.74525 1.316 1.708 2.060 2.485 2.787 3.72526 1.315 1.706 2.056 2.479 2.779 3.70727 1.314 1.703 2.052 2.473 2.771 3.69028 1.313 1.701 2.048 2.467 2.763 3.67429 1.311 1.699 2.045 2.462 2.756 3.65930 1.310 1.697 2.042 2.457 2.750 3.64640 1.303 1.684 2.021 2.423 2.704 3.55160 1 .296 1 .671 2.000 2.390 2.660 3.460

120 1.289 1.658 1.980 2.358 2.617 3.3731 .282 1 .645 1 .960 2.326 2.576 3.291

10Display of the Rejection Region, Critical Value, and the computed

Test Statistic

Display of the Rejection Region, Critical Value, and the computed

Test Statistic

Region ofrejection

t

Critical value1.833

0.05

df = 9

11To 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.

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.

COMPARING TWO POPULATIONS MEANS

COMPARING TWO POPULATIONS MEANS

12POOLED SAMPLE VARIANCE AND TEST STATISTIC

POOLED SAMPLE VARIANCE AND TEST STATISTIC

Pooled Sample Variance

sn s n s

n n

Test Statistic

tX X

s n n

p

p

=

1

2 1 12

2 22

1 2

2

21 2

1 12

1 1

( ) ( )

(9 -3)

(9 -2)

13EXAMPLE 2EXAMPLE 2

A recent Environmental Protection Agency (EPA) study compared the highway fuel economy of domestic and imported passengers cars. A sample of 15 domestic cars revealed a mean of 33.7 mpg with a standard deviation of 2.4 mpg. A sample of 12 imported cars revealed a mean of 35.7 mpg with a standard deviation of 3.9. At the 0.05 significance level can the EPA conclude that the miles per gallon is higher on the imported cars? (Let subscript 1 be associated with the domestic cars).

A recent Environmental Protection Agency (EPA) study compared the highway fuel economy of domestic and imported passengers cars. A sample of 15 domestic cars revealed a mean of 33.7 mpg with a standard deviation of 2.4 mpg. A sample of 12 imported cars revealed a mean of 35.7 mpg with a standard deviation of 3.9. At the 0.05 significance level can the EPA conclude that the miles per gallon is higher on the imported cars? (Let subscript 1 be associated with the domestic cars).

14Step 1: State the null and the alternative

hypotheses.H0: H1:

Step 2: State the decision rule.H0 is rejected if 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.

Step 1: State the null and the alternative hypotheses.

H0: H1:

Step 2: State the decision rule.H0 is rejected if 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.

EXAMPLE 2 (continued)EXAMPLE 2 (continued)

15STUDENT t DISTRIBUTION

Level of significance for one-tailed testdf .10 .05 .025 .01 .005 .0005

Level of significance for two-tailed test.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 12.9244 1.533 2.132 2.776 3.747 4.604 8.6105 1.476 2.015 2.571 3.365 4.032 6.8696 1.440 1.943 2.447 3.143 3.707 5.9597 1.415 1.895 2.365 2.998 3.499 5.4088 1.397 1.860 2.306 2.896 3.355 5.0419 1.383 1.833 2.262 2.821 3.250 4.781

10 1.372 1.812 2.228 2.764 3.169 4.58711 1.363 1.796 2.201 2.718 3.106 4.43712 1.356 1.782 2.179 2.681 3.055 4.31813 1.350 1.771 2.160 2.650 3.012 4.22114 1.345 1.761 2.145 2.624 2.977 4.14015 1.341 1.753 2.131 2.602 2.947 4.07316 1.337 1.746 2.120 2.583 2.921 4.01517 1.333 1.740 2.110 2.567 2.898 3.96518 1.330 1.734 2.101 2.552 2.878 3.92219 1.328 1.729 2.093 2.539 2.861 3.88320 1.325 1.725 2.086 2.528 2.845 3.85021 1.323 1.721 2.080 2.518 2.831 3.81922 1.321 1.717 2.074 2.508 2.819 3.79223 1.319 1.714 2.069 2.500 2.807 3.76824 1.318 1.711 2.064 2.492 2.797 3.74525 1.316 1.708 2.060 2.485 2.787 3.72526 1.315 1.706 2.056 2.479 2.779 3.70727 1.314 1.703 2.052 2.473 2.771 3.69028 1.313 1.701 2.048 2.467 2.763 3.67429 1.311 1.699 2.045 2.462 2.756 3.65930 1.310 1.697 2.042 2.457 2.750 3.64640 1.303 1.684 2.021 2.423 2.704 3.55160 1 .296 1 .671 2.000 2.390 2.660 3.460

120 1.289 1.658 1.980 2.358 2.617 3.3731 .282 1 .645 1 .960 2.326 2.576 3.291

16Sampling Distribution for the Statistic t for a Two-Tailed Test, 0.05 Level of

Significance

Sampling Distribution for the Statistic t for a Two-Tailed Test, 0.05 Level of

Significance

Criticalvalue2.06

Criticalvalue-2.06

0.95

Do notreject H0

Region ofrejectionRegion of

rejection

0.025 0.025

t-2.06 2.06

df = 25

0

17Use 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 the two 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.

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 the two 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.

HYPOTHESIS TESTING INVOLVING PAIRED OBSERVATIONS

HYPOTHESIS TESTING INVOLVING PAIRED OBSERVATIONS

18TEST STATISTICTEST STATISTIC

(9 - 4) (9 - 4)

d-bar is the average of the differences

sd is the standard deviation of the differences

n is the number of pairs (differences)

t ds nd

/

19An independent testing agency is comparing the

daily rental cost for renting a compact car from Hertz and Avis. A random 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.

An independent testing agency is comparing the daily rental cost for renting a compact car from Hertz and Avis. A random 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.

EXAMPLE 3EXAMPLE 3

20

EXAMPLE 3 (continued)EXAMPLE 3 (continued)

21State the null and the alternative hypotheses:H0:d = H1:d

State the decision rule.H0 is rejected if t < -2.365 or t > 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.

State the null and the alternative hypotheses:H0:d = H1:d

State the decision rule.H0 is rejected if t < -2.365 or t > 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.

EXAMPLE 3 (continued)EXAMPLE 3 (continued)

22STUDENT t DISTRIBUTION

Level of significance for one-tailed testdf .10 .05 .025 .01 .005 .0005

Level of significance for two-tailed test.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 12.9244 1.533 2.132 2.776 3.747 4.604 8.6105 1.476 2.015 2.571 3.365 4.032 6.8696 1.440 1.943 2.447 3.143 3.707 5.9597 1.415 1.895 2.365 2.998 3.499 5.4088 1.397 1.860 2.306 2.896 3.355 5.0419 1.383 1.833 2.262 2.821 3.250 4.781

10 1.372 1.812 2.228 2.764 3.169 4.58711 1.363 1.796 2.201 2.718 3.106 4.43712 1.356 1.782 2.179 2.681 3.055 4.31813 1.350 1.771 2.160 2.650 3.012 4.22114 1.345 1.761 2.145 2.624 2.977 4.14015 1.341 1.753 2.131 2.602 2.947 4.07316 1.337 1.746 2.120 2.583 2.921 4.01517 1.333 1.740 2.110 2.567 2.898 3.96518 1.330 1.734 2.101 2.552 2.878 3.92219 1.328 1.729 2.093 2.539 2.861 3.88320 1.325 1.725 2.086 2.528 2.845 3.85021 1.323 1.721 2.080 2.518 2.831 3.81922 1.321 1.717 2.074 2.508 2.819 3.79223 1.319 1.714 2.069 2.500 2.807 3.76824 1.318 1.711 2.064 2.492 2.797 3.74525 1.316 1.708 2.060 2.485 2.787 3.72526 1.315 1.706 2.056 2.479 2.779 3.70727 1.314 1.703 2.052 2.473 2.771 3.69028 1.313 1.701 2.048 2.467 2.763 3.67429 1.311 1.699 2.045 2.462 2.756 3.65930 1.310 1.697 2.042 2.457 2.750 3.64640 1.303 1.684 2.021 2.423 2.704 3.55160 1 .296 1 .671 2.000 2.390 2.660 3.460

120 1.289 1.658 1.980 2.358 2.617 3.3731 .282 1 .645 1 .960 2.326 2.576 3.291

23

2.3650

Computedt = 0.89

Rejectionregion

-2.365

Rejectionregion

0.89

df = 7

t

24

Chapter 11 : CD-ROM

Chapter 12 : CD-ROM

Chapter 13 : CD-ROM

Chapter 11 : CD-ROM

Chapter 12 : CD-ROM

Chapter 13 : CD-ROM

Homework for Chapter 11,12,13 Homework for Chapter 11,12,13