Statistics for Business and Economics Inferences Based on Two Samples: Confidence Intervals & Tests of Hypotheses Chapter 9

Learning Objectives 1. Solve Hypothesis Testing Problems for Two Populations Mean Proportion Variance 2. Distinguish Independent & Related Populations 3. Explain the F Distribution As a result of this class, you will be able to ...

Thinking Challenge How Would You Try to Answer These Questions? D O S Who Gets Higher Grades: Males or Females? Which Programs Are Faster to Learn: Windows or DOS? D O S

Two Population Tests

Independent Sampling & Paired Difference Experiments Testing Two Means Independent Sampling & Paired Difference Experiments 9 26

Two Population Tests

Independent & Related Populations Matching Match according to some characteristic of interest. Repeated Measures Assumes the same individual behaves similarly under both treatments except for treatment effect. Any difference will be due to treatment effect.

Independent & Related Populations 1. Different Data Sources Unrelated Independent Matching Match according to some characteristic of interest. Repeated Measures Assumes the same individual behaves similarly under both treatments except for treatment effect. Any difference will be due to treatment effect.

Independent & Related Populations 1. Different Data Sources Unrelated Independent 1. Same Data Source Paired or Matched Repeated Measures (Before/After) Matching Match according to some characteristic of interest. Repeated Measures Assumes the same individual behaves similarly under both treatments except for treatment effect. Any difference will be due to treatment effect.

Independent & Related Populations 1. Different Data Sources Unrelated Independent 2. Use Difference Between the 2 Sample Means X1 -X2 1. Same Data Source Paired or Matched Repeated Measures (Before/After) Matching Match according to some characteristic of interest. Repeated Measures Assumes the same individual behaves similarly under both treatments except for treatment effect. Any difference will be due to treatment effect.

Independent & Related Populations 1. Different Data Sources Unrelated Independent 2. Use Difference Between the 2 Sample Means X1 -X2 1. Same Data Source Paired or Matched Repeated Measures (Before/After) 2. Use Difference Between Each Pair of Observations Di = X1i - X2i Matching Match according to some characteristic of interest. Repeated Measures Assumes the same individual behaves similarly under both treatments except for treatment effect. Any difference will be due to treatment effect.

Two Independent Populations Examples 1. An economist wishes to determine whether there is a difference in mean family income for households in 2 socioeconomic groups. 2. An admissions officer of a small liberal arts college wants to compare the mean SAT scores of applicants educated in rural high schools & in urban high schools. These are comparative studies. The general purpose of comparative studies is to establish similarities or to detect and measure differences between populations. The populations can be (1) existing populations or (2) hypothetical populations.

Two Related Populations Examples 1. Nike wants to see if there is a difference in durability of 2 sole materials. One type is placed on one shoe, the other type on the other shoe of the same pair. 2. An analyst for Educational Testing Service wants to compare the mean GMAT scores of students before & after taking a GMAT review course.

Thinking Challenge Are They Independent or Paired? 1. Miles per gallon ratings of cars before & after mounting radial tires 2. The life expectancy of light bulbs made in 2 different factories 3. Difference in hardness between 2 metals: one contains an alloy, one doesn’t 4. Tread life of two different motorcycle tires: one on the front, the other on the back 1. Related. Same car should be used since car weights vary. 2. Independent. Can’t match light bulbs. 3. Independent. No relationship between two alloys. Only differ in alloy. 4. Related. Tires are matched to same motorcycle. Should alternate tire types between front and back and motorcycle.

Testing 2 Independent Means 9 26

Two Population Tests

Two Independent Populations Hypotheses for Means 9

Two Independent Populations Hypotheses for Means 30

Two Independent Populations Hypotheses for Means 30

Sampling Distribution In this diagram, do the populations have equal or unequal variances? Unequal. 31

Sampling Distribution  Population 1 1  1 In this diagram, do the populations have equal or unequal variances? Unequal. 32

Sampling Distribution   Population 1 2 Population 1 2    2 1 In this diagram, do the populations have equal or unequal variances? Unequal. 33

Sampling Distribution   Population 1 2 Population 1 2    2 1 In this diagram, do the populations have equal or unequal variances? Unequal. Select simple random sample, size n . 1 Compute  X 1 34

Sampling Distribution   Population 1 2 Population 1 2    2 1 In this diagram, do the populations have equal or unequal variances? Unequal. Select simple random Select simple random sample, size n . sample, size n . 1 2 Compute  X Compute  X 1 2 35

Sampling Distribution   Population 1 2 Population 1 2    2 1 In this diagram, do the populations have equal or unequal variances? Unequal. Select simple random Compute  X -  X Select simple random 1 2 sample, size n . for every pair sample, size n . 1 2 Compute  X of samples Compute  X 1 2 36

Sampling Distribution   Population 1 2 Population 1 2    2 1 In this diagram, do the populations have equal or unequal variances? Unequal. Select simple random Compute  X -  X Select simple random 1 2 sample, size n . for every pair sample, size n . 1 2 Compute  X of samples Compute  X 1 2 Astronomical number of  X -  X values 1 2 37

Sampling Distribution In this diagram, do the populations have equal or unequal variances? Unequal. 38

Large-Sample Z Test for 2 Independent Means 9 26

Two Population Tests

Large-Sample Z Test for 2 Independent Means

Large-Sample Z Test for 2 Independent Means 1. Assumptions Independent, Random Samples Populations Are Normally Distributed If Not Normal, Can Be Approximated by Normal Distribution (n1  30 & n2  30 )

Large-Sample Z Test for 2 Independent Means 1. Assumptions Independent, Random Samples Populations Are Normally Distributed If Not Normal, Can Be Approximated by Normal Distribution (n1  30 & n2  30 ) 2. Two Independent Sample Z-Test Statistic  X n    1 2   X n    1 2 s Z

Large-Sample Z Test Example You’re a financial analyst for Charles Schwab. You want to find out if there is a difference in dividend yield between stocks listed on NYSE & NASDAQ. You collect the following data: NYSE NASDAQ Number 121 125 Mean 3.27 2.53 Std Dev 1.30 1.16 Is there a difference in average yield ( = .05)? © 1984-1994 T/Maker Co.

Large-Sample Z Test Solution

Large-Sample Z Test Solution H0: Ha:   n1 = , n2 = Critical Value(s): Test Statistic: Decision: Conclusion:

Large-Sample Z Test Solution H0: 1 - 2 = 0 (1 = 2) Ha: 1 - 2  0 (1  2)   n1 = , n2 = Critical Value(s): Test Statistic: Decision: Conclusion:

Large-Sample Z Test Solution H0: 1 - 2 = 0 (1 = 2) Ha: 1 - 2  0 (1  2)   .05 n1 = 121, n2 = 125 Critical Value(s): Test Statistic: Decision: Conclusion:

Large-Sample Z Test Solution H0: 1 - 2 = 0 (1 = 2) Ha: 1 - 2  0 (1  2)   .05 n1 = 121, n2 = 125 Critical Value(s): Test Statistic: Decision: Conclusion: Reject H Reject H .025 z -1.96 1.96

Large-Sample Z Test Solution H0: 1 - 2 = 0 (1 = 2) Ha: 1 - 2  0 (1  2)   .05 n1 = 121, n2 = 125 Critical Value(s): Test Statistic: Decision: Conclusion: 3 . 27  2 . 53 z    4 . 69 1 . 698 1 . 353  121 125 Reject H Reject H .025 z -1.96 1.96

Large-Sample Z Test Solution H0: 1 - 2 = 0 (1 = 2) Ha: 1 - 2  0 (1  2)   .05 n1 = 121, n2 = 125 Critical Value(s): Test Statistic: Decision: Conclusion: 3 . 27  2 . 53 z    4 . 69 1 . 698 1 . 353  121 125 Reject H Reject H Reject at  = .05 .025 .025 z -1.96 1.96

Large-Sample Z Test Solution H0: 1 - 2 = 0 (1 = 2) Ha: 1 - 2  0 (1  2)   .05 n1 = 121, n2 = 125 Critical Value(s): Test Statistic: Decision: Conclusion: 3 . 27  2 . 53 z    4 . 69 1 . 698 1 . 353  121 125 Reject H Reject H Reject at  = .05 .025 .025 There is Evidence of a Difference in Means z -1.96 1.96

Large-Sample Z Test Thinking Challenge You’re an economist for the Department of Education. You want to find out if there is a difference in spending per pupil between urban & rural high schools. You collect the following: Urban Rural Number 35 35 Mean $ 6,012 $ 5,832 Std Dev $ 602 $ 497 Is there any difference in population means ( = .10)? Allow students about 15 minutes to solve.

Large-Sample Z Test Solution* H0: 1 - 2 = 0 (1 = 2) Ha: 1 - 2  0 (1  2)   .10 n1 = 35, n2 = 35 Critical Value(s): Test Statistic: Decision: Conclusion: 6012  5832  z    1 . 36 2 2 602 497  35 35 Reject H Reject H Do Not Reject at  = .10 .05 .05 There is No Evidence of a Difference in Means z -1.645 1.645

Small-Sample t Test for 2 Independent Means 9 26

Two Population Tests

Small-Sample t Test for 2 Independent Means 1. Tests Means of 2 Independent Populations Having Equal Variances 2. Assumptions Independent, Random Samples Both Populations Are Normally Distributed Population Variances Are Unknown But Assumed Equal

Small-Sample t Test Test Statistic Hypothesized difference 39

Small-Sample t Test Example You’re a financial analyst for Charles Schwab. Is there a difference in dividend yield between stocks listed on the NYSE & NASDAQ? You collect the following data: NYSE NASDAQ Number 21 25 Mean 3.27 2.53 Std Dev 1.30 1.16 Assuming normal populations, is there a difference in average yield ( = .05)? © 1984-1994 T/Maker Co.

Small-Sample t Test Solution 41

Small-Sample t Test Solution H0: Ha:   df  Critical Value(s): Test Statistic: Decision: Conclusion: 41

Small-Sample t Test Solution H0: 1 - 2 = 0 (1 = 2) Ha: 1 - 2  0 (1  2)   df  Critical Value(s): Test Statistic: Decision: Conclusion: 41

Small-Sample t Test Solution H0: 1 - 2 = 0 (1 = 2) Ha: 1 - 2  0 (1  2)   .05 df  21 + 25 - 2 = 44 Critical Value(s): Test Statistic: Decision: Conclusion: 41

Small-Sample t Test Solution H0: 1 - 2 = 0 (1 = 2) Ha: 1 - 2  0 (1  2)   .05 df  21 + 25 - 2 = 44 Critical Value(s): Test Statistic: Decision: Conclusion: 41

Small-Sample t Test Solution 42

Small-Sample t Test Solution H0: 1 - 2 = 0 (1 = 2) Ha: 1 - 2  0 (1  2)   .05 df  21 + 25 - 2 = 44 Critical Value(s): Test Statistic: Decision: Conclusion: 41

Small-Sample t Test Solution H0: 1 - 2 = 0 (1 = 2) Ha: 1 - 2  0 (1  2)   .05 df  21 + 25 - 2 = 44 Critical Value(s): Test Statistic: Decision: Conclusion: Reject at  = .05 41

Small-Sample t Test Solution H0: 1 - 2 = 0 (1 = 2) Ha: 1 - 2  0 (1  2)   .05 df  21 + 25 - 2 = 44 Critical Value(s): Test Statistic: Decision: Conclusion: Reject at  = .05 There is evidence of a difference in means 41

Small-Sample t Test Thinking Challenge You’re a research analyst for General Motors. Assuming equal variances, is there a difference in the average miles per gallon (mpg) of two car models ( = .05)? You collect the following: Sedan Van Number 15 11 Mean 22.00 20.27 Std Dev 4.77 3.64 Allow students about 15 minutes to solve. 43

Small-Sample t Test Solution* H0: 1 - 2 = 0 (1 = 2) Ha: 1 - 2  0 (1  2)   .05 df  15 + 11 - 2 = 24 Critical Value(s): Test Statistic: Decision: Conclusion: Do not reject at  = .05 There is no evidence of a difference in means 45

Small-Sample t Test Solution* 46

Paired Difference Experiments Paired-Sample t Test Paired Difference Experiments 9 26

Two Population Tests

Paired-Sample t Test for Mean Difference 1. Tests Means of 2 Related Populations Paired or Matched Repeated Measures (Before/After) 2. Eliminates Variation Among Subjects 3. Assumptions Both Population Are Normally Distributed If Not Normal, Can Be Approximated by Normal Distribution (n1  30 & n2  30 )

Paired-Sample t Test Hypotheses Research Questions No Difference Pop 1  Pop 2 Pop 1  Pop 2 Hypothesis Any Difference Pop 1 < Pop 2 Pop 1 > Pop 2 H    = 0   D D D H     < 0 > 0 1 D D D Note: Di = X1i - X2i for ith observation

Paired-Sample t Test Data Collection Table Observation Group 1 Group 2 Difference 1 x x D = x -x 11 21 1 11 21 2 x x D = x -x 12 22 2 12 22     i x x D = x - x 1i 2i i 1i 2i     n x x D = x - x 1n 2n n 1n 2n

Paired-Sample t Test Test Statistic xD  D t  df  n  1 D S D n Just take the mean and standard deviation of the difference. SD is simply the standard deviation. The formula is the computational formula. D Sample Mean Sample Standard Deviation n n   D (Di - xD)2 i x  i  1 S  i  1 D D n n  1 D D

Paired-Sample t Test Example You work in Human Resources. You want to see if a training program is effective. You collect the following test score data: Name Before (1) After (2) Sam 85 94 Tamika 94 87 Brian 78 79 Mike 87 88 At the .10 level, was the training effective?

Computation Table Observation Before After Difference Sam 85 94 -9 Tamika 94 87 7 Brian 78 79 -1 Mike 87 88 -1 Total - 4

Null Hypothesis Solution 1. Was the training effective? 2. Effective means ‘After’ > ‘Before’. 3. Statistically, this means A > B. 4. Rearranging terms gives 0 B - A. 5. Defining D = B - A & substituting into (4) gives 0 D or D . 6. The alternative hypothesis is Ha: D 0.

Paired-Sample t Test Solution

Paired-Sample t Test Solution H0: Ha:  = df = Critical Value(s): Test Statistic: Decision: Conclusion:

Paired-Sample t Test Solution H0: D = 0 (D = B - A) Ha: D < 0  = df = Critical Value(s): Test Statistic: Decision: Conclusion:

Paired-Sample t Test Solution H0: D = 0 (D = B - A) Ha: D < 0  = .10 df = 4 - 1 = 3 Critical Value(s): Test Statistic: Decision: Conclusion:

Paired-Sample t Test Solution H0: D = 0 (D = B - A) Ha: D < 0  = .10 df = 4 - 1 = 3 Critical Value(s): Test Statistic: Decision: Conclusion: Reject .10 -1.6377 t

Paired-Sample t Test Solution H0: D = 0 (D = B - A) Ha: D < 0  = .10 df = 4 - 1 = 3 Critical Value(s): Test Statistic: Decision: Conclusion: x  D  1  D t     . 306 S 6 . 53 D n 4 D Reject .10 -1.6377 t

Paired-Sample t Test Solution H0: D = 0 (D = B - A) Ha: D < 0  = .10 df = 4 - 1 = 3 Critical Value(s): Test Statistic: Decision: Conclusion: x  D  1  D t     . 306 S 6 . 53 D n 4 D Reject Do Not Reject at  = .10 .10 -1.6377 t

Paired-Sample t Test Solution H0: D = 0 (D = B - A) Ha: D < 0  = .10 df = 4 - 1 = 3 Critical Value(s): Test Statistic: Decision: Conclusion: x  D  1  D t     . 306 S 6 . 53 D n 4 D Reject Do Not Reject at  = .10 .10 There Is No Evidence Training Was Effective -1.6377 t

Paired-Sample t Test Thinking Challenge (1) (2) Store Client Competitor 1 $ 10 $ 11 2 8 11 3 7 10 4 9 12 5 11 11 6 10 13 7 9 12 8 8 10 You’re a marketing research analyst. You want to compare a client’s calculator to a competitor’s. You sample 8 retail stores. At the .01 level, does your client’s calculator sell for less than their competitor’s Why related populations? Control for differences in store price. Some stores might be higher priced in terms of all goods. Allow students about 15 minutes to solve this.

Paired-Sample t Test Solution* H0: D = 0 (D = 1 - 2) Ha: D < 0  = .01 df = 8 - 1 = 7 Critical Value(s): Test Statistic: Decision: Conclusion: x  D  2 . 25  D t     5 . 486 S 1 . 16 D n 8 D Reject Reject at  = .01 .01 There Is Evidence Client’s Brand (1) Sells for Less -2.998 t

Z Test for Differences in Two Proportions 9 26

Two Population Tests

Z Test for Difference in Two Proportions

Z Test for Difference in Two Proportions 1. Assumptions Populations Are Independent Populations Follow Binomial Distribution Normal Approximation Can Be Used Does Not Contain 0 or n

Z Test for Difference in Two Proportions 1. Assumptions Populations Are Independent Populations Follow Binomial Distribution Normal Approximation Can Be Used Does Not Contain 0 or n 2. Z-Test Statistic for Two Proportions

Hypotheses for Two Proportions 9

Hypotheses for Two Proportions 9

Hypotheses for Two Proportions 9

Hypotheses for Two Proportions 9

Hypotheses for Two Proportions 9

Hypotheses for Two Proportions 9

Z Test for Two Proportions Example As personnel director, you want to test the perception of fairness of two methods of performance evaluation. 63 of 78 employees rated Method 1 as fair. 49 of 82 rated Method 2 as fair. At the .01 level, is there a difference in perceptions? To check assumptions, use sample proportions as estimators of population proportion: n1·p = 78·63/78 = 63 n1·(1-p) = 78·(1-63/78) = 15

Z Test for Two Proportions Solution 11

Z Test for Two Proportions Solution H0: Ha:  = n1 = n1 = Critical Value(s): Test Statistic: Decision: Conclusion: 11

Z Test for Two Proportions Solution H0: p1 - p2 = 0 Ha: p1 - p2  0  = n1 = n1 = Critical Value(s): Test Statistic: Decision: Conclusion: 11

Z Test for Two Proportions Solution H0: p1 - p2 = 0 Ha: p1 - p2  0  = .01 n1 = 78 n1 = 82 Critical Value(s): Test Statistic: Decision: Conclusion: 11

Z Test for Two Proportions Solution H0: p1 - p2 = 0 Ha: p1 - p2  0  = .01 n1 = 78 n1 = 82 Critical Value(s): Test Statistic: Decision: Conclusion: 11

Z Test for Two Proportions Solution 12

Z Test for Two Proportions Solution H0: p1 - p2 = 0 Ha: p1 - p2  0  = .01 n1 = 78 n1 = 82 Critical Value(s): Test Statistic: Decision: Conclusion: 11

Z Test for Two Proportions Solution H0: p1 - p2 = 0 Ha: p1 - p2  0  = .01 n1 = 78 n1 = 82 Critical Value(s): Test Statistic: Decision: Conclusion: Reject at  = .01 11

Z Test for Two Proportions Solution H0: p1 - p2 = 0 Ha: p1 - p2  0  = .01 n1 = 78 n1 = 82 Critical Value(s): Test Statistic: Decision: Conclusion: Reject at  = .01 There is evidence of a difference in proportions 11

Z Test for Two Proportions Thinking Challenge You’re an economist for the Department of Labor. You’re studying unemployment rates. In MA, 74 of 1500 people surveyed were unemployed. In CA, 129 of 1500 were unemployed. At the .05 level, does MA have a lower unemployment rate? MA CA

Z Test for Two Proportions Solution* 15

Z Test for Two Proportions Solution* H0: Ha:  = nMA = nCA = Critical Value(s): Test Statistic: Decision: Conclusion: 15

Z Test for Two Proportions Solution* H0: pMA - pCA = 0 Ha: pMA - pCA < 0  = nMA = nCA = Critical Value(s): Test Statistic: Decision: Conclusion: 15

Z Test for Two Proportions Solution* H0: pMA - pCA = 0 Ha: pMA - pCA < 0  = .05 nMA = 1500 nCA = 1500 Critical Value(s): Test Statistic: Decision: Conclusion: 15

Z Test for Two Proportions Solution* H0: pMA - pCA = 0 Ha: pMA - pCA < 0  = .05 nMA = 1500 nCA = 1500 Critical Value(s): Test Statistic: Decision: Conclusion: 15

Z Test for Two Proportions Solution* 16

Z Test for Two Proportions Solution* H0: pMA - pCA = 0 Ha: pMA - pCA < 0  = .05 nMA = 1500 nCA = 1500 Critical Value(s): Test Statistic: Decision: Conclusion: Z = -4.00 15

Z Test for Two Proportions Solution* H0: pMA - pCA = 0 Ha: pMA - pCA < 0  = .05 nMA = 1500 nCA = 1500 Critical Value(s): Test Statistic: Decision: Conclusion: Z = -4.00 Reject at  = .05 15

Z Test for Two Proportions Solution* H0: pMA - pCA = 0 Ha: pMA - pCA < 0  = .05 nMA = 1500 nCA = 1500 Critical Value(s): Test Statistic: Decision: Conclusion: Z = -4.00 Reject at  = .05 There is evidence MA is less than CA 15

F-Test for Two Population Variances 9 26

Two Population Tests

F-Test for Two Variances 1. Tests for Differences in 2 Population Variances 2. Assumptions Both Populations Are Normally Distributed Test Is Not Robust to Violations Independent, Random Samples Key words that show a test on variances, not on means or proportions: variance, standard deviation, variability, variation, scatter, dispersion.

F-Test for Variances Hypotheses H0: 12 = 22 OR H0: 12  22 (or ) Ha: 12  22 Ha: 12 22 (or >) 2. Test Statistic F = s12 /s22 Two Sets of Degrees of Freedom 1 = n1 - 1; 2 = n2 - 1 Follows F Distribution F ratio is a statistic defined as the ratio of 2 independent estimates of a normally distributed population’s variance. Note: degrees of freedom refer to numerator and denominator

F Distribution The sampling distribution is a function of the sample sizes upon which the sample variances are based. Hint: Recall the formula for variance! s2 = (x -x)2/(n-1) 58

F Distribution   Population 1 1 1 The sampling distribution is a function of the sample sizes upon which the sample variances are based. Hint: Recall the formula for variance! s2 = (x -x)2/(n-1) 59

F Distribution     Population Population 1 2 1 2 1 2 The sampling distribution is a function of the sample sizes upon which the sample variances are based. Hint: Recall the formula for variance! s2 = (x -x)2/(n-1) 60

F Distribution     Population Population 1 2 The sampling distribution is a function of the sample sizes upon which the sample variances are based. Hint: Recall the formula for variance! s2 = (x -x)2/(n-1) Select simple random sample, size n 1 Compute S 2 1 61

F Distribution     Population Population 1 2 The sampling distribution is a function of the sample sizes upon which the sample variances are based. Hint: Recall the formula for variance! s2 = (x -x)2/(n-1) Select simple random Select simple random sample, size n sample, size n 1 2 Compute S 2 Compute S 2 1 2 62

F Distribution     Population Population 1 2 The sampling distribution is a function of the sample sizes upon which the sample variances are based. Hint: Recall the formula for variance! s2 = (x -x)2/(n-1) Select simple random Compute F =  S 2 /S 2 Select simple random 1 2 sample, size n for every pair of sample, size n 1 2 Compute S 2 n & n size samples Compute S 2 1 1 2 2 63

F Distribution     Population Population 1 2 The sampling distribution is a function of the sample sizes upon which the sample variances are based. Hint: Recall the formula for variance! s2 = (x -x)2/(n-1) Select simple random Compute F =  S 2 /S 2 Select simple random 1 2 sample, size n for every pair of sample, size n 1 2 Compute S 2 n & n size samples Compute S 2 1 1 2 2 Astronomical number of S 2 /S 2 values 1 2 64

F Distribution The sampling distribution is a function of the sample sizes upon which the sample variances are based. Hint: Recall the formula for variance! s2 = (x -x)2/(n-1) 65

F-Test for 2 Variances Critical Values The lower F is not the reciprocal of the upper F. What do you do if equal sample sizes? /2 /2 Note! 66

F-Test for Variances Example You’re a financial analyst for Charles Schwab. You want to compare dividend yields between stocks listed on the NYSE & NASDAQ. You collect the following data: NYSE NASDAQ Number 21 25 Mean 3.27 2.53 Std Dev 1.30 1.16 Is there a difference in variances between the NYSE & NASDAQ at the .05 level? © 1984-1994 T/Maker Co.

F-Test for 2 Variances Solution 68

F-Test for 2 Variances Solution H0: Ha:   1  2  Critical Value(s): Test Statistic: Decision: Conclusion: 68

F-Test for 2 Variances Solution H0: 12 = 22 Ha: 12  22   1  2  Critical Value(s): Test Statistic: Decision: Conclusion: 68

F-Test for 2 Variances Solution H0: 12 = 22 Ha: 12  22   .05 1  20 2  24 Critical Value(s): Test Statistic: Decision: Conclusion: 68

F-Test for 2 Variances Solution H0: 12 = 22 Ha: 12  22   .05 1  20 2  24 Critical Value(s): Test Statistic: Decision: Conclusion: 68

F-Test for 2 Variances Solution /2 = .025 /2 = .025 69

F-Test for 2 Variances Solution H0: 12 = 22 Ha: 12  22   .05 1  20 2  24 Critical Value(s): Test Statistic: Decision: Conclusion: 68

F-Test for 2 Variances Solution H0: 12 = 22 Ha: 12  22   .05 1  20 2  24 Critical Value(s): Test Statistic: Decision: Conclusion: Do not reject at  = .05 68

F-Test for 2 Variances Solution H0: 12 = 22 Ha: 12  22   .05 1  20 2  24 Critical Value(s): Test Statistic: Decision: Conclusion: Do not reject at  = .05 There is no evidence of a difference in variances 68

F-Test for Variances Thinking Challenge You’re an analyst for the Light & Power Company. You want to compare the electricity consumption of single-family homes in 2 towns. You compute the following from a sample of homes: Town 1 Town 2 Number 25 21 Mean $ 85 $ 68 Std Dev $ 30 $ 18 At the .05 level, is there evidence of a difference in variances between the two towns? Allow students about 15 minutes to solve this.

F-Test for 2 Variances Solution* H0: 12 = 22 Ha: 12  22   .05 1  24 2  20 Critical Value(s): Test Statistic: Decision: Conclusion: Reject at  = .05 There is evidence of a difference in variances 72

Critical Values Solution* /2 = .025 /2 = .025 73

Conclusion 1. Solved Hypothesis Testing Problems for Two Populations Mean Proportion Variance 2. Distinguished Independent & Related Populations 3. Explained the F Distribution As a result of this class, you will be able to ...

Any blank slides that follow are blank intentionally. End of Chapter Any blank slides that follow are blank intentionally.