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Chapter Thirteen Hypothesis Testing

Copyright © Houghton Mifflin Company. All rights reserved.13 | 3 Interesting Hypotheses Bankers assumed high-income earners are more profitable than low-income earners Clients who carefully balance their checkbooks every month and minimize fees due to overdrafts are unprofitable checking account customers Old clients were more likely to diminish CD balances by large amounts compared to younger clients –This was nonintutive because conventional wisdom suggested that older clients have a larger portfolio of assets and seek less risky investments

Copyright © Houghton Mifflin Company. All rights reserved.13 | 4 Data Analysis Descriptive –Computing measures of central tendency and dispersion,as well as constructing one-way tables Inferential –Data analysis aimed at testing specific hypotheses is usually called inferential analysis

Copyright © Houghton Mifflin Company. All rights reserved.13 | 5 Null and Alternative Hypotheses H 0 -> Null Hypotheses H a -> Alternative Hypotheses Hypotheses always pertain to population parameters or characteristics rather than to sample characteristics. It is the population, not the sample, that we want to make an infernece about from limited data

Copyright © Houghton Mifflin Company. All rights reserved.13 | 6 Steps in Conducting a Hypothesis Test Step 1. Set up H 0 and H a Step 2. Identify the nature of the sampling distribution curve and specify the appropriate test statistic Step 3. Determine whether the hypothesis test is one-tailed or two-tailed

Copyright © Houghton Mifflin Company. All rights reserved.13 | 7 Steps in Conducting a Hypothesis Test (Cont’d) Step 4. Taking into account the specified significance level, determine the critical value (two critical values for a two-tailed test) for the test statistic from the appropriate statistical table Step 5. State the decision rule for rejecting H 0 Step 6. Compute the value for the test statistic from the sample data Step 7. Using the decision rule specified in step 5, either reject H 0 or reject H a

Copyright © Houghton Mifflin Company. All rights reserved.13 | 8 Launching a Product Line Into a New Market Area Karen, product manager for a line of apparel, to introduce the product line into a new market area Survey of a random sample of 400 households in that market showed a mean income per household of \$30,000. Karen strongly believes the product line will be adequately profitable only in markets where the mean household income is greater than \$29,000. Should Karen introduce the product line into the new market?

Copyright © Houghton Mifflin Company. All rights reserved.13 | 9 Karen’s Criterion for Decision Making To reach a final decision, Karen has to make a general inference (about the population) from the sample data Criterion: mean income across across all households in the market area under consideration If the mean population household income is greater than \$29,000, then Karen should introduce the product line into the new market

Copyright © Houghton Mifflin Company. All rights reserved.13 | 10 Karen’s Hypothesis Karen’s decision making is equivalent to either accepting or rejecting the hypothesis: –The population mean household income in the new market area is greater than \$29,000

Copyright © Houghton Mifflin Company. All rights reserved.13 | 11 One-Tailed Hypothesis Test The term one-tailed signifies that all - or z- values that would cause Karen to reject H 0, are in just one tail of the sampling distribution  -> Population Mean H 0 :   \$29,000 H a :   \$29,000

Copyright © Houghton Mifflin Company. All rights reserved.13 | 12 Type I and Type II Errors Type I error occurs if the null hypothesis is rejected when it is true Type II error occurs if the null hypothesis is not rejected when it is false

Copyright © Houghton Mifflin Company. All rights reserved.13 | 13 Significance Level  -> Significance level—the upper-bound probability of a Type I error 1 -  ->confidence level—the complement of significance level

Copyright © Houghton Mifflin Company. All rights reserved.13 | 15 Level of Risk Two firms considering introducing a new product that radically differs from their current product line –Firm ABC Well-established customer base, distinct reputation for its existing product line –Firm XYZ No loyal clientele, no distinct image for its present products Which of these two firms should be more cautious in making a decision to introduce the new product?

Copyright © Houghton Mifflin Company. All rights reserved.13 | 16 Scenario - Firms ABC & XYZ Firm ABC –ABC should be more cautious Firm XYZ –XYZ should be less cautious

Copyright © Houghton Mifflin Company. All rights reserved.13 | 17 Sample mean (x) values greater than \$29,000--that is x-values on the right-hand side of the sampling distribution centered on µ = \$29,000--suggest that H 0 may be false. More important the farther to the right x is, the stronger is the evidence against H 0 Exhibit 13.1 Identifying the Critical Sample Mean Value – Sampling Distribution

Copyright © Houghton Mifflin Company. All rights reserved.13 | 18 Karen’s Decision Rule for Rejecting the Null Hypothesis Reject H 0 if the sample mean exceeds x c

Copyright © Houghton Mifflin Company. All rights reserved.13 | 19 Every mean x has a corresponding equivalent standard Normal Deviate: The expression for z x-  Z = --------- s x x =  + zs x Substituting x c for x and z c for z x c =  + z c s x where z c is standard normal deviate corresponding to the critical sample mean, x c. Criterion Value

Copyright © Houghton Mifflin Company. All rights reserved.13 | 20 Standard deviation for the sample of 400 households is \$8,000. The standard error of the mean (s x ) is given by S s = ----= \$400  n Critical mean household income x c through the following two steps: 1. Determine the critical z-value, z c. For  =.05, From Appendix 1, z c = 1.645. 2. Substitute the values of z c, s, and  (under the assumption that H 0 is "just" true ), x c =  + z c s = \$29,658. x Computing the Criterion Value

Copyright © Houghton Mifflin Company. All rights reserved.13 | 21 Karen’s Decision Rule If the sample mean household income is greater than \$29,658, reject the null hypothesis and introduce the product line into the new market area.

Copyright © Houghton Mifflin Company. All rights reserved.13 | 22 The value of the test statistic is simply the z- value corresponding to = \$30,000. x-  Z = ------ = 2.5 s Test Statistic

Copyright © Houghton Mifflin Company. All rights reserved.13 | 24 P - Value – Actual Significance Level The probability of obtaining an x-value as high as \$30,000 or more when  is only \$29,000 =.0062 This value is sometimes called the actual significance level, or the p-value The actual significance level of.0062 in this case means the odds are less than 62 out of 10,000 that the sample mean income of \$30,000 would have occurred entirely due to chance (when the population mean income is \$29,000 or less)

Copyright © Houghton Mifflin Company. All rights reserved.13 | 25 Conduct T-Test when sample is small Let the sample size, n = 25 X = \$30,000, s = \$8,000 From the t-table in Appendix 3, t c = 1.71 for  =.05 and d.f. = 24. Decision rule: “Reject H 0 if t  1.7l.” T-test

Copyright © Houghton Mifflin Company. All rights reserved.13 | 26 The value of t from the sample data: S =8000/  25= \$1,600 x-  t = ------ = 0.625 s x The computed value of t is less than 1.71, H 0 cannot be rejected. Karen should not introduce the product line into the new market area. T-test (Cont’d)

Copyright © Houghton Mifflin Company. All rights reserved.13 | 27 Two-Tailed Hypothesis Test Two-tailed test is one in which values of the test statistic leading to rejectioin of the null hypothesis fall in both tails of the sampling distribution curve H 0 :  = \$29,000 H a :   \$29,000

Copyright © Houghton Mifflin Company. All rights reserved.13 | 28 Test of Two Means A health service agency has designed a public service campaign to promote physical fitness and the importance of regular exercise. Since the campaign is a major one, the agency wants to make sure of its potential effectiveness before running it on a national scale –To conduct a controlled test of the campaign’s effectiveness, the agency needs two similar cities –The agency identified two similar cities city 1 will serve as the test city city 2 will serve as a control city

Copyright © Houghton Mifflin Company. All rights reserved.13 | 29 Test of Two Means (Cont’d) Random survey of was conducted to measure the average time per day a typical adult in each city spent on some form of exercise –300 adults in city 1, –200 adults in city 2 Results of the survey : –average was 30 minutes per day (with a standard deviation of 22 minutes) in city 1 –Average was 35 minutes per day (with a standard deviation of 25 minutes) in city 2 Question –From these results, can the agency conclude confidently that the two cities are well matched for the controlled test?

Copyright © Houghton Mifflin Company. All rights reserved.13 | 30 City 1: n 1 = 300 x 1 = 30 s 1 = 22 City 2: n 2 = 200 x 2 = 35 s 2 = 25 The hypotheses are H 0 :  1 =  2 or  1 -  2 = 0 H a :  1   2 or  1 -  2  0 Basic Statistics and Hypotheses

Copyright © Houghton Mifflin Company. All rights reserved.13 | 31 Test statistic is the z-statistic, given by (x 1 - x 2 ) - (  1 -  2 ) z = -------------------------------  s 1 2 /n 1 + s 2 2 /n 2 n 1 and n 2 are greater than 30. The z-statistic can therefore be used as the test statistic. Test Statistic

Copyright © Houghton Mifflin Company. All rights reserved.13 | 32 Decision – Two-Tailed Test For Two-Tailed tests –Identify two critical values of z, one for each tail of the sampling distribution –The probability corresponding to each tail is.025, since  =.05 –From the Normal Table, the z-value, for  /2 =.025 is 1.96 Decision rule : “Reject H 0 if z  -1.96 or if z  1.96.”

Copyright © Houghton Mifflin Company. All rights reserved.13 | 33 Computing the value of z from the survey results and under the customary assumption that the null hypothesis is true (i.e.,  1 -  2 = 0): (30 - 35) - (0) z = --------------------------------- = -2.29  (22) 2 /300 + (25) 2 /200 Since z  -1.96, we should reject H 0. Computing Z-value – Two-Tailed Test

Copyright © Houghton Mifflin Company. All rights reserved.13 | 35 Test statistic (x 1 - x 2 ) - (  1 -  2 ) t = ------------------------- s * (  1/n 1 + 1/n 2 ) with d.f. = n 1 + n 2 - 2. In this expression, s * is the pooled standard deviation, given by (n 1 – 1)s 1 2 + (n 2 – 1)s 2 2 s * = --------------------------------- n 1 + n 2 - 2 T- Test for Independent Samples

Copyright © Houghton Mifflin Company. All rights reserved.13 | 36 n 1 = 20 x 1 = 30 s 1 = 22 n 2 = 10 x 2 = 35 s 2 = 25 The degrees of freedom for the t ‑ statistic are d.f. = 28 Critical value of t with 28 d.f for a tail probability of.025 is 2.05. Decision rule : “Reject H 0 if t  -2.05 or if t  2.05." The pooled standard deviation is s * =  529 (approximately) = 23 T- Test for Independent Samples - Two Cities

Copyright © Houghton Mifflin Company. All rights reserved.13 | 37 The test statistic is t = -.56 Since t is neither less than -2.05 nor greater than 2.05, we cannot reject H 0 The sample evidence is not strong enough to conclude that the two cities differ in terms of levels of exercising activity of their residents. T- Test for Independent Samples

Copyright © Houghton Mifflin Company. All rights reserved.13 | 38 National Insurance Company Study – Perceived Service Quality Differences Between Males and Females Test of Two Means Using the SPSS T-TEST Program –On the 10-point scale, males gave a mean rating of approximately 7.87 females gave a mean rating of approximately 7.83.

Copyright © Houghton Mifflin Company. All rights reserved.13 | 39 National Insurance Company Study – Perceived Service Quality Differences Between Males and Females In SPSS, 1. Select ANALYZE from the menu, 2. Click COMPARE MEANS 3. Select INDEPENDENT-SAMPLES T -TEST 4. Move “OQ – Over all Service Quality” to the “TEST VARIABLES(S)” box 5. Move “gender” to “GROUPING VARIABLE” box 6. DEFINE GROUPS (SEX = 1 for male and 2 for female) 7. Click OK.

Copyright © Houghton Mifflin Company. All rights reserved.13 | 40 OQ – Overall Perceived Service Quality Gender – Sex = 1 for male Sex = 2 for female National Insurance Company Study – Perceived Service Quality Differences Between Males and Females

Copyright © Houghton Mifflin Company. All rights reserved.13 | 42 F-Test--to see if the variance of the 2 groups are assumed to be equal p-value =.210 --> null hypothesis cannot be rejected at  = 0.05 P-value >  = 0.05 -- Do not Reject, Equal variance assumed is correct Use this row when the null hypothesi s of equality of variance is rejected National Insurance Company Study – Perceived Service Quality Differences Between Males and Females

Copyright © Houghton Mifflin Company. All rights reserved.13 | 43 P-value=.88 is greater than the  = of 0.05. Do not reject Ho. The p-value implies that the odds are 88 to 100 that a difference of magnitude.04 (i.e., 7.87 - 7.83) could have occurred from chance. The null hypothesis cannot be rejected at the customary significance level of.05. National Insurance Company Study – Perceived Service Quality Differences Between Males and Females

Copyright © Houghton Mifflin Company. All rights reserved.13 | 44 Test of Two Means When Samples Are Dependent The need to check for significant differences between two mean values when the samples are not independent

Copyright © Houghton Mifflin Company. All rights reserved.13 | 45 Test of Two Means When Samples Are Dependent (Cont’d) A retail chain ran a special promotion in a representative sample of 10 of its stores to boost sales Weekly sales per store before and after the introduction of the special promotion are shown Did the special promotion lead to a significant increase in sales?

Copyright © Houghton Mifflin Company. All rights reserved.13 | 47 One-Tailed Hypothesis Test: H 0 :  d  0; H a :  d  0. The sample estimate of  d is x d, given by n  X di i=1 x d = ----- n where n is the sample size. x d = 50/10 = 5 Test of Two Means When Samples Are Dependent (Cont’d)

Copyright © Houghton Mifflin Company. All rights reserved.13 | 48 Test statistic is x d -  t = ----------- = 2.10 s/  n Test of Two Means When Samples Are Dependent (Cont’d)

Copyright © Houghton Mifflin Company. All rights reserved.13 | 49 Standard deviation (s) = 7.53,  = 0.05, t c for 9 d.f = 1.83 from the Appendix 3 Decision rule: “Reject H 0 if t  1.83.” Test Statistic, t  1.83, we reject H 0 and conclude that the mean change in sales per store was significantly greater than zero. The special promotion was indeed effective. Test of Two Means When Samples Are Dependent (Cont’d)

Copyright © Houghton Mifflin Company. All rights reserved.13 | 51 Test for a Single Proportion Ms.Jones wants to substantially increase her firm's advertising budget. The firm sells a variety of personal computer accessories Random sample : 20/100 (20%) know the brand name True awareness rate for the brand name across all personal computer owners is less than.3 Should Ms. Jones increase the advertising budget on the basis of survey results?

Copyright © Houghton Mifflin Company. All rights reserved.13 | 52 Test for a Single Proportion (Cont’d) Need to test the population proportion of personal computer owners who are aware of the brand: H 0 :  .3 H a :  .3 (  is the symbol for population proportion)

Copyright © Houghton Mifflin Company. All rights reserved.13 | 53 The test statistic: p -  Z = ---------------------   (1-  )/n where p is the sample proportion. From the Normal Table, z c, = -1.645 for  =.05. Decision rule here is: “Reject H o if z  - 1.645.” p =.2,  =.3, and n = 100, z = -2.174 Test for a Single Proportion (Cont’d)

Copyright © Houghton Mifflin Company. All rights reserved.13 | 54 Since -2.174  -1.645, we reject H 0 ; The sample awareness rate of.2 is too low to support the hypothesis that the population awareness rate is.3 or more. The actual significance level (p-value) corresponding to z = -2.174 is approximately.015 (from Appendix 1). Level of significance implies that the odds are lower than 15 in 1,000 that the sample awareness rate of.2 would have occurred entirely by chance (that is, when the population awareness rate is.3 or higher). Test for a Single Proportion

Copyright © Houghton Mifflin Company. All rights reserved.13 | 56 Test of Two Proportions: Choosing Between Commercial X & Commercial Y For a New Product Tom, advertising manager for a frozen-foods, company, is in the process of deciding between two TV commercials (X and Y) for a new frozen food to be introduced –Commercial X Runs for 20 seconds Random sample: 20 % awareness out of 200 respondents –Commercial Y Runs for 30 seconds Random sample: 25 % awareness out of 200 respondents

Copyright © Houghton Mifflin Company. All rights reserved.13 | 57 Test of Two Proportions (Cont’d) Question –Can Tom conclude that commercial Y will be more effective in the total market for the new product?

Copyright © Houghton Mifflin Company. All rights reserved.13 | 58 Criterion for Decision Making To reach a final decision, Tom has to make a general inference (about the population) from the sample data Criterion: relative degrees of awareness likely to be created by the 2 commercials in the population of all adult consumers Tom should conclude that commercial Y is more effective than commercial X only if the anticipated population awareness rate for commercial Y is greater than that for X

Copyright © Houghton Mifflin Company. All rights reserved.13 | 59 Hypothesis Tom’s decision-making is equvalent to either accepting or rejecting the hypothesis –The potential awareness rate that commercial Y can generate among the population of consumers is greater than that which commercial X can generate

Copyright © Houghton Mifflin Company. All rights reserved.13 | 60 Commercial Commercial X Y Sample sizes: n 1 = 200n 2 = 200 Sample proportions: p 1 =.25 p 2 =.20 The hypotheses are: H 0 :  1   2 or  1 -  2  0 H a :  1   2 or  1 -  2  0 Null and Alternative Hypotheses

Copyright © Houghton Mifflin Company. All rights reserved.13 | 61 (p 1 – p 2 ) - (  1 -  2 ) z = ------------------------  p1 - p2 -- is estimated by the sample standard error formula Sample Standard Error s p1 - p2 = PQ ( 1/n 1 + 1/n 2 ) n 1 p 1 + n 2 p 2 P = ------------------- n 1 + n 2 Q = 1 - P Test of Two Proportions – Sample Standard Error

Copyright © Houghton Mifflin Company. All rights reserved.13 | 62 For  =.05, the critical value of z (from Appendix 1) is 1.645. Decision rule: “Reject H 0 if z  1.645.” First compute P and Q, then s p1 - p2 and z: 200(.25) + 200(.2) P =----------------------- =.225 200 + 200 Q = 1 -.225 =.775 Test of Two Proportions

Copyright © Houghton Mifflin Company. All rights reserved.13 | 63 s p1 - p2 =  (.225)(.775) (1/200 + 1/200) =0.042 (.25 -.20) - (0) z = ---------------------- = 1.19.042 Since z  1.645, we cannot reject H 0. The sample evidence is not strong enough to suggest that commercial Y will be more effective than commercial X. Test of Two Proportions

Copyright © Houghton Mifflin Company. All rights reserved.13 | 65 Cross-Tabulations: Chi-square Contingency Test Technique used for determining whether there is a statistically significant relationship between two categorical (nominal or ordinal) variables

Copyright © Houghton Mifflin Company. All rights reserved.13 | 66 Telecommunications Company Marketing manager of a telecommunications company is reviewing the results of a study of potential users of a new cell phone –Random sample of 200 respondents A cross-tabulation of data on whether target consumers would buy the phone (Yes or No) and whether the cell phone had Bluetooth wireless technology (Yes or No) Question –Can the marketing manager infer that an association exists between Bluetooth technology and buying the cell phone?

Copyright © Houghton Mifflin Company. All rights reserved.13 | 68 H 0 : There is no association between wireless technology and buying the cell phone (the two variables are independent of each other). H a : There is some association between the Bluetooth feature and buying the cell phone (the two variables are not independent of each other). Cross Tabulations - Hypotheses

Copyright © Houghton Mifflin Company. All rights reserved.13 | 69 Conducting the Test Test involves comparing the actual, or observed, cell frequencies in the cross- tabulation with a corresponding set of expected cell frequencies (E ij )

Copyright © Houghton Mifflin Company. All rights reserved.13 | 70 n i n j E ij = ----- n Where n i and n j are the marginal frequencies, that is, the total number of sample units in category i of the row variable and category j of the column variable, respectively Expected Values

Copyright © Houghton Mifflin Company. All rights reserved.13 | 71 The expected frequency for the first-row, first-column cell is given by 100  100 E 11 =------------ = 50 200 Computing Expected Values

Copyright © Houghton Mifflin Company. All rights reserved.13 | 73 Where r and c are the number of rows and columns, respectively, in the contingency table. The number of degrees of freedom associated with this chi ‑ square statistic are given by the product (r - 1)(c - 1). = 72.00 Chi-square Test Statistic

Copyright © Houghton Mifflin Company. All rights reserved.13 | 74 For d.f. = 1, Assuming  =.05, from Appendix 2, the critical chi ‑ square value (  2 c ) = 3.84. Decision rule is: “Reject H 0 if  2  3.84.” Computed  2 = 72.00 Since the computed Chi-square value is greater than the critical value of 3.84, reject H 0. The apparent relationship between “Bluetooth technology"and "would buy the cellular phone" revealed by the sample data is unlikely to have occurred because of chance Chi-square Test Statistic in a Contingency Test

Copyright © Houghton Mifflin Company. All rights reserved.13 | 75 Interpretation The actual significance level associated with a chi-square value of 72 is less than.001 (from Appendix 2). Thus, the chances of getting a chi-square value as high as 72 when there is no relationship between Bluetooth technology and purchase of cell phones are less than 1 in 1,000.

Copyright © Houghton Mifflin Company. All rights reserved.13 | 76 Cross-Tabulation Using SPSS for National Insurance Company One crucial issue in the customer survey of National Insurance Company was how a customer's education was associated with whether or not she or he would recommend National to a friend.

Copyright © Houghton Mifflin Company. All rights reserved.13 | 77 Need to Conduct Chi-square Test to Reach a Conclusion The hypotheses are –H 0 :There is no association between educational level and willingness to recommend National to a friend (the two variables are independent of each other) –H a :There is some association between educational level and willingness to recommend National to a friend (the two variables are not independent of each other)

Copyright © Houghton Mifflin Company. All rights reserved.13 | 78 For two-way tabulation: 1. Select ANALYZE on the SPSS menu, 2. Click on DESCRIPTIVE STATISTICS, 3. Select CROSS-TABS. 4. Move the “highest level of schooling” to ROW(S) box, 5. Move “rec” variable to “COLUMN(S) box. 6. Click on CELLS, 7. Select OBSERVED, and ROW PERCENTAGES. 8. Click CONTINUE and 9. Click OK. Association Between Education and Customer’s Willingness to recommend National to a Friend

Copyright © Houghton Mifflin Company. All rights reserved.13 | 80 COUNT represents the actual number of customers in each cell. The percentages are based on the corresponding Association Between Education and Customer’s Willingness to recommend National to a Friend

Copyright © Houghton Mifflin Company. All rights reserved.13 | 82 National Insurance Company Study - Chi-Square Test For Chi-Square Assessment: 1. Select ANALYZE 2. Click on DESCRIPTIVE STATISTICS 3. Select CROSS-TABS 4. Move the variable “highest level of schooling” to ROW(s) box 5. Move “rec” to COLUMN(s) box; 6. Click on “STATISTICS” 7. Select CHI-SQUARE, CONTINGENCY COEFFICIENT, and CRAMER’S V 8. Click on CELLS, 9. Select OBSERVED and EXPECTED FREQUENCIES 10.Click CONTINUE 11.Click OK.

Copyright © Houghton Mifflin Company. All rights reserved.13 | 86 National Insurance Company Study – P-Value Significance The actual significance level (p-value) = 0.019 The chances of getting a chi-square value as high as 10.007 when there is no relationship between education and recommendation are less than 19 in 1000 The apparent relationship between education and recommendation revealed by the sample data is unlikely to have occurred because of chance Jill and Tom can safely reject null hypothesis

Copyright © Houghton Mifflin Company. All rights reserved.13 | 87 Precautions in Interpreting Cross Tabulation Results Two-way tables cannot show conclusive evidence of a causal relationship Watch out for small cell sizes Increases the risk of drawing erroneous inferences when more than two variables are involved

Copyright © Houghton Mifflin Company. All rights reserved.13 | 88 Is there a causal relationship between patients who jog and patients with hearth disease? Two-way Table Based on a Survey of 200 Hospital Patients: