An MBA program was experiencing problems scheduling its courses. Demand for the program's optional courses and specializations varied yearly. Within a year, students seem to want marketing courses; in other years, accounting or finance are in demand. In desperation, the business school dean turned to a statistics professor for help. The statistician professor believes that the problem may be the variability in students' academic background and that the bachelor's degree affects career choice. To get started, he took a random sample of last year's MBA students and recorded the selected undergraduate degree and specialization in the MBA program. The undergraduate degrees were Bachelor of Arts, Bachelor of Engineering, Bachelor of Business Administration, and others. There are three possible careers for MBA students: accounting, finance, and marketing. The results were summarized in a cross-classification table(Xm16-02).

• Can the statistician conclude that undergraduate degrees affect the choice of MBA concentration? (Xm16-02)

Degree W-AN- W -NA- - - - W - N W W W - N - - W N - A . W - A A W A W W N N - W - N - - - 3 MBA Major N - N - W W - W N N W - W - - - - W N N - N N N - N N - - N W - - W W - - - N- - WN - XM/6 - OZ ANA-ANWAD W W N - - - N W - - W N N W N - - N - - W - WA- . W W W - - W - - N - W W W N N - W N N W W N W - N W W N - N N W N W - N N - - - W W - - W N W W - N W - - W - - - - - NW XXM16- 02 B\fW - NIW W .- W - N - W - X M16 - 02 D\fChi-Square for Chi-square 1. The objective of the chi-square test is to determine the relationship between two variables of nominal type, through the existence or not of independence between the two variables. That two variables are independent means that they have no relation, and that therefore one does not depend on the other, nor vice versa. he two v are indepe 2. With the study of independence, a method is also originated to verify if the frequencies observed in each category are compatible with the independence between both variables . 3. A contingency table is constructed to establish that there is sufficient evidence to infer that two nominal variables are related and to infer that there are differences between two or more populations of nominal variables. 4. The notation used to represent chi-square is X2Hypothesis for Chi-square 1. The null hypothesis of the chi-square analysis technique specifies that there is no relationship between the two variables and they are proposed as follows: Ho: The two variables are independent H1: The two variables are dependentTest Statistic for Chi-square 1. The test statistic to determine the value of: ( fi - ei ) 2 Y2 =2 i=1 2. Expected frequencies for a contingency table: ; = _ raw i total x column j total sample sizeTest Statistic for Chi-square The following data summary contains nominal type data with the use of the following codes: (Undergraduate) (MBA Concentration) 1 = Bachelor of Arts 1 = Marketing 2 = Bachelor of Engineering 2 = Finance 3 = Bachelor of Business Administration 3 = Accounting 4 =Other MBA Concentration Total Bachelor Total 152Test Statistic for Chi-square The following data summary contains nominal type data with the use of the following codes: (Undergraduate) (MBA Concentration) 1 = Bachelor of Arts 1 = Marketing 2 = Bachelor of Engineering 2 = Finance 3 = Bachelor of Business Administration 3 = Accounting 4 = Other MBA Concentration Accounting Finance Marketing Total Bachelor of Arts Bachelor of Engineering N Bachelor of Business Administration 8 Other LO Y TotalTest Statistic for Chi-square MBA Concentration Accounting Finance Marketing Total 61 60x 44 -= 17.37 60x 41 =18 55 Bachelor of Arts 60x 157- 24.08 152 152 44 Bachelor in 31x 152 31x 1 =12.44 Engineering 152 = 8.97 31 X 152 9.59 N Bachelor in Business Administration 39x 152 - 15.65 39x 44 -11.29 39x 47 - 12.06 152 152 on 44 22x 61 - 8 83 31x 22x - = 6.80 Other 52 152 152 =6.37 Total 152Test Statistic for Chi-square MBA Concentration Accounting Bachelor of Arts Finance Marketing 31 (24.08) Bachelor of 13 ( 17.37 ) 16 ( 18.55 ) Engineering 8(12.44) 16(8.97) 7(9.59 ) Bachelor in Business Administration 12 (15.65) 10(11.29) 17(12.06) Other 10 (8.83 ) 5 (6.37 ) 7(6.80) x2 =0k ( fi- ei ) 2 = (31-24.08 ) 2 + (31-17.37) 2 + . (16-18.55)2 + (8-12.44) 2 + (16 - 8.97 ) 2 + (7 - 9.59 ) 2 24.08 17.37 18.55 + (12-15.65) 2 (10-11.29)2 12.44 8.97 (17-12.06)2 9.59 (10-8.33) 2 15.65 + + + (5-6.37) 2 (7-6.80)2 11.29 = 14.70 12.06 8.33 6.37 6.80Rejection Region and p-value To determine the rejection region, one must know the degrees of freedom associated with the Chi-square statistic. The degrees of freedom for the contingency table with r rows and c columns is v = (r - 1) (c - 1). For exercise, the degrees of freedom are v=r-1 c-1 = 4-1 3 -1 = 6. If the significance level is 5%, the rejection region is X} > (2 =)(2 ) 12.5916 . 05 ,6 Chi-square table https://people.richland.edu/james/lecture/m170/tbl-chi.html As the statistical Chi-square value (14.70) is greater than the critical chi-square value (12.5916). Therefore, the null hypothesis is rejected and it is concluded that there is evidence of a relationship between the undergraduate degree and the MBA concentration. Unfortunately, you cannot determine the p-value manually. 8Application of Chi-square Technique Through Excel Mega- Stat The exercise Xm16-02 shows the dataset of two nominal variables to apply the chi-square technique. Click on the Add-ins menu and in MegaStat select Crosstabulation to create the contingency table, critical x2 , statistics x2 and p-value. AutoSave . on 9 Xm 16-02 (16) - Compatibility Mode File Home Insert Draw Page Layout Formulas Data Review View Add-ins File Home Insert Draw Page Layout Formulas Data Review w View Add-ins MegaStat MegaStat ~ Descriptive Statistics. Frequency Distributions Probability Menu Comm Confidence Intervals / Sample Size.. AA26 X Hypothesis Tests E G H 1 Degree MBA Major BINK-Degree BINMBA Analysis of Variance Correlation / Regression Time Series / Forecasting Chi-Square / Crosstab Contingency Table. Nonparametric Tests Crosstabulation .. Quality Control Process Charts.. Goodness of Fit Test. Repeat Last Option Generate Random Numbers .Application of Chi-square Technique Through Excel Mega- Stat The Data range includes the dataset of the nominal variables and in specification range the BIN or the basic range. Select in the output options (chi-square) and OK to display the results. The value of the statistics x2 is 14.70 and p-value is .0227 File Home Page Layout Formulas Data Review View Add-ins Help Acrobat B C WU Crosstabulation Crosstabulation MBA Major Row varlet Total Sheetl'$451-545153 Data range OK 6 Degree Sheet'SC$1:$CSA -Specification range Clear Sheet:$251:$85153 Data range Sheetl 'SD$1:$053 - Specification range Total 61 ONOOPENMYY Output Options chi-square Expected values Phi coefficient So of som TO-E Coefficient of contingency 14.70 chi-square so of column (0-EX/E Cramer's y 6 df So of total So of chi-square Fisher Exact Test 0227 p-value Output Sheet1 Sheet2 Sheet?Application of Chi-square Technique Through Excel Mega- Stat The Data range includes the dataset of the nominal variables and in specification range the BIN or the basic range. Select in the output options (chi-square) and OK to display the results. The value of the statistics x2 is 14.70 and p-value is .0227 A LL There is evidence to infer that having a 2 Crosstabulation previous undergraduate degree is related to MBA Major choosing an MBA concentration. Total Degree 1 60 to 39 m T 22 Total 61 152 14.70 chi-square 6 df .0227 p-valueApplication of Mann-Whitney Technique Through Excel Mega-Stat Exercise Xm21-02 shows the dataset representing the level of effectiveness of two types of drugs that was administered to 30 people. 15 people were ingested a new drug and another 15 people were ingested aspirin. There is a significant difference between the levels of effectiveness of both drugs. New Aspirin AutoSave Con - patibility Mode Search 5 = extremely effective 4 File Home Insert Draw Page Layout Formulas Data Review View Add-ins = quite effective MegaStat~ 3 = something effective 2 Descriptive Statistics. = slightly effective Frequency Distributions 1 = nothing effective Probability Confidence Intervals / Sample Size. Hypothesis Tests E F G H In the add-ins menu in Analysis of Variance MegaStat you select Correlation / Regression Time Series / Forecasting Nonparametric Test in Wilcoxon Chi-Square / Crosstab -Mann/Whitney Test to apply the Nonparametric Tests Sign Test. statistical technique. Quality Control Process Charts . Runs Test for Random Sequence. Repeat Last Option Wilcoxon - Mann/ Whitney Test. Generate Random Numbers . Wilcoxon Signed Ranks Test.Application of Mann-Whitney Technique Through Excel Mega-Stat Exercise Xm21-02 shows the dataset representing the level of effectiveness of two types of drugs that was administered to 30 people. 15 people were ingested a new drug and another 15 people were ingested aspirin. There is a significant difference between the levels of effectiveness of both drugs. New Aspirin D LU Hypothesis 2 Wilcoxon - Mann/Whitney Test Ho: the two drugs are equal sum of ranks 276.5 New H1 : the two drugs are different. 188.5 Aspirin 465 total 232.500 expected value Interpretation 24.109 standard deviation 1.825 Z 0680 p-value (two-tailed) At 95% confidence, the p-value found (.0680) is not less than .05 it is concluded that the two drugs have the same effectiveness.Application of Sign Technique Through PSPP In an investigation it is desired to determine which of two cars is perceived as the most comfortable, 25 people traveled (separately) in the back seat of a European vehicle and also in the back seat of an American car. Each person was asked to rate the trip on the following scale: 1 = very uncomfortable, 2 = quite uncomfortable, 3 = neither uncomfortable nor comfortable, 4 = quite comfortable, 5 = very comfortable. European American In the menu of Analyze in Non- Parametric Tests and in 2 Paired Samples the Sign technique is appliedApplication of Sign Technique Through PSPP In an investigation it is desired to determine which of two cars is perceived as the most comfortable, 25 people traveled (separately) in the back seat of a European vehicle and also in the back seat of an American car. Each person was asked to rate the trip on the following scale: 1 = very uncomfortable, 2 = quite uncomfortable, 3 = neither uncomfortable nor comfortable, 4 = quite comfortable, 5 = very comfortable. European American Hypothesiss NPAR TESTS NPAR TEST Ho: both vehicles are the same SIGN European with American (PAIRED) Frequencies H1: the vehicles are different. N European-American Negative Differences NO WW AWD A NA NW NAW NAA -UINWUN Interpretation Positive Differences Links Total There is strong evidence indicating that Statistical Tests people perceive that the European car European-American provides a more comfortable ride than the Exact Sig. (2 tails) .011 Exact Sig. (1 tail) .005 American car. Probability point .004Application of Kruskal-Wallis Technique Via Excel Mega- Stat A restaurant manager is interested in knowing how his customers value the quality of the food and the service and cleanliness of the restaurant in the three shifts. 10 customers selected from each shift responded: 4 = excellent, 3 = good, 2 = fair, 1 = poor 4:00-mid Mid-8:00 8:00-4:00 Draw Page Layout Formulas Data Review View Add-ins there Como In the add-ins menu in MegaStat 1 . 400 mid 4:00-mid Kid.8:00 8:00-4:00 Kruskal- Walls Test you select Nonparametric Test in N N m N Sheet1 15451:5C$14/ Kruskal-Wallis Test and select P Output rocked dota the data and mark in Correct of Correct for thes ties to apply the statistical my m m m m N m technique. Output Sheet1 Sheet2 Sheet3Application of Kruskal-Wallis Technique Via Excel Mega-Stat A restaurant manager is interested in knowing how his customers value the quality of the food and the service and cleanliness of the restaurant in the three shifts. 10 customers selected from each shift responded: 4 = excellent, 3 = good, 2 = fair, 1 = poor Hypothesis 4:00-mid Mid-8:00 8:00-4:00 B C D G Ho: the three shifts are Kruskal-Wallis Test equal + m n Median 10 1 Avg. Part 40-mid 3.00 10 15.60 Mid- 8:00 H1 : at least two of the shifts + + 10 250 12.25 8:00-4:00 Total differ m 3.013 H (corrected for ties ) 12 2 d.f 2217 p-value N multiple comparison values for avg. ranks 9.43 (05) 11.56 (.01) m 17 Kruskal-Wallis Test Interpretation median n mm mm 3.00 Avg. Rans_mid T 3.00 10 15.60 Mid-8:00 2 50 There is not enough evidence 10 3 00 30 2.25 8:00-4:00 + m m N Total N to infer that there is a 2.645 H 2 d.f. .2665 p-value difference in quality of m multiple comparison values for avg. ranks 9.43 (.05) 11.56 ( 01) service between the three shifts..Application of Friedman Technique Through Mega-Stat of Excel The personnel supervisor of an accounting firm has been receiving complaints from senior managers about the quality of recent hires. All new accountants are hired through a process whereby four managers interview the candidate and qualify them in various dimensions, including academic credential, previous work experience, and staff suitability. Then, each manager summarizes the results and produces an evaluation of the candidate under the following possibilities: 1 = The candidate is in the top 5% of applicants. 2 = The candidate is in the top 10% of applicants, but not in the top 5%. 3 = Manager 1 Manager 2 Manager 3 Manager 4 The candidate is in the top 25% of applicants, but not in the top 10%. 4 = The candidate is in the top 50% of applicants, but not in the top 25%. 5 = The candidate is in the bottom 50% of applicants. Y N M M N T Hypothesis Ho: all four managers are the same HI : at least two of the managers differApplication of Friedman technique Through Mega-Stat of Excel Autosme a Xm21 05 (1) - Comprobity Mode. P se File Home Insert Draw Page Layout Formulas Data Review Vien Add-ins Hesp Mogestal - Manager 1 Manager 2 Manager 3 Manager 4 Descriptive Statistic Frequency Distributions Probability Confidence intervals / Sam Hypothesis EF m Briolysis of Gerente xion / Regression N Chi Square / m etric Tests Quality Control Process Charts. Pures Test for Random Se Bruskat - Walls Test . help / intormet Friedman Test N t m in in Kendal Confident of Concor N in m Fisher Bract Test Output Sheet! Sheetz sheets Manager: Gerente In the add-ins menu in MegaStat you select Nonparametric Test in Friedman Test and select the data and mark in Output Ranked Data to apply the statistical technique.Application of Friedman Technique Through Mega-Stat of Excel B C D E LL Interpretation Friedman Test Manager 1 Manager 2 Manager 3 Manager 4 Sum of Ranks Avg. Rank Manager: Gerente 21.00 2.63 Gerente 1 There is enough evidence to 10.00 1.25 Gerente 2 24.50 3.06 Gerente 3 infer that managers' 24.50 3.06 Gerente 4 80.00 2.50 Total assessments differ because the p-value (.0140) is less than 8 n 10.613 chi-square .05. 3 d.f. 0140 p-value multiple comparison values for avg. ranks LO 1.70 (.05) 2.03 (.01) LOApplication of Spearman Technique Through Mega-Stat of Excel The production manager of a company wants to examine the relationship between the aptitude test score given before the hiring of production line workers and the performance ratings received by employees 3 months after starting work. The results of the study would allow the company to decide how much weight these aptitude tests have in relation to other work history information obtained, including references. Aptitude test results range from 0 to 100: Aptitude Performance 1 = Employee has performed well below average . AutoSave .0 File Home Insert Draw Page Layout Formulas Data View Add-ins Help Acrobat 2 = The employee has performed somewhat below average . 3 = Megastar The employee has performed at a medium level. Descriptive Statistics Erquincy Distributions 4 = Employee has performed somewhat above average. 5 = Probability Confidence Intervals / Sample Site The employee has performed well above average . Hypothesis Tests D F GH Analysis of Variance Correlation In the add-ins menu in Time Series / Forecasting Che- Square / Crosstab MegaStat you select m N t m N t m m n r m m N m m t N m m D Nonpurany tric Tests Sign Test Quality Cont Buns Test Random Sequence Nonparametric Test in Repeat Last Option on - Mann/Whitney Test Generate Random Numbers- Wikaxon Squied Ranks Test Spearman Coefficiente of Utilities uskal-Wallis Test Rank Correlation and select Help/ Information Fledmen Test Kendall Coefficient of Concordance. Spearman Coefficient of Rank Comelation the data and mark in Correct of Fisher Exact Test Output Sheet1 Sheetz Sheets ties to apply the statistical technique. 9Aptitude Performance Hypothesis Spearman Coefficient of Rank Correlation Ho: Ps = 0 Aptitude Performance HI: Ps # 0 Aptitude 1.000 Performance 0.379 1.000 Interpretation 20 sample size m N t m N t m m n r m m N m m t N m m + .444 critical value .05 (two tail) There is no evidence to infer + .561 critical value .01 (two tail) that there is a relationship between aptitude and performance