MG 620 Research and statistics for Managers Final Exam Instructions: Answer all questions Part I : Problem 1. Fitting a straight line to a set of data yields the following prediction line: Y = 4 + 3X i) Interpret the meaning of the Y intercept, b0. ii) Interpret the meaning of the slope, b1 iii) Predict the mean value of Y for X = 5 Dr. D Rawana Final Winter 2016 Page 1 2. What is the difference between the standard deviation, standard error of the mean, and standard error of the estimate? Discuss The standard deviation measures the amount of variability or dispersion for a subject set of data from the mean, while the standard error of the mean measures how far the sample mean of the date is likely to be from the true population mean. The standard error is an estimate of the standard deviation of a statistic. , it is used to compute the confidence intervals and margins of error. 3. In a sample of size of 5, the sum of all values is 200. What is the sample mean a. 20 b. 40 SHOW YOUR WORK! Dr. D Rawana c. d. 800 None of the above Final Winter 2016 Page 2 The following are the durations in minutes of a sample of long-distance phone calls within the continental United States reported by one long-distance carrier Table 1 Time (in minutes) Relative Frequency 0 but less than 5 5 but less than 10 10 but less than 15 15 but less than 20 20 but less than 25 25 but less than 30 30 or more 0.37 0.22 0.15 0.10 0.07 0.07 0.02 4. Referring to Table 1, what is the width of each class? a. 1 minute b. 2% SHOW YOUR WORK! c. d. 5 minutes 100% 1. Width refers to the difference between the lower limit of any two consecutive classes 2. Therefore, 5-0 = 5 or 10-5 = 5 5. Referring to Table 1, what is the cumulative relative frequency for the percentage of calls that lasted under 10 minutes? a. 0.10 b. 0.76 SHOW YOUR WORK! c. d. 0.59 0.84 Cumulative relative frequency is the sum of all relative frequency from the top to the last values: for example, the sum of all calls under 10 minutes is found by adding (.37 + .22) = .59). for under 20 minutes, it is .84 as shown above for all values in the third column. Dr. D Rawana Final Winter 2016 Page 3 6. The managers of a real estate firm are interested in finding out if the number of new clients a broker brings into the firm affects the sales generated by the broker. They sample 5 brokers and determine the number of new clients they have enrolled in the last year and their sales amounts in thousands of dollars. These data are presented in the table that follows. Broker Clients (X) 1 2 3 4 5 1 2 1 0 1 Sales (Y) 4 6 3 1 1 a. Construct a scatter diagram for these data. Does the scatter diagram show a linear relationship between sales and number of new clients? Explain and show all work including which variable is independent and dependent. b. Y 6 5 4 3 2 1 0 1 2 3 X b) Estimate the intercept (b0). Show your work c) Estimate the slope (b1). Show your work c) Draw the regression line Dr. D Rawana Final Winter 2016 Page 4 d) Is the relationship positive or negative? Explain 7. If SSR = 76 and SST = 98, and n = 12 a. Determine the coefficient of determination , r2, and interpret its meaning. b. Compute the standard error of the estimate. c. How useful do you think this regression model is for predicting? 8. The Management at Ohio National Bank does not want its customers to wait in line for service for too long. The manager of a branch of this bank estimated that the customers currently have to wait an average of 8 minutes for service. Assume that the waiting times for all customers at this branch have a normal distribution with a mean of 8 minutes and a standard deviation of 2 minutes. Find the probability that randomly selected customer will have to wait for less than 4 minutes? Instructions: Show all steps: 1. Draw the normal curve and indicate the mean, standard deviation, and the X bar scale 2. Identify the area of interest (that is shade the area under the curve that you will compute the probability). 3. Covert the X bar values in Z scores 4. Look up the Z standardized table for the cumulative area(s). 5. Now, make your decision ( that a customer will wait for less than 4 minutes) Dr. D Rawana Final Winter 2016 Page 5 9. The Management at Ohio National Bank does not want its customers to wait in line for service for too long. The manager of a branch of this bank estimated that the customers currently have to wait an average of 8 minutes for service. Assume that the waiting times for all customers at this branch have a normal distribution with a mean of 8 minutes and a standard deviation of 2 minutes. What is the probability that customers have to wait for 4 to 8 minutes? Instructions: Show all steps: a. Draw the normal curve and indicate the mean, standard deviation, and the X bar scale b. Identify the area of interest (that is shade the area under the curve that you will compute the probability). c. Covert the X bar values in Z scores d. Look up the Z standardized table for the cumulative area(s). e. Now, make your decision ( that a customer will have to wait for 4 to 8 minutes) Dr. D Rawana Final Winter 2016 Page 6 10. Five hundred employees were selected from a city's large private companies, and they were asked whether or not they have any retirement benefits provided by their companies. Based on this information, the following two-way contingency table was prepared. Have Retirement Benefits Yes No Men 225 75 Women 150 Total 50 Total If one employee is selected at random from these 500 employees, a) Find the probability that this employee is a woman b) Has retirement benefits c) Has retirement benefits given the employee is a man d) Is a woman given that she does not have retirement benefits Dr. D Rawana Final Winter 2016 Page 7 11. Use the given data to construct a frequency distribution. Lori asked 24 students how many hours they had spent doing homework during the previous week. The results are shown below: Number of students 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 28 19 20 21 22 23 24 Hours of study 11 10 11 9 11 11 15 12 11 8 13 10 10 13 11 10 13 11 10 12 10 13 12 9 Construct a frequency distribution, using 4 classes and a class width of 2 hours, and a lower limit of 8 for class 1. Dr. D Rawana Final Winter 2016 Page 8 12. An entrepreneur is considering the purchase of a coin-operated laundry. The current owner claims that over the past 10 years, the average daily revenue was $675 with a standard deviation of $75. A sample of 30 days reveals daily revenue of $625. Consider H0: = 675 versus Ha: < 675 a. At the .05 level of significance , is there evidence that the average daily revenue is less than 675 b. Compute the p-value and interpret its meaning c. Construct a 95% confidence interval estimate of the population mean revenue of coin-operated laundry. d. Compare the results of (a) , (b), and (c). What conclusions do you reach? Show all steps! Instructions: Show all steps such as 1. State the Hypothesis 2. Draw the normal distribution and identify the acceptance and rejection regions 3. Make your decision with respect to each part of the question. Dr. D Rawana Final Winter 2016 Page 9 MG 620 Research and statistics for Managers Final Exam Instructions: Answer all questions Part I : Problem 1. Fitting a straight line to a set of data yields the following prediction line: Y = 4 + 3X i) Interpret the meaning of the Y intercept, b0. b0=4 is the value of Y when X is 0 ii) Interpret the meaning of the slope, b1 b1=3 mean that a unit change in X will increase Y by 3 units. iii) Predict the mean value of Y for X = 5 Y=4+3X Y=4+5(5)=9 2. What is the difference between the standard deviation, standard error of the mean, and standard error of the estimate? Discuss The standard deviation measures the amount of variability or dispersion for a subject set of data from the mean, while the standard error of the mean measures how far the sample mean of the date is likely to be from the true population mean. The standard error is an estimate of the standard deviation of a statistic. , it is used to compute the confidence intervals and margins of error. Dr. D Rawana Final Winter 2016 Page 1 3. In a sample of size of 5, the sum of all values is 200. What is the sample mean a. 20 b. 40 SHOW YOUR WORK! c. d. 800 None of the above Mean =1/nXi Mean=1/5(200)=40 The following are the durations in minutes of a sample of long-distance phone calls within the continental United States reported by one long-distance carrier Table 1 Time (in minutes) Relative Frequency 0 but less than 5 5 but less than 10 10 but less than 15 15 but less than 20 20 but less than 25 25 but less than 30 30 or more 0.37 0.22 0.15 0.10 0.07 0.07 0.02 4. Referring to Table 1, what is the width of each class? a. 1 minute b. 2% SHOW YOUR WORK! Dr. D Rawana c. d. 5 minutes 100% Final Winter 2016 Page 2 1. Width refers to the difference between the lower limit of any two consecutive classes Therefore, 5-0 = 5 or 10-5 = 5 5. Referring to Table 1, what is the cumulative relative frequency for the percentage of calls that lasted under 10 minutes? a. 0.10 b. 0.76 SHOW YOUR WORK! c. d. 0.59 0.84 Cumulative relative frequency is the sum of all relative frequency from the top to the last values: for example, the sum of all calls under 10 minutes is found by adding (.37 + .22) = .59). for under 20 minutes, it is .84 as shown above for all values in the third column. 6. The managers of a real estate firm are interested in finding out if the number of new clients a broker brings into the firm affects the sales generated by the broker. They sample 5 brokers and determine the number of new clients they have enrolled in the last year and their sales amounts in thousands of dollars. These data are presented in the table that follows. Broker Clients (X) 1 2 3 4 5 1 2 1 0 1 Sales (Y) 4 6 3 1 1 a. Construct a scatter diagram for these data. Does the scatter diagram show a linear relationship between sales and number of new clients? Explain and show all work including which variable is independent and dependent. Dr. D Rawana Final Winter 2016 Page 3 b. Y X b) Estimate the intercept (b0). Show your work X Y XY XX 1 4 4 1 2 6 12 4 1 3 3 1 0 1 0 0 1 1 1 1 X=5 , Y=15 , XY=20 , X2=7 b0=0.5 is the value of Y when X=0 c) Estimate the slope (b1). Show your work Dr. D Rawana Final Winter 2016 Page 4 c) Draw the regression line d) Is the relationship positive or negative? Explain Positive as Y increases with increase in X (Regression line is upward sloping) 7. If SSR = 76 and SST = 98, and n = 12 a. Determine the coefficient of determination , r2, and interpret its meaning. r r 2 2 = SSR/SST =76/98=0.7755 b. Compute the standard error of the estimate. Dr. D Rawana Final Winter 2016 Page 5 Standard Error is calculated by taking the square root of the average prediction error. SSE=SST-SSR=98-76=22 Standard error= SSE = nk 22 =2 121 Standard error=2 c. How useful do you think this regression model is for predicting? R2 value is quite high. This means the model is a good fit and can be used for predicting sales. 8. The Management at Ohio National Bank does not want its customers to wait in line for service for too long. The manager of a branch of this bank estimated that the customers currently have to wait an average of 8 minutes for service. Assume that the waiting times for all customers at this branch have a normal distribution with a mean of 8 minutes and a standard deviation of 2 minutes. Find the probability that randomly selected customer will have to wait for less than 4 minutes? Instructions: Show all steps: 1. Draw the normal curve and indicate the mean, standard deviation, and the X bar scale =8 Xbar=4 2. Identify the area of interest (that is shade the area under the curve that you will compute the probability). 3. Covert the X bar values in Z scores 4. P(X<4)=P(Z< 48 ) 2 =P (Z<-2) Z<-2 5. Look up the Z standardized table for the cumulative area(s). P(Z<-2)= 0.0228 6. Now, make your decision ( that a customer will wait for less than 4 minutes) 0.0228<0.05 Probability of customer being served with less than 4 minutes is insignificant Dr. D Rawana Final Winter 2016 Page 6 9. The Management at Ohio National Bank does not want its customers to wait in line for service for too long. The manager of a branch of this bank estimated that the customers currently have to wait an average of 8 minutes for service. Assume that the waiting times for all customers at this branch have a normal distribution with a mean of 8 minutes and a standard deviation of 2 minutes. What is the probability that customers have to wait for 4 to 8 minutes? Instructions: Show all steps: a. Draw the normal curve and indicate the mean, standard deviation, and the X bar scale Xbar=4 =8 b. Identify the area of interest (that is shade the area under the curve that you will compute the probability). c. Covert the X bar values in Z scores P(4X4)=P( 48 88 Z ) 2 2 =P (-2Z0) -2Z<0 d. e. look up the z standardized table for cumulative area(s). p(-2z<0) =0.5-0.0228=0.4772 f. now, make your decision ( that a customer will have to wait 4 8 minutes) 0.4772>0.05 Therefore customer will have to wait for 4 to 8 minutes is significant with probability of 0.4772 Dr. D Rawana Final Winter 2016 Page 7 10. Five hundred employees were selected from a city's large private companies, and they were asked whether or not they have any retirement benefits provided by their companies. Based on this information, the following two-way contingency table was prepared. Have Retirement Benefits Yes No Men 225 75 Women 150 Total 50 Total If one employee is selected at random from these 500 employees, a) Find the probability that this employee is a woman (200/(200+300)=0.4 b) Has retirement benefits (125+150)/(500)=0.55 Dr. D Rawana Final Winter 2016 Page 8 c) Has retirement benefits given the employee is a man 225/500=0.45 d) Is a woman given that she does not have retirement benefits 50/(50+75)=0.4 11. Use the given data to construct a frequency distribution. Lori asked 24 students how many hours they had spent doing homework during the previous week. The results are shown below: Number of students 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 28 19 20 21 22 23 24 Hours of study 11 10 11 9 11 11 15 12 11 8 13 10 10 13 11 10 13 11 10 12 10 13 12 9 Construct a frequency distribution, using 4 classes and a class width of 2 hours, and a lower limit of 8 for class 1. Dr. D Rawana Final Winter 2016 Page 9 Class (hour) 8-12 12-14 14-16 16-18 Frequency (No. of students) 19 4 1 0 12. An entrepreneur is considering the purchase of a coin-operated laundry. The current owner claims that over the past 10 years, the average daily revenue was $675 with a standard deviation of $75. A sample of 30 days reveals daily revenue of $625. Consider H0: = 675 versus Ha: < 675 a. At the .05 level of significance , is there evidence that the average daily revenue is less than 675 P(X<625)=P( Z < 625675 ) 75 / 30 =P(Z<-3.65)= 0.0001 0.0001<0.05 H0 is tot rejected. There is no enough evidence that the average daily revenue is less than 675 b. Compute the p-value and interpret its meaning P(Z<-3.65)= 0.0001 this is the probability that mean is less than 675 c. Construct a 95% confidence interval estimate of the population mean revenue of coin-operated laundry. 6251.96*75/ 30 62518.98 (606.02, 643.98) d. Compare the results of (a) , (b), and (c). What conclusions do you reach? Show all steps! There is no enough evidence that the average daily revenue is less than 675 Instructions: Show all steps such as 1. State the Hypothesis 2. Draw the normal distribution and identify the acceptance and rejection regions 3. Make your decision with respect to each part of the question. H0: = 675 versus Ha: < 675 Dr. D Rawana Final Winter 2016 Page 10 X=625 =675 There is no enough evidence that the average daily revenue is less than 675 Dr. D Rawana Final Winter 2016 Page 11 MG 620 Research and statistics for Managers Final Exam Instructions: Answer all questions Part I : Problem 1. Fitting a straight line to a set of data yields the following prediction line: Y = 4 + 3X i) Interpret the meaning of the Y intercept, b0. b0=4 is the value of Y when X is 0 ii) Interpret the meaning of the slope, b1 b1=3 mean that a unit change in X will increase Y by 3 units. iii) Predict the mean value of Y for X = 5 Y=4+3X Y=4+5(5)=9 2. What is the difference between the standard deviation, standard error of the mean, and standard error of the estimate? Discuss The standard deviation measures the amount of variability or dispersion for a subject set of data from the mean, while the standard error of the mean measures how far the sample mean of the date is likely to be from the true population mean. The standard error is an estimate of the standard deviation of a statistic. , it is used to compute the confidence intervals and margins of error. Dr. D Rawana Final Winter 2016 Page 1 3. In a sample of size of 5, the sum of all values is 200. What is the sample mean a. 20 b. 40 SHOW YOUR WORK! c. d. 800 None of the above Mean =1/nXi Mean=1/5(200)=40 The following are the durations in minutes of a sample of long-distance phone calls within the continental United States reported by one long-distance carrier Table 1 Time (in minutes) Relative Frequency 0 but less than 5 5 but less than 10 10 but less than 15 15 but less than 20 20 but less than 25 25 but less than 30 30 or more 0.37 0.22 0.15 0.10 0.07 0.07 0.02 4. Referring to Table 1, what is the width of each class? a. 1 minute b. 2% SHOW YOUR WORK! Dr. D Rawana c. d. 5 minutes 100% Final Winter 2016 Page 2 1. Width refers to the difference between the lower limit of any two consecutive classes Therefore, 5-0 = 5 or 10-5 = 5 5. Referring to Table 1, what is the cumulative relative frequency for the percentage of calls that lasted under 10 minutes? a. 0.10 b. 0.76 SHOW YOUR WORK! c. d. 0.59 0.84 Cumulative relative frequency is the sum of all relative frequency from the top to the last values: for example, the sum of all calls under 10 minutes is found by adding (.37 + .22) = .59). for under 20 minutes, it is .84 as shown above for all values in the third column. 6. The managers of a real estate firm are interested in finding out if the number of new clients a broker brings into the firm affects the sales generated by the broker. They sample 5 brokers and determine the number of new clients they have enrolled in the last year and their sales amounts in thousands of dollars. These data are presented in the table that follows. Broker Clients (X) 1 2 3 4 5 1 2 1 0 1 Sales (Y) 4 6 3 1 1 a. Construct a scatter diagram for these data. Does the scatter diagram show a linear relationship between sales and number of new clients? Explain and show all work including which variable is independent and dependent. Dr. D Rawana Final Winter 2016 Page 3 b. Y X b) Estimate the intercept (b0). Show your work X Y XY XX 1 4 4 1 2 6 12 4 1 3 3 1 0 1 0 0 1 1 1 1 X=5 , Y=15 , XY=20 , X2=7 b0=0.5 is the value of Y when X=0 c) Estimate the slope (b1). Show your work Dr. D Rawana Final Winter 2016 Page 4 c) Draw the regression line d) Is the relationship positive or negative? Explain Positive as Y increases with increase in X (Regression line is upward sloping) 7. If SSR = 76 and SST = 98, and n = 12 a. Determine the coefficient of determination , r2, and interpret its meaning. r r 2 2 = SSR/SST =76/98=0.7755 b. Compute the standard error of the estimate. Dr. D Rawana Final Winter 2016 Page 5 Standard Error is calculated by taking the square root of the average prediction error. SSE=SST-SSR=98-76=22 Standard error= SSE = nk 22 =2 121 Standard error=2 c. How useful do you think this regression model is for predicting? R2 value is quite high. This means the model is a good fit and can be used for predicting sales. 8. The Management at Ohio National Bank does not want its customers to wait in line for service for too long. The manager of a branch of this bank estimated that the customers currently have to wait an average of 8 minutes for service. Assume that the waiting times for all customers at this branch have a normal distribution with a mean of 8 minutes and a standard deviation of 2 minutes. Find the probability that randomly selected customer will have to wait for less than 4 minutes? Instructions: Show all steps: 1. Draw the normal curve and indicate the mean, standard deviation, and the X bar scale =8 Xbar=4 2. Identify the area of interest (that is shade the area under the curve that you will compute the probability). 3. Covert the X bar values in Z scores 4. P(X<4)=P(Z< 48 ) 2 =P (Z<-2) Z<-2 5. Look up the Z standardized table for the cumulative area(s). P(Z<-2)= 0.0228 6. Now, make your decision ( that a customer will wait for less than 4 minutes) 0.0228<0.05 Probability of customer being served with less than 4 minutes is insignificant Dr. D Rawana Final Winter 2016 Page 6 9. The Management at Ohio National Bank does not want its customers to wait in line for service for too long. The manager of a branch of this bank estimated that the customers currently have to wait an average of 8 minutes for service. Assume that the waiting times for all customers at this branch have a normal distribution with a mean of 8 minutes and a standard deviation of 2 minutes. What is the probability that customers have to wait for 4 to 8 minutes? Instructions: Show all steps: a. Draw the normal curve and indicate the mean, standard deviation, and the X bar scale Xbar=4 =8 b. Identify the area of interest (that is shade the area under the curve that you will compute the probability). c. Covert the X bar values in Z scores P(4X4)=P( 48 88 Z ) 2 2 =P (-2Z0) -2Z<0 d. e. look up the z standardized table for cumulative area(s). p(-2z<0) =0.5-0.0228=0.4772 f. now, make your decision ( that a customer will have to wait 4 8 minutes) 0.4772>0.05 Therefore customer will have to wait for 4 to 8 minutes is significant with probability of 0.4772 Dr. D Rawana Final Winter 2016 Page 7 10. Five hundred employees were selected from a city's large private companies, and they were asked whether or not they have any retirement benefits provided by their companies. Based on this information, the following two-way contingency table was prepared. Have Retirement Benefits Yes No Men 225 75 Women 150 Total 50 Total If one employee is selected at random from these 500 employees, a) Find the probability that this employee is a woman (200/(200+300)=0.4 b) Has retirement benefits (125+150)/(500)=0.55 Dr. D Rawana Final Winter 2016 Page 8 c) Has retirement benefits given the employee is a man 225/500=0.45 d) Is a woman given that she does not have retirement benefits 50/(50+75)=0.4 11. Use the given data to construct a frequency distribution. Lori asked 24 students how many hours they had spent doing homework during the previous week. The results are shown below: Number of students 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 28 19 20 21 22 23 24 Hours of study 11 10 11 9 11 11 15 12 11 8 13 10 10 13 11 10 13 11 10 12 10 13 12 9 Construct a frequency distribution, using 4 classes and a class width of 2 hours, and a lower limit of 8 for class 1. Dr. D Rawana Final Winter 2016 Page 9 Class (hour) 8-12 12-14 14-16 16-18 Frequency (No. of students) 19 4 1 0 12. An entrepreneur is considering the purchase of a coin-operated laundry. The current owner claims that over the past 10 years, the average daily revenue was $675 with a standard deviation of $75. A sample of 30 days reveals daily revenue of $625. Consider H0: = 675 versus Ha: < 675 a. At the .05 level of significance , is there evidence that the average daily revenue is less than 675 P(X<625)=P( Z < 625675 ) 75 / 30 =P(Z<-3.65)= 0.0001 0.0001<0.05 H0 is tot rejected. There is no enough evidence that the average daily revenue is less than 675 b. Compute the p-value and interpret its meaning P(Z<-3.65)= 0.0001 this is the probability that mean is less than 675 c. Construct a 95% confidence interval estimate of the population mean revenue of coin-operated laundry. 6251.96*75/ 30 62518.98 (606.02, 643.98) d. Compare the results of (a) , (b), and (c). What conclusions do you reach? Show all steps! There is no enough evidence that the average daily revenue is less than 675 Instructions: Show all steps such as 1. State the Hypothesis 2. Draw the normal distribution and identify the acceptance and rejection regions 3. Make your decision with respect to each part of the question. H0: = 675 versus Ha: < 675 Dr. D Rawana Final Winter 2016 Page 10 X=625 =675 There is no enough evidence that the average daily revenue is less than 675 Dr. D Rawana Final Winter 2016 Page 11