DSCI 346 Yamasaki Practice for Final Exam

DSCI Practice Final (24 pages) PROBLEM 1 (11.46 pg. 564) Mike is a restaurant owner and has read about the effect that music can have on the amount of time that customers stay at their tables for dinner. To test this effect, Mike makes arrangements to play slow-paced music at his establishment on a Saturday night and recorded the time customers at the randomly selected tables spent in the restaurant during five months of the year. Mike did the same on subsequent Saturday nights with fast-paced music and no music to adjust for any seasonal effects. Data appears below: Perform the appropriate test(s). Let a=.05. What do you conclude (in terms of the problem) DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) ANOVA table Source SS df MS F Between SSB k-1 MSB MSB/MSW Blocks SSBL b-1 MBBL MSBL/MSW Error SSW (k-1)(b-1) MSE Total SST N-1 ANOVA table Source SS df MS F Between 230.5335 2 115.2668 4.716 Blocks 11,101.6830 4 2,775.4208 113.562 Error 195.5168 8 24.4396 Total 11,527.7333 14 F2,8,05 = 4.459 Since 4.716 >4.459, sufficient evidence to conclude not all average times are the same. DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) Since 113.5 >3.838, sufficient evidence to conclude blocking factor was effective. b. If appropriate, find which pairs of means are significantly different (a = .05) DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) Find absolute differences between pairs of sample means |x1 – x2| = |97.40-106.80| = 9.40 |x1 – x3| = |97.40-100.40| = 3.00 |x2 – x3| = |106.80-100.40| = 6.40 CR = Qa  MSE/b D1 = k =3 D2 = (b-1)*(k-1) = )(5-1)*(3-1) = 4*2 = 8 Qa = 4.04 CR = 4.04 √24.4396/5 = 8.93 , so mean times customers stay at their tables for dinner is different when no music is played relative to when slow music is played DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) PROBLEM 2 (11.48 pg. 565) You have been assigned to test the hypothesis that the average number of hours worked per week by employees differs between the United States, China, and Sweden. A random sample of four employees from each country is chosen and the number of hours they worked last week is recorded. Data appears below Perform the appropriate test(s). Let a=.05. What do you conclude (in terms of the problem) DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) Typical way of reporting (ANOVA table) Source SS df MS F Between SSB k-1 MSB MSB/MSW (a.k.a. among) Within SSW N-k MSW (a.k.a. error) Total SST N-1 DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) Source SS df MS F Between 288 2 144 4.946 Within 262 9 29.1111 Total 716.6668 11 From F table F2,9,.05 = 4.256 H0: m1 = m2 = m3 HA: not all equal a = .05 Since F=4.946 is larger than 4.256 we can reject null hypothesis. Conclude sufficient evidence to indicate not all three countries work the same average number of hours. b. If appropriate, find which pairs of means are significantly different (a = .05) DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) Technique is same as for One Way ANOVA Find absolute differences between pairs of sample means |x1 – x2| = |36-30| = 6 |x1 – x3| = |36-42| = 6 |x2 – x3| = |30-42| = 12 CRij = QaMSW/2)(1/ni +1/nj) D1 = k =3 D2 = N-k =12-3 = 9 Q.05 = 3.95 CR = 3.95 √29.111/4 = 10.656 , so the average number of hours worked per week is significantly different between Sweden and China DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) PROBLEM 3 (14.50 pg 695) Comcast’s customer service department asks its customers to rate their over-the phone service on a scale of 1-20 immediately after the service has been completed. The company then matches each customer ‘s ratings with the number of minutes the person waited on hold. The company would like to predict the customers’ ratings based on the number of minutes waited and uses data from 10 randomly selected customers. Data appears below: Sxx = 84 Sxy = -41 Syy = 70.1 Sx = 50 Sy = 147 n = 10 Calculate the least squares regression line DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) b1 = Sxy/Sxx = -41/84 = -.4881 b0 = y- b1* x = (147/10) - (-.4881)*(50/10) = 17.1405 ⌃ Y = 17.1405 - .4481x b. What would be the predicted rating if a customer was on hold for 5 mins DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) ⌃ Y = 17.1405 - .4481(5) = 17.1405 – 2.2405 = 14.90 c. Is there sufficient evidence to indicate that the population slope of the regression line is less than 0? Let a = .05 Use Sy/x = 2.5022. Find the p value of your test statistic. DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) H0: b = 0 HA: b<0 =.05 T = b1-0 = -.4881 = -.4881 = -1.788 Sy/x/√Sxx 2.5022/√84 .2730 -t8,05 = -1.86. Since -1.788 is not < -1.86, insufficient evidence to indicate population slope <0 -1.86 < -1.788 < -1.397 so .05 < p value < .10 d. Calculate the estimate of the correlation coefficient DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) r = Sxy = -41 = -41 = -.5343 √Sxx √Syy √84 * √70.1 76.7359 DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) PROBLEM 4 Table 15.4 pg. 732 David is a street vendor who sells hot dogs in the city and would like to predict daily demand for his product in order to improve inventory control. He believes the three main variables affecting sales are his price per hot dog, the high temperature during business hours that day and whether or not the day falls on a weekend. David records the data for 12 randomly selected days, coding weekdays as 1 and weekends as 0. Complete the table below Source SS DF MS F Regression 10220.4191 Residual Total 12321.0000 DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) Source SS DF MS F Regression 10,220.4191 3 3,406.8064 12.9747 Residual 2,100.5809 8 262.5726 Total 12,321.0000 11 b. Based on your results to part a, what do you conclude DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) Since F3,8,05 = 4.066 and F=12.9747> 4.066 sufficient evidence to indicate a significant linear relationship between temperature, price, day and the demand for his hot dogs. c. Calculate the coefficient of determination DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) R2 = SSregression/SStotal = 10,220.4191 / 12321 = .8295 d. Based on the following plot do you think the assumption of equal variance is violated? DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) No, approximately a “horizontal sleeve” e. Based on the following plot do you think the assumption of normality is violated DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) No, approximate straight line DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) PROBLEM 5 Data from 16.5 pg. 782 Bob is a photographer who sells his 13 x 9 prints on consignment at Island Art in Stone Harbor, New Jersey. Each week during the summer Bob checks the inventory so that he can replenish the prints tht have been sold. The following data show the number of prints sold each week during the last eight weeks. a. Forecast the demand for Week nine using a two-period moving average. DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) (SMA p=2) = (11+7)/2 = 18/2 = 9 b. Compute MAD DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) c. Forecast Week nine demand using a three period weighted moving average with weights 3 (most recent), 2, 1 (oldest) DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) (WMA p=3 (3,2,1)) = (3*7+2*11+1*15)/6 = 58/6 = 9.67 d. Forecast demand for Week nine using exponential smoothing with a = .5. DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) Ft = F t-1 + a(At-1 – Ft-1) Let a = 0.5 F1 = A1 = 9 F2 = F 1 + (.5)(A1 – F1) = 9 + (.5)*(9-9) = 9 F3 = F 2 + (.5)(A2 – F2) = 9 + (.5)*(12-9) = 9+ 1.5 = 10.5 F4 = F 3 + (.5)(A3 – F3) = 10.5 + (.5)*(10-10.5) = 10.5 - 0.25 = 10.25 F5 = F 4 + (.5)(A4 – F4) = 10.25 + (.5)*(6-10.25) = 10.25 – 2.125 = 8.125 F6 = F 5 + (.5)(A5 – F5) = 8.125 + (.5)*(9-8.125) = 8.125+ .4375 = 8.5625 F7 = F 6 + (.5)(A6– F6) = 8.5625 + (.5)*(15-8.5625) = 8.5625 + 3.2188 = 11.7813 F8 = F 7 + (.5)(A7 – F7) = 11.7813 + (.5)*(11-11.7813 = 11.7813 - .3907 = 11.3907 F9 = F 8+ (.5)(A8– F8) = 11.3907 + (.5)*(7-11.3907) = 11.3907 – 2.1954 = 9.1954 DSCI Practice Final (24 pages)

DSCI Practice Final (24 pages) Week Sold SMA(p=2) AD WMA(3,2,1) ES (.5) 1 9.00   2 12.00 3.00 3 10.00 10.50 0.5 0.50 4 6.00 11.00 5 4.50 10.25 4.25 8.00 8.33 0.67 8.13 0.88 6 15.00 7.50 7.5 8.17 6.83 8.56 6.44 7 11.50 11.78 0.78 8 7.00 13.00 5.00 11.39 2.64 9 9.67 9.20 MAD 3.5 DSCI Practice Final (24 pages)