Consider the following given below: Week Sales('000) 1 39 44 3 40 4 45 38 43 39 What is the Mean Absolute Percent error (MAPE)?
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- Discuss the basic differences between the mean absolute deviation and mean absolute percent error.Chart and Regression analysis : What does the intercept predict? X: C16 (number of cars on the sales lot) versus Y: C17 (cars sold per day) Equation: y=2.9x + 14.5 Slope:2.9 Intercept:14.5 Does the intercept mean the intercept is 14.5 means that the cars sold per day( Y) predicted number of cars on sale lot(X) to be 14.5, but this intercept has no meaning. So, I will not use to predict cars sold per day?standard Deviation (σ) = SQRT(Variance) = SQRT(6) = 2.45 This is not clear for me. How did you do it?
- ns Management || fall20 Random variation is also known as common cause variation. Select one: O a. TRUE O b. FALSEConsider the ANOVA table that follows. Analysis of Variance Source DF SS MS F p Regression 2 77.907 38.954 4.14 0.021 Residual Error 62 583.693 9.414 Total 64 661.600 a-1. Determine the standard error of estimate. (Round your answer to 3 decimal places.) a-2. About 95% of the residuals will be between what two values? (Round your answers to 3 decimal places.) b-1. Determine the coefficient of multiple determination. (Round your answer to 3 decimal places.) b-2. Interpret the coefficient of multiple determination. (Round your answer to 1 decimal place.) Determine the coefficient of multiple determination, adjusted for the degrees of freedom. (Round your answer to 3 decimal places.) Prev Question 5 of 30 Total5 of 3The following data was recorded while observing the increase in the number of butterflies in two different ponds, Pond A and Pond B. Number of Butterflies Present Over 5 Years Year Pond A Pond B 0 50 65 85 1 2 3 4 5 100 O 110 121 133 146 161 110 143 186 What is the rate of growth, r, for each pond and which pond has the fastest average rate of growth over the five years? TA = 1.1, TB = 1.3, mg = 3.72 butterflies/yr TA=0.9, TB=0.8, mg = 3.72 butterflies/yr TA=0.9, TB=0.8, mg=27.2 butterflies/yr TA = 1.1, TB = 1.3, mg=27.2 butterflies/yr
- Bill Kime's bowling ball factory makes bowling balls of adult size and weight only. The standard deviation in the weight of a bowling ball produced at the factory is known to be 0.39 pounds. Each day for 24 days, the average weight, in pounds, of 9 of the bowling balls produced that day has been assessed as follows Day 1 2 3 4 5 6 Average (lb) 9.9 9.9 10.1 10.1 99 99 Day 7 8 9 10 11 12 Average (lb) Day Average (b) 13 14 15 16 17 18 10.1 10.0 10.0 10.1 10.1 10.0 9.9 10.1 10.1 10.0 10.1 10.1 Day 19 20 21 22 23 24 Average (lb) 99 10.0 10.1 9.9 10.0 10.0 a) Establish a control chart for monitoring the average weights of the bowling balls in which the upper and lower control limits are each two standard deviations from the mean What are the values of the control limits? Upper Control Limit (UCL)=b (Round your response to two decimal places)Given the attached data. Answer the following questions for a 7 period moving average. t At 1 5477 2 3338 3 4799 4 2852 5 6332 6 8385 7 4382 8 4184 9 14229 10 10385 11 11676 12 5271 13 8938 14 9279 15 13990 16 2890 17 13027 18 8970 19 6700 20 7065 21 12491 22 8748 23 8396 24 6102 25 7559 26 7997 27 11981 28 10680 29 5783 30 2501 31 7620 32 10194 33 7908 34 11985 35 5863 36 3024 37 8758 38 1640 39 6569 40 11728 41 6659 42 2830 43 10018 44 8448 45 12409 46 2083 47 6374 48 7438 49 4576 50 4795 MAD = Average(|A-F|) TS =SUM(A-F)/MAD MSE = Average(A-F)2 1. Compute your forecast for period 51. The potential answers are: A: 11221.6 units. B: 12461 units. C: 8020.83 units. D: 4414 units. E: 6589 units.Bo Using data from 50 workers, a researcher estimates Wage = ẞe + B₁Education + B2Experience + B3Age + ε, where Wage is the hourly wage rate and Education, Experience, and Age are the years of higher education, the years of experience, and the age of the worker, respectively. A portion of the regression results is shown in the following table. Intercept Education Experience Age Coefficients Standard Error t Stat p-Value 7.17 4.26 1.68 0.0991 1.81 0.35 5.17 0.0000 0.45 0.10 4.50 0.0000 -0.01 0.06 -0.17 0.8684 a-1. Interpret the point estimate for ẞ1. As Education increases by 1 year, Wage is predicted to increase by 1.81/hour. As Education increases by 1 year, Wage is predicted to increase by 0.45/hour. As Education increases by 1 year, Wage is predicted to increase by 1.81/hour, holding Age and Experience constant. As Education increases by 1 year, Wage is predicted to increase by 0.45/hour, holding Age and Experience constant. a-2. Interpret the point estimate for $2. As Experience…
