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# A 4500 b 7125 c 4950 d 125 e 450 6 18 a statistician

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A. 45.00 B. 71.25 C. 49.50 D. 1.25 E. 4.50 6

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18. A statistician working for a car manufacturer developed a statistical model for predicting delivery time (the number of days between ordering a car and actual delivery) of a particular model for which there is a range of factory-fitted options. Use the output below to forecast the difference in delivery times for cars with 6 options and 9 options, rounded to two decimal places. A. 4.38 B. 3.28 C. 6.58 D. 9.84 E. 29.73 SUMMARY OUTPUT Regression Statistics Multiple R 0.92 R Square 0.85 Adjusted R Square 0.83 Standard Error 4.38 Observations 10 Analysis of Variance df SS Regression 1 890.19 Residual 8 153.41 Total 9 1043.59 Coefficients Standard Error Intercept 29.73 2.99 Factory-Fitted Options 3.28 0.48 7
19. Given the table below what is the MAD and the MSE? Period Actual Forecast 95 100 108 110 123 120 130 130 MAD = ___________________ MSE = ___________________________ 20. The Acme Computer Company has recorded sales of one of its products for a six-week period: Using the three-week simple moving-average method, forecast sales for week 7. A. 20 B. 21 C. 22 D. 23 F. 24 21. Shown below are data that reflect the number of daily traffic accidents at a dangerous city intersection. The regression equation is: number of accidents = 5.3 + 0.5 t What is the forecasted number of accidents for day 6? Day (t) number of accidents 1 5 2 7 3 8 4 6 5 8 A. 8.0 B. 5.8 C. 8.3 D. 0.6 E. 9.0 8

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22. The following table contains the number of consumer complaints received in a Publix supermarket in Florida. Use exponential smoothing with a constant of α = 0.33 to forecast the number of complaints in March, round to the nearest whole number of complaints. Month Number of Complaints January 36 February 45 March 81 April 90 May 108 June 144 A. 39 B. 42 C. 45 D. 53 E. 64 For the next 3 questions, consider the following summary output from Microsoft Excel for a simple regression of Halliburton (ticker symbol HAL) weekly stock prices on the S&P500 stock index, for the period January 2004 through August 2006. SUMMARY OUTPUT Regression Statistics Multiple R 0.93 R Square 0.87 Adjusted R 0.86 Standard Error 3.20 Observations 140 ANOVA Df SS MS F Significanc Regression 1 9067.88 9067.8 884.90 6.84E-62 Residual 138 1414.13 10.24 Total 139 10482.0 Coefficient Standard t Stat P-value Lower 95% Upper Intercept -121.65 4.914 -24.74 1.37E- -131.37 -111.93 S&P500 Index 0.122 0.004 29.75 6.84E- 0.11 0.13 23. According to the regression model, the correlation between Halliburton’s stock price and the S&P500 index value is: A. Positive. B. Negative. C. Significant and either positive or negative. D. Significant, but it cannot be determined whether the correlation is positive or negative. E. Not significant. 9
24. The regression model ABOVE is: A. Not significant, because the standard error of the intercept is high compared to the standard error of the independent variable. B. Significant at the 0.01 level, because the coefficient of determination is less than 0.05. C. Significant at the 0.05 level, because the p-value is less than 0.05. D. Not significant, because the coefficient of determination is larger than 1.0. E. Significant at the 0.01 level, because the t-statistic is negative.

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A 4500 B 7125 C 4950 D 125 E 450 6 18 A statistician...

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