BUSS 230 Midterm Solution, Fall 2013

BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 1 of 32 BUSS 230 Managerial Economics Fall 2013 – 2014 Dr. George G. Namur Midterm Review Problems Solution Problem 1 a) The profit function (Q) is given by: ( ) ( ) ( ) ( ) ( ) ( ) b) Profit is maximized where the first derivative of the profit function with respect to output equals zero, while the second derivative is negative, or: √ ( )( ) ( ) The other solution is negative. Verify that at Q = 4. ( ) Profit is therefore maximized at Q* = 4. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 2 of 32 Problem 2 Marginal cost is minimized where and Therefore, marginal cost is minimal where Q* = 4.5. Problem 3 Solving the above two equations simultaneously yields: X* = 10 and Y* = 15. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 3 of 32 Problem 4 Substitution method ( ) ( ) ( ) Maximizing profit means setting the first derivative of the profit function equal to zero. Lagrangian method ( ) BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 4 of 32 First equation minus twice the second: So Plugging Y into third equation, Plugging X and Y into second equation yields BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 5 of 32 Problem 5 Using PX = 6, PY = 12 and I = 4.5 ( ) ( ) ( ) a) . b) Demand is inelastic since |EP| < 1. c) At this point on the demand for product X (inelastic portion), the sign of the MR is negative. d) Decreasing the price of X from 6 to 5 will cause TR to decrease because demand is inelastic. e) At PX = 7, QX = 59. The arc price elasticity of demand between PX = 6 and PX = 7 is equal to f) . Y is a substitute product to X because a higher price for product Y yields a higher demand for X. g) . X is a normal good because a higher income level yields a higher demand for X. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 6 of 32 Problem 6 a) An increase in the price will lead to an increase in total revenues (as Terry would want) only when the prices lie on the inelastic portion of the demand. There are two alternative approaches to answering this question. The first consists of using marginal revenue to obtain the range of prices for which demand is elastic or inelastic. That is, first obtain the total revenue function as: ( ) ( ) This implies that the marginal revenue function is: ( ) We know that demand is elastic when MR > 0, inelastic when MR < 0 and unitary elastic (midpoint of the demand curve) where MR = 0. ( ) The corresponding price is ( ) . That is, at Q = 50 and P = $7.5, demand is such that |EP| = 1. This implies that for Q < 50, or P > $7.5, demand is elastic and |EP| > 1. This means that a price of $9 is on the elastic portion of demand. If Terry raises her price above $9, this will lead to decrease in total revenues. Therefore she should not increase the price. Alternatively, one can directly compute the point price elasticity of demand at P = $9. At P = $9, the point price elasticity of demand is given by: BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 7 of 32 At P = $9, Q = 40. . Given that |EP| > 1 (a price of $9 is on the elastic portion of demand), a price increase will lead to a decrease in total revenues since the new price will also lie over the elastic portion of demand. b) Only the first approach will be used here. Given P = 22- 0.22Q, the marginal revenue function is MR(Q) = 22-0.44Q. MR = 0 for Q = 50 and P = $11 and this corresponds to the unitary elastic point (midpoint) |EP| = 1 of the demand. If Terry raises her price anywhere from $9 to $11, she would be operating over the inelastic portion of the demand and total revenues will increase. Terry should therefore increase her price for haircuts. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 8 of 32 Problem 7 a) The inverse demand function is given by expressing P as a function of Q ( ) b) The total revenue function is given by: ( ) c) The marginal revenue function is given by: d) The quantity, Q*, at which total revenue is maximized is obtained by setting MR equal to zero. Therefore, Q* = 10. e) No, total revenue maximization does not qualify as an example of constrained optimization, since there is no constraint to account for (a cost constraint for example). BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 9 of 32 Problem 8 a) The total revenue function is given by: ( ) The marginal revenue function is equal to the first derivative of the total revenue function with respect to output, or: b) The demand for the firm’s product is price elastic when MR > 0, i.e. when: Plugging Q < 37.5 back into the demand function yields: c) Total revenue is maximized when marginal revenue equals zero, i.e., when Q = 37.5 and P = $1,500. Problem 9 a) These goods are extremely close substitutes. A 1% increase (decrease) in the price of one good results in a 1.2% increase (decrease) in the quantity demanded for the other. b) . Therefore, %Q = 6%. This implies that the demand for the good rises by 6% as a result of a 5% increase in the price of the substitute. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 10 of 32 Problem 10 a) By plugging the value of P into the demand function we get: for P = $10, Q = 5,000 – 50(10) = 4,500 for P = $20, Q = 5,000 – 50(20) = 4,000 for P = $30, Q = 5,000 – 50(30) = 3,500 b) The arc price elasticity is given by: The arc price elasticity between $10 and $20 is equal to: The arc price elasticity between $20 and $30 is equal to: c) The point price elasticity is given by: At P = $10: At P = $20: At P = $10: BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 11 of 32 Problem 11 The arc price elasticity can be computed using: ( ) ( ) The total and marginal revenues are equal to: and Note that the changes considered here are changes of 10 units, so that you have to divide changes in TR by 10 to get the MR. Quantity Arc Total Marginal Price Qd (pounds Elasticity Revenue Revenue P ($) of steak) Ed ($) ($/unit) $12 30 NA 360 NA $11 40 -3.290 440 8 $10 50 -2.330 500 6 $9 60 -1.727 540 4 $8 70 -1.308 560 2 $7 80 -1.000 560 0 $6 90 -0.765 540 -2 $5 100 -0.579 500 -4 $4 110 -0.429 440 -6 BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 12 of 32 Problem 12 a) Plugging the values P = $200, M = $60,000 and PR = $100 yields: ( ) ( ) ( ) b) The price elasticity of demand is given by: The above equation uses the fact that ΔQ/ΔP, the slope of the demand function with respect to P, is equal to -500. c) The income elasticity of demand is given by: Again, the above equation uses the fact that ΔQ/ΔM, the slope of the demand function with respect to M, is equal to -1.5. Good X is inferior since an increase in M, leads to a decrease in Q. The effect of a 4% increase in income can be calculated using: EM = %ΔQ/%ΔM = -2.5. Therefore, %ΔQ/4% = -2.5, or %ΔQ = -10%. This implies that a 4% increase in come leads to a 10% decrease in demand. d) The cross-price elasticity is given by: The above computation uses the fact that ΔQ/ΔPR, the slope of the demand function with respect to R, is equal to -240. The negative sign of the cross-price elasticity between X and R indicates that they are complementary goods, since an increase in PR leads to a decrease in the demand for X. To assess the effect of a 5% decrease in PR on the demand for X we use: EXR = %ΔQ/%ΔPR = -0.6667 so that %ΔQ/-5% = -0.6667 implying %ΔQ = 3.335%. This means that a 5% decrease in the price of R leads to 3.335% increase in the demand for X. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 13 of 32 Problem 13 The estimated market demand for good X is: ̂ Where ̂ is the estimated number of units of good X demanded, P is the price of the good, M is income, and PZ is the price of the related good Z. (all parameter estimates are statistically significant at the 1% level.) a) X is an inferior good. The negative sign of ΔQ/ ΔM = -0.6 implies that an increase in income, M, leads to decrease in demand Q. b) X and Z are substitutes. The positive sign of ΔQ/ΔPZ = 4 indicates that the goods are substitutes. An increase in PZ leads to an increase in the demand for X. c) Plugging P = 10, M = 30, and PZ = 6 into the demand function we get: ( ) ( ) ( ) Therefore, BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 14 of 32 Problem 14 a) Law of demand stipulates that b should be negative (there’s a negative relationship between the price of the good and quantity demanded). ̂ is indeed negative as is predicted theoretically and equals -6.50. b) ̂ = ΔQ/ΔM=0.13926 > 0 which implies that the good is normal since demand increases as income increases. c) The goods are complementary since ̂ = -10.77= ΔQ/ΔPR < 0. This implies that an increase in the price of R leads to a decrease in the demand for X. d) First, using p-values: (On an exam, you should state the null and alternative hypotheses and provide details similar to the ones provided here) Decision rule: Reject H0 if p-value < α α= 5% = 0.05 First, starting with the intercept, the null and alternative hypotheses are: H0: a = 0 H1: a ≠ 0 p-value for , ̂ is 0.0001 < 0.05 This implies that we reject H0, ̂ is significant. To test the significance of own price of the good, we state the null and alternative hypotheses: H0: b = 0 H1: b ≠ 0 p-value is 0.0492 < 0.05. This implies we reject H0, so that the price of the good significantly affects quantity demanded. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 15 of 32 To test the significance of income, we state the null and alternative hypotheses: H0: c = 0 H1: c ≠ 0 p-value is 0.0001 < 0.05. This implies we reject H0, so that income significantly affects demand. Regarding the price of the related good R, the null and alternative hypotheses are: H0: d = 0 H1: d ≠ 0 p-value is 0.0002 < 0.05. This implies we reject H0, so that the price of related good R significantly affects demand. Now, we test the same hypotheses stated above, using t-statistics: Decision rule: Reject H0 if the calculated t-statistic, in absolute value, exceed the critical value. The calculated t-statistics are: ̂ ̂ ( ̂) ̂ ̂ ( ̂) ̂ ̂ ( ̂) ̂ ̂ ( ̂) Critical value at α = 5% with degrees of freedom df = (n – k) = 24 – 4 = 20 is obtained from the t-table as: c/2,n-k = 2.086 (Note n = 24 as shown next to observations on the computer output whereas k = 4). BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 16 of 32 | ̂| = 5.41 > 2.08 so that we reject H0, meaning that the intercept is significant. | ̂| =2.06 < 2.08 meaning we do not reject H0. This implies that the price of the good does not affect the quantity demanded significantly. | ̂| =10.63 > 2.08. We reject H0, meaning that income affect demand significantly. | ̂| =4.40 > 2.08. We reject H0, meaning that the price of good R affects demand significantly. Note that using p-values and t-tests leads to different conclusions regarding the significance of the price of the good. This is an inconsistent result and should not typically occur. e) The null and alternative hypotheses are: H0: a = b = c = d =0 H1: H0 is not true The regression output shows that R2 = 0.8118. We also have k = 4. To test the null of joint insignificance, we need to compute the F-statistic as: ( ) ⁄ ( ) ( ) ⁄ ( ) ⁄ ( ) ( ) ⁄ The appropriate critical value is obtained from the F-table at a significance level α = 5% as Fc = 3.10 (3 degrees of freedom for the numerator and 20 degrees of freedom for the denominator). The decision rule is to reject H0 if F > Fc. Since 28.7872 > 3.10, the null hypothesis is rejected. This implies that the variables included in this regression are jointly significant. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 17 of 32 The decision rule using p-values is to reject the null if p-value < α. Given that the p-value associated with the F-statistic is 0.0001 < 0.05, H0 is rejected using p-values and the variables are jointly significant. f) By plugging P = 225, M = 24,000 and PR = 60 into the demand function we get the quantity: 1) ̂ 2) ̂ 3) ̂ BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 18 of 32 Problem 15 a) First, using p-values: Decision rule: Reject H0 if p-value < α  = 5% = 0.05 (since it has not been specified) First, starting with the intercept, the null and alternative hypotheses are: H0: a = 0 H1: a ≠ 0 p-value for ̂ is 0.0716 > 0.05. This implies that we do not reject H0, or that the intercept is not significant. To test the significance of own price of the good, we state the null and alternative hypotheses: H0: b = 0 H1: b ≠ 0 p-value for ̂ is 0.0093 < 0.05. This implies we reject H0, so that the price of the good significantly affects quantity demanded. To test the significance of income, we state the null and alternative hypotheses: H0: c = 0 H1: c ≠ 0 p-value for ̂ is 0.0009 < 0.05. This implies we reject H0, so that income significantly affects demand. Regarding the price of tennis rackets, the null and alternative hypotheses are: H0: d = 0 H1: d ≠ 0 p-value for ̂ is 0.006 < 0.05. This implies we reject H0, so that the price of tennis rackets significantly affects demand. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 19 of 32 Note that the same conclusions would hold if we use a level of significance of 1%. At the 10% level, all the variables including the intercept would be significant. The signs of the parameter estimates are consistent with demand theory: ̂ is negative as the law of demand predicts, ̂ is positive indicating tennis balls are a normal good and we expect for most goods, and ̂ is negative as it should be for complements (tennis balls and tennis rackets). Now, we test the same hypotheses stated above, using t-statistics: Decision rule: Reject H0 if the calculated t-statistics, in absolute value, exceed the critical value. The calculated t-statistics are: ̂ ̂ ( ̂) ̂ ̂ ( ̂) ̂ ̂ ( ̂) ̂ ̂ ( ̂) Critical value at α = 5% with degrees of freedom df = (n – k) = 20 – 4 = 16 is obtained from the t-table as: c/2,n-k = 2.12 (Note n = 20 as shown next to observations on the computer output whereas k = 4). | ̂| = 1.92 < 2.12 so we do not reject H0, meaning that the intercept is not significant. | ̂| = 2.96 > 2.12 so we reject H0, meaning that the price of the good affects quantity demanded significantly. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 20 of 32 | ̂| =4.0811 > 2.12 so we reject H0, meaning that income affect demand significantly. | ̂| =3.16 > 2.12 so we reject H0, meaning that the price of tennis rackets affects demand significantly. The decisions regarding the significance of the variables that we get from p-values and t-statistics are consistent. b) Denote by c/2,n-k the critical value corresponding to a 99% confidence interval (that is, at an α = 1%). Then, the 99% confidence intervals for b, c and d are given respectively by: [ ̂ ( ̂) ̂ ( ̂) ] [ ̂ ( ̂) ̂ ( ̂) ] [ ̂ ( ̂) ̂ ( ̂) ] Here α = 1% and n - k = 16 so that c/2,n-k = 2.921. Therefore, for the population parameter b, we have: or Similarly, for the population parameter c, we have: or Similarly, for the population parameter d, we have: or The decision rule using confidence intervals is to reject H0 if the hypothesized value lies outside the interval. In our case, the hypothesized value is zero. Since zero lies outside the BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 21 of 32 99% confidence interval for b, H0 is rejected. Similarly, the null hypothesis H0 is rejected using the other two 99% confidence intervals for c and d since the hypothesized value lies outside the 99% confidence interval. c) The null and alternative hypotheses are: H0: a = b = c = d =0 H1: H0 is not true The regression output shows that R2 = 0.8435. We also have k = 4. To test the null of joint insignificance, we need to compute the F-statistic as: ( ) ⁄ ( ) ( ) ⁄ ( ) ⁄ ( ) ( ) ⁄ The appropriate critical value is obtained from the F-table at a significance level α = 5% as Fc = 3.24 (3 degrees of freedom for the numerator and 16 degrees of freedom for the denominator). The decision rule is to reject H0 if F > Fc. Since 28.6939 > 3.24, the null hypothesis is rejected. This implies that the variables included in this regression are jointly significant. d) As indicated by R2, 84.35% of the variation of cans of tennis balls sold quarterly is “explained” by the independent variables in this regression. The unexplained variation is therefore 15.65%. e) The adjusted R2 of this regression can be computed as: ̅̅̅̅ ( )( ) ( )( ) f) Plugging those values into the demand equation we get: ( ) ( ) ( ) 134 cans per quarter. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 22 of 32 g) The elasticities can be computed as: h) % ΔQ /-15% = -0.256; implying that sales rise by 3.84% when the price of tennis balls decreases by 15%. i) % ΔQ / 20% = -0.153; implying that sales rise by 2.98% due to an increase in average household income by 10%. j) % ΔQ / 25% = -0.667; implying that sales fall by 16.67% due to an increase in the average price of tennis rackets by 25%. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 23 of 32 Problem 16 ̂ a) Yes, the sign of the estimated coefficient is consistent with theory. The law of demand holds that price and quantity demanded are inversely related and this is what we observe in this estimated demand function. b) Given that the estimated coefficient on income is -0.12, X is an inferior good. An increase in income leads to a decrease in the demand for X. c) Goods X and G are complements. This is indicated by the sign of the coefficient associated with PG (the coefficient is -30). This, in turn, indicates that an increase in the price of G leads to a decrease in the demand for X. d) The predicted quantity of good X is equal to: ̂ e) 1) Own price elasticity 2) Cross-price elasticity 3) Income elasticity f) 65% of the variation in quantity demanded is “explained” by variation in the independent variables included in this regression. The remaining 35% of the variation remains unexplained. g) Given that n = 25 and k = 4, and R2 = 0.65: ( ) ⁄ ( ) ( ) ⁄ ( ) ⁄ ( ) ( ) ⁄ BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 24 of 32 h) The appropriate critical value at the 5% level is Fc = 3.07. The null hypothesis is: H0: a = b = c = d =0 The alternative hypothesis is: H1: H0 is not true We reject H0 if F-statistic > critical value. In this case 12.976 > 3.07 so the null hypothesis is rejected at the 5%, and the coefficients are jointly significant. i) A 20% increase in income will lead to 20% 1.38= 27.6% decrease in the demand for X. j) A 15% increase in the price of G leads to 15% 0.58 = 8.7% decrease in the demand for X. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 25 of 32 Problem 17 a) In order to check whether shoe sales exhibit an upward trend we need to consider the sign and the significance of the coefficient associated with the time trend, namely b. The estimated coefficient ̂ is positive, meaning there is an upward trend in sales. Next, we need to check the significance of the population coefficient. In other words, we need to test the following null hypothesis: H0: b = 0 H1: b ≠ 0 The critical value of t for 28 – 5 = 23 degrees of freedom and a 5% significance level is 2.069 (from the t-table), while the t-ratio for ̂ is 2100/340 = 6.1765 and is much greater than the critical value. (The rejection rule is such that the null is rejected when the absolute value of the t-statistic exceeds the critical value). This leads us to reject the null hypothesis which implies that ̂ is significant or equivalently that the time trend enters the regression significantly. Using p-values leads to a similar result. The p-value for the time trend (0.0001) indicates strong statistical significance as it is much smaller than  = 0.05. Thus, the regression analysis provides very strong statistical evidence of an upward trend in shoe sales. b) Looking at the t-tests (at α = 5% = 0.05 significance level, the critical value being 2.069) we have: For D1, the null and alternative hypotheses are: H0: c1 = 0 H1: c1 ≠ 0 The t-statistic is: t = 3280/1510 = 2.17 BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 26 of 32 Since t –statistic = 2.17 > 2.069 we reject the null hypothesis meaning that the dummy variable associated with the first quarter, D1, is significant. For D2, the null and alterative hypotheses are: H0: c2 = 0 H1: c2 ≠ 0 The t-statistic is t = 2.82 > 2.069 meaning that we reject the null and that D2 is significant. For D3, the null and alterative hypotheses are: H0: c3 = 0 H1: c3 ≠ 0 The t-statistic is = 4.44 > 2.069 meaning that D3 is significant. Testing the joint significance of all the variables included in the regression amounts to testing the null and alternative hypotheses that follow: H0: a = b = c1 = c2 = c3 = 0 H1: H0 is not true To do that, we can compute the F-statistic as: ( ) ⁄ ( ) ( ) ⁄ With n = 28, k = 5 and R2 = 0.9651. Doing the computations yields an F- statistic of 159.01 which far exceeds the critical value of 2.80 coming from the F-table with α= 5%. This implies that the variables are jointly significant. Since all seasonal dummy variables are significantly different from zero, the data suggest that athletic shoe sales tend to exhibit a seasonal BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 27 of 32 pattern. Individual and joint significance tests point to an important seasonal pattern in sales. Ceteris paribus, sales are expected to be 3,280 units higher in quarter 1 than in quarter 4, 6,250 units higher in quarter 2 than in quarter 4, and 7,010 units higher in quarter 3 than in the quarter 4. c) For 2008Q3 and 2009Q2, t = 31 and 34, respectively. ( ) ( ) ( ) ( ) units ( ) ( ) ( ) ( ) units d) You might be able to improve this forecast equation by adding some additional explanatory variables that affect sales such as the price of the good, income, or a measure of aggregate economic activity. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 28 of 32 Problem 18 3-Year Exponential Actual Moving Smoothing Year Demand Average (w = 0.3) 1995 800 NA NA 1996 925 NA 800.00 1997 900 NA 837.50 1998 1,025 875 856.25 1999 1,150 950 906.87 2000 1,160 1025 979.81 2001 1,200 1112 1033.87 2002 1,150 1170 1083.71 a) The first forecast, for 1998, is obtained as F(1998)= (800+925+900)/3 = 875. The forecast for 1999, is F(1999) = (1025+900+925)/3 = 950. The remaining forecasts are obtained similarly. See table above for the other values of the forecasts. b) The exponential smoothing forecasts are computed as: ( ) When w = 0.3 and the first forecast is set to the actual, the forecast for 1996 is simply 800 (the value of the actual demand from the previous year) and the recursion can be started there to get as an example a forecast for 1997 of: ( ) and so on. See the table for the other values. c) The criterion used to compare forecast accuracy is the root mean squared error (RMSE) given by: √∑ ( ) BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 29 of 32 where At denotes the actual (realized) value of the series, Ft denotes a forecast coming from one of the competing methods (moving averages, exponential smoothing) and n denotes the number of forecasts. The forecasting method yielding the lowest RMSE would be the best. In other words, when comparing two forecasting methods, a lower RMSE implies a better forecasting method. d) For the 3-year moving average forecasting method: 3-Year Actual Moving Year Demand Average (A – F)2 1995 800 NA NA 1996 925 NA NA 1997 900 NA NA 1998 1,025 875 22,500 1999 1,150 950 40,000 2000 1,160 1025 18,225 2001 1,200 1112 7,744 2002 1,150 1170 400 RMSE = 133.318 BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 30 of 32 For the exponential smoothing forecasting method: Exponential Actual Smoothing Year Demand (w = 0.3) (A – F)2 1995 800 NA NA 1996 925 800.00 15,625 1997 900 837.50 3,906.25 1998 1,025 856.25 28,476.56 1999 1,150 906.87 59,112.20 2000 1,160 979.81 32,468.44 2001 1,200 1033.87 27,599.18 2002 1,150 1083.71 4,394.36 RMSE = 156.56 Since the RMSE from the 3-year moving average is lower than the RMSE from the exponential smoothing method, the 3-year moving average forecasting method does better. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 31 of 32 Problem 19 a) The expected quantity demanded for the coming year is 10,200,000 pounds of tea. b) The expected quantity demanded for the coming year is 11,055,000 pounds of tea. BUSS 230 | Fall 2013-2014 | Dr. George G. Namur Page 32 of 32 Problem 20 Quarter 1, 2006: t =21: D1=1, D2= 0, D3=0 Y2006Q1 =8.25 + .125(21) − 2.75(1) + 2.25(0) + 3.50(0) = $8.125 (million) Quarter 2, 2006: t = 22: D1=0, D2=1, D3=0 Y2006Q2 = 8.25 + .125(22) − 2.75(0) + 2.25(1) + 3.50(0) = $13.25 (million) Quarter 3, 2006: t = 23: D1=0, D2=0, D3=1 Y2006Q3 = 8.25 + .125(23) − 2.75(0) + 2.25(0) + 3.50(1) = $14.625 (million) Quarter 4, 2006: t = 24: D1=0, D2=0, D3=0 Y2006Q4 =8.25 + .125(24) −2.75(0) + 2.25(0) + 3.50(0) = $11.25 (million)