"1) The statistical Summary of the data is shown below: Year January FebruaryMarch April May June July August September October November December Total Sales1 242 235 232 178 184 140 145 152 110 130 152 206 21062 263 238 247 193 193 149 157 161 122 130 167 230 22503 282 255 265 205 210 160 166 174 126 148 173 235 2399Total:787 728 744 576 587 449 468 487 358 408 492 671 6755Mean:262.3333333 242.66667 248 192 195.6667 149.6667 156 162.3333 119.33333 136 164 223.6667 2251.666667Variance: 400.3333333 116.33333 273 183 174.3333 100.3333 111 122.3333 69.333333 108 117 240.3333 21464.33333StDev:20.0083316 10.785793 16.52271 13.52775 13.20353 10.01665 10.53565 11.06044 8.326664 10.3923 10.81665 15.50269 146.5071102The graph of the time series is as shown below. 2)Seasonal Indices are Calculated in the Matrix below: Month Jan Feb March April May June July Aug Sept Oct Nov Decyear1 242 235 232 178 184 140 145 152 110 130 152 2062 263 238 247 193 193 149 157 161 122 130 167 2303 282 255 265 205 210 160 166 174 126 148 173 235No of 3 3 3 3 3 3 3 3 3 3 3 3EntriesTotal 787 728 744 576 587 449 468 487 358 408 492 671Mean 262.33 242.66 248 192 195.66 149.66 156 162.33 119.33 136 164 223.66Grand 187.639 MeanIndex 1.398 1.293 1.322 1.023 1.043 0.798 0.831 0.865 0.636 0.725 0.874 1.192The following is a hypothesis test of seasonality:Null Hypothesis: There is no significant seasonal component in the time series, versus Alternative Hypothesis: There is a significant seasonal component in the time seriesMonth Jan Feb March April May June July Aug Sept Oct Nov DecyearSeasonal 1.398 1.293 1.322 1.023 1.043 0.798 0.831 0.865 0.636 0.725 0.874 1.192IndexChi 69.72245 squareP Value 0 Conclusion: There is very strong evidence against the null hypothesis, hence the nullhypothesis is rejected. There is sufficient evidence to conclude that there existssignificant seasonal component in the time series.We now perform linear regression model with the following data: Year Total Sales1 21062 22503 2399y = 146.5x + 1958.7 Linear Regression2 R = 0.9999 2450 2400 2350 2300 2250 2200 2150 2100 2050 0 0.5 1 1.5 2 2.5 3 3.5 Year ---> 2 This regression is extremely good with an value of 0.999. The regression equationR is given as: y = 146.5x + 1958.7For the fourth year, the total sales are obtained by plugging in x = 4 in the aboveequation.Total Sales in 1000's --->?y(x ? 4) ? 146.5*4 ?1958.7 ? 2544.7(in thousands of dollars)The average monthly sales during the fourth year, therefore, is 2544.7/12 = 212.058(in thousands of dollars).The forecast for a particular month (say July) is calculated by multiplying the averagemonthly sales forecast by that month’s (July’s) seasonal index. For the month of July,it will be 0.831*212.058 = 176.22 (in thousands of dollars).The monthly forecasts for the 12 months of the fourth year are as shown below:Month (yr.4) January FebruaryMarch April May June July August September October November DecemberS. Index 1.398 1.293 1.322 1.023 1.043 0.798 0.831 0.865 0.636 0.725 0.874 1.192Forecast 296.45755 274.19143 280.3411 216.9357 221.1768 169.2226 176.2205 183.4305 134.8691 153.7423 185.339 252.7735Suppose the actual January sales for the fourth year turn out to be $295,000. Theforecasted January sales are $296,458.Error between actual and forecasted sales = $296,458 - $295,000 = $1458Error 1458 Percentage Error = *100 ? *100 ? 0.49% ActualSales 295000 This is an extremely small percentage error. Karen does not have to worry about thiserror and she can be assured that her forecast model is extremely good."