Determine the forecast for period 11 using the exponential smoothing technique with alpha=0.4.
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Given the following units sold:
| time | Units sold |
| 1 | 147 |
| 2 | 148 |
| 3 | 151 |
| 4 | 145 |
| 5 | 155 |
| 6 | 152 |
| 7 | 155 |
| 8 | 157 |
| 9 | 160 |
| 10 | 165 |
Determine the forecast for period 11 using the exponential smoothing technique with alpha=0.4.
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- Table 6 shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 2012. a. Let x represent time in years starting with x=0 for the year 1997. Let y represent the number of seals in thousands. Use logistic regression to fit a model to these data. b. Use the model to predict the seal population for the year 2020. c. To the nearest whole number, what is the limiting value of this model?The following fictitious table shows kryptonite price, in dollar per gram, t years after 2006. t= Years since 2006 0 1 2 3 4 5 6 7 8 9 10 K= Price 56 51 50 55 58 52 45 43 44 48 51 Make a quartic model of these data. Round the regression parameters to two decimal places.Cable TV The following table shows the number C. in millions, of basic subscribers to cable TV in the indicated year These data are from the Statistical Abstract of the United States. Year 1975 1980 1985 1990 1995 2000 C 9.8 17.5 35.4 50.5 60.6 60.6 a. Use regression to find a logistic model for these data. b. By what annual percentage would you expect the number of cable subscribers to grow in the absence of limiting factors? c. The estimated number of subscribers in 2005 was 65.3million. What light does this shed on the model you found in part a?
- Olympic Pole Vault The graph in Figure 7 indicates that in recent years the winning Olympic men’s pole vault height has fallen below the value predicted by the regression line in Example 2. This might have occurred because when the pole vault was a new event there was much room for improvement in vaulters’ performances, whereas now even the best training can produce only incremental advances. Let’s see whether concentrating on more recent results gives a better predictor of future records. (a) Use the data in Table 2 (page 176) to complete the table of winning pole vault heights shown in the margin. (Note that we are using x=0 to correspond to the year 1972, where this restricted data set begins.) (b) Find the regression line for the data in part ‚(a). (c) Plot the data and the regression line on the same axes. Does the regression line seem to provide a good model for the data? (d) What does the regression line predict as the winning pole vault height for the 2012 Olympics? Compare this predicted value to the actual 2012 winning height of 5.97 m, as described on page 177. Has this new regression line provided a better prediction than the line in Example 2?Consider the following time series data Month 1 2 3 4 5 6 7 8 9 10 11 12 Value 90 89 86 91 90 91 88 86 91 93 90 88 a) Construct a time series plot in Excel. Label the axes and graph. c) Develop a smoothing model forecast with α = 0.35. Compute MSE and forecast for month 13 d) Which model is the better predictor? Why?For exponential smoothing forecasting with an alpha of 0.1, the MSE (mean squared error) is 2.98. Using an alpha of 0.05, the MSE is 1.32. Using the MSE, which value of alpha provides the best estimate? 1.) alpha = 0.1 2.) alpha = 0.05
- Given is a historical time series for job services demand in the prior 6 months. Month Demand 1 812 2 799 3 777 4 811 5 765 6 791 The Period 4 forecast by using exponential smoothing with 0.33 alpha = (in 2 decimal places)Consider the following time series data. Month 1 2 3 4 5 6 7 Value 24 13 20 12 19 23 15 Using the Exponential Smoothing method to forecast the next period, compute the MSE (Mean square Error). Consider: the smoothing constant (α) = 0.5 Which of the following options is the answer? Select one:a. 41.24b. 39.95c. 40.05d. 41.02You estimated a regression with the following output. Source | SS df MS Number of obs = 325 -------------+---------------------------------- F(1, 323) = 42850.36 Model | 285905003 1 285905003 Prob > F = 0.0000 Residual | 2155111.65 323 6672.17228 R-squared = 0.9925 -------------+---------------------------------- Adj R-squared = 0.9925 Total | 288060115 324 889074.429 Root MSE = 81.683 ------------------------------------------------------------------------------ Y | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- X | 11.83842 .0571895 207.00 0.000 11.72591 11.95093 _cons | 52.14457 5.949458 8.76 0.000 40.43999 63.84915…
- Consider the following sample regressions for the linear and the logarithmic models. Linear Logarithmic Intercept 6.7904 −5.6712 x 1.0607 NA In(x) NA 10.5447 * se 2.4935 1.5231 R2 0.8233 0.9341 Adjusted R2 0.8013 0.9259 b. Use the selected model to predict y for x = 10. (Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) predicted y=The following plots have been obtained for a time series. a) Suggest an appropriate ARIMA model. b) The following ARIMA output has been obtained from R. Based on this output, which model would you recommend for forecasting?You estimated a regression with the following output. Source | SS df MS Number of obs = 115 -------------+---------------------------------- F(1, 113) = 5454.39 Model | 186947380 1 186947380 Prob > F = 0.0000 Residual | 3873036.62 113 34274.6603 R-squared = 0.9797 -------------+---------------------------------- Adj R-squared = 0.9795 Total | 190820417 114 1673863.3 Root MSE = 185.13 ------------------------------------------------------------------------------ Y | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+---------------------------------------------------------------- X | 28.58986 .3871141 73.85 0.000 27.82292 29.3568 _cons | 10.54686 26.92706 0.39 0.696 -42.80051 63.89423…
