A time series model has been fit and checked with historical data yielding Yt=50

Question

A time series model has been fit and checked with historical data yielding Yt=50+0.45 Y t-1+et Suppose at time t=60 the observation is Y60=57. A. determine forecasts for periods 61,62, and 63 from origin 60. B. suppose the observed value of Y61 is 59. Update teh forecast for periods 62 and 63. C. Suppose the estimate for the variance of the error term is s^2=3.2. Compute a 95% prediction interval about the forecast for period 61. Please answer ALL parts to this question and show work. PLEASE HELP. Thank you

Explanation / Answer

Given model : Yt = 50 + (0.45Yt-1) + et

Part A.) At t=60, Y60=57
We know that expected value of the error component is always zero.
So, E(Y61) = 50+(0.45 x 57) + 0 = 75.65
Similarly, E(Y62) = 50+(0.45 x 75.65) + 0 = 84.04
Also, E(Y63) = 50 + (0.45 x 84.04) + 0 = 87.82

Part B.) Given that Y61 = 59
So, E(Y62) = 50+ (0.45 x 59) +0 = 76.55
And E(Y63) = 50 + (0.45 x 76.55) + 0 = 84.45

Part C.) Given that Var(et) = 3.2 and Y60 = 57
Assuming that error component is a normal variate with parameters 0 and s^2,
Var(Y61) = Var (50 + (0.45 x 57) + e61) = 0 + Var(e61)
Assuming that the observations come from normal population, we have

z= (Y61 - E(Y61)) / (Var(Y61)) = N (0,1)

Since we know that value of a standard normal variate at 5% level of significance is 1.96,

95% Confidence / Prediction interval is given by :

E(Y61) ± 1.96 x (3.2) = 75.65 ± (1.96 x 3.2) = 75.65 ± 3.5062 = (72.1438 , 79.1562)

So, (72.1438 , 79.1562) is the required 95% confidence interval.

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