Business
Answer:
a.
Compute the 95% confidence interval for the autocorrelation coefficient at any lag.
Consider the confidence level is 95%, and
.
The formula for confidence interval is given below:
Here, the confidence level is 0.95.
For two tailed test,
From Table B-2, the required
value for 95% confidence level is 1.96. Thus,
.
Substitute 1.96 for
, 100 for n in the above confidence interval formula.
Thus, the 95% confidence interval for the autocorrelation coefficient at any lag is
.
b.
Conclusion:
Here, the series is random, because there is no particular pattern occurs in the plot and also all the autocorrelation coefficients lie within 95% confidence intervals.
c.
Conclusion:
The series may possibly stationary autoregressive process or non-stationary autoregressive process, because the inference depends on how rapid the autocorrelations decline to zero for the given time period.
d.
Conclusion:
Here, the series is seasonal with the period is 4, because the autocorrelations
,
, and
are significantly different from zero and also the process is quarterly.
Answer:
Compute the forecasts for periods 5, 6, 7.
From the given information, the first four observations are
,
,
and
. Assume that
and
.
The model is given below:
The forecast value for periods 1, 2, 3, and 4 are calculated as follows:
The forecast value for period 5 is,
Thus, the forecast value for period 5 is
.
The forecast value for period 6 is,
Thus, the forecast value for period 6 is
.
The forecast value for period 7 is,
Thus, the forecast value for period 7 is
.
Answer:
a.
Compute forecasts for periods 61, 62, and 63 from origin 60.
Assume that
and
.
The time series model is given below:
The forecast value for period 61 is,
Thus, the forecast value for period 61 is
.
The forecast value for period 62 is,
Thus, the forecast value for period 62 is
.
The forecast value for period 63 is,
Thus, the forecast value for period 63 is
.
b.
Compute forecasts for periods 62 and 63.
Assume that
and
.
The time series model is given below:
The forecast value for period 62 is,
Thus, the forecast value for period 62 is
.
The forecast value for period 63 is,
Thus, the forecast value for period 63 is
.
c.
Compute the 95% prediction interval about the forecast for period 61.
From part a.,
.
The formula for prediction interval is given below:
Substitute 75.65 for
and 3.2 for
in the above prediction interval formula.
Thus, the 95% prediction interval about the forecast for period 61 is
.