Question

A k-period moving average uses __________.

the mean of the time series values for the most recent k periods to forecast the time series value for the next period

the mean of the time series values for the most recent k - 1 periods to forecast the time series value for the next period

the mean of the time series values for the most recent k + 1 periods to forecast the time series value for the next period

None of these

Can a 3-period moving average forecast be made for period 3?

Yes

No

It depends on one dataset to another.

Cannot be determined

An exponential smoothing forecast is ___________.

α times the most recent period's time series value, plus (1 - α) times the most recent period's forecast value, where α is a smoothing constant
between 0 and 1

α times the most recent period's time series value, plus (1 - α) times the most recent period's forecast value, where α is a smoothing constant
between 0 and 10

α times the most recent period's time series value, plus (1 - α) times the most recent period's forecast value, where α is any positive smoothing constant

α times the most recent period's time series value, plus (1 - α) times the most recent period's forecast value, where α is any negative smoothing constant

In the binomial probability distribution, the binomial coefficient gives the number of ways to arrange items when the order doesn’t matter. That is, suppose we have four letters {A, B, C, D} and we choose two letters, therefore, there are 6 ways to have a two-letter word. Now, suppose we have five letters {A, B, C, D, E}. How many ways to choose three-letter words?

131

12 ways.

14 ways.dd

10 ways.

Answer 1

A k-period moving average uses:

None of these

Can a 3-period moving average forecast be made for period 3? No

An exponential smoothing forecast is α times the most recent period's time series value, plus (1 - α) times the most recent period's forecast value, where α is a smoothing constant
between 0 and 1

In the binomial probability distribution, the binomial coefficient gives the number of ways to arrange items when the order doesn’t matter. That is, suppose we have four letters {A, B, C, D} and we choose two letters, therefore, there are 6 ways to have a two-letter word. Now, suppose we have five letters {A, B, C, D, E}. How many ways to choose three-letter words?