$$a$)$ Using a$2-$year moving$$ average, the forecast for year $6=$

$$

$3750$ miles $($round your response to the nearest whole$$ number$).$

Part $3$

$$b$)$ If a$2-$year moving average is used to make the$$ forecast, the MAD based on this$=$

$$

$83\mathrm{.}3$ miles $($round your response to one decimal$$ place$).($Hint: You will have only $3$ years of matched$$ data.$)$

Part $4$

$$c$)$ The forecast for year $6$ using a weighted$2-$year moving average with weights of $0\mathrm{.}40$ and $0\mathrm{.}60($the weight of $0\mathrm{.}60$ is for the most recent$$ period$)=$

$$

$3740$ miles $($round your response to the nearest whole$$ number$).$

Part $5$

The MAD for the forecast developed using a weighted$2-$year moving average with weights of $0\mathrm{.}40$ and $0\mathrm{.}60=$

$$

$123\mathrm{.}3$ miles $($round your response to one decimal$$ place$).($Hint: You will have only $3$ years of matched$$ data.$)$

Part $6$

$$d$)$ Using exponential smoothing with alpha $=0\mathrm{.}30$ and the forecast for year $1$ being $3$ comma $050,$ the forecast for year $6=$

$$

$3640$ miles $($round your response to the nearest whole$$ number$).$The Carbondale Hospital is considering the purchase of a new ambulance. The decision will rest partly on the anticipated mileage to be driven next year. The miles driven during the past $5$ years are as follows:

a$)$ Using a $2-$year moving average, the forecast for year $6=3750$ miles $($round your response to the nearest whole number$).$

b$)$ If a $2-$year moving average is used to make the forecast, the MAD based on this $=|,$ miles $($round your response to one decimal place$).($Hint: You will have only $3$ years of matched data.$)$

c$)$ The forecast for year $6$ using a weighted $2-$year moving average with weights of $0\mathrm{.}40$ and $0\mathrm{.}60($the weight of $0\mathrm{.}60$ is for the most recent period$)=$ miles $($round your response to the nearest whole number$).$

The MAD for the forecast developed using a weighted $2-$year moving average with weights of $0\mathrm{.}40$ and $0\mathrm{.}60=$ miles $($round your response to one decimal place$).($Hint: You will have only $3$ years of matched data.$)$

d$)$ Using exponential smoothing with $=0\mathrm{.}30$ and the forecast for year $1$ being $3,050,$ the forecast for year $6=$ miles $($round your response to the nearest whole number$).$