1. ## Mean Absolute Deviation

I understand MAD, but I cannot figure out this question. Here's the information.

Daily high temperatures in St. Louis for the last week were as follows: 92, 91, 94, 95, 96, 88,90.
The mean absolute deviation based on a 2-day moving average = 3.20

How in the world is it 3.20? I've tried 100's of combinations to try to repeat this answer, but I can't do it. I'm 100% sure this is the correct answer. Can anyone please show me a step by step walk through on this problem? I've spent all day on this...

2. I've got something but am not 100% sure. They may be talking about a 2-day moving average forecasts and measuring these forecasts using MAD. Then the situation is as follows.

$\displaystyle \begin{array}{lrrrrrrr} \mbox{Actual temperature} & 92 & 91 & 94 & 95 & 96 & 88 & 90\\ \mbox{2-day average forecast} & & & 91.5 & 92.5 & 94.5 & 95.5 & 92\\ \mbox{Absolute deviation} & & & 2.5 & 2.5 & 1.5 & 7.5 & 2 \end{array}$

The average of the last line is 3.2.

3. Originally Posted by smashedpumpkins
I understand MAD, but I cannot figure out this question. Here's the information.

Daily high temperatures in St. Louis for the last week were as follows: 92, 91, 94, 95, 96, 88,90.
The mean absolute deviation based on a 2-day moving average = 3.20

How in the world is it 3.20? I've tried 100's of combinations to try to repeat this answer, but I can't do it. I'm 100% sure this is the correct answer. Can anyone please show me a step by step walk through on this problem? I've spent all day on this...
What is the question exactly?! do The values from last week ( 7 days) have something to do with MAD of those two days? Or 3.20 have been obtained from 2 new temperatures?

4. Originally Posted by emakarov
I've got something but am not 100% sure. They may be talking about a 2-day moving average forecasts and measuring these forecasts using MAD. Then the situation is as follows.

$\displaystyle \begin{array}{lrrrrrrr} \mbox{Actual temperature} & 92 & 91 & 94 & 95 & 96 & 88 & 90\\ \mbox{2-day average forecast} & & & 91.5 & 92.5 & 94.5 & 95.5 & 92\\ \mbox{Absolute deviation} & & & 2.5 & 2.5 & 1.5 & 7.5 & 2 \end{array}$

The average of the last line is 3.2.
Yep this is for forecasting. Ohhh wow. I guess I didn't understand 2-day moving average. I was trying everything I could with the last two numbers. I was even trying to divide them by 5. Now I see why it's 5 observations... I guess the keyword is "moving". Thanks a million!

Randy

,

,

### daily high temperatures in st. louis for the last week

Click on a term to search for related topics.