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November 01, 2025, 10:47:57 am

Author Topic: maths quest excercise 4C  (Read 805 times)  Share 

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sam0001

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maths quest excercise 4C
« on: January 26, 2012, 06:44:30 pm »
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question 8, shouldnt the answer be 364.2? not 419?

linle

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Re: maths quest excercise 4C
« Reply #1 on: January 27, 2012, 12:31:06 am »
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It's actually 419. I did it with Excel.

Here's the table:

2   141   
3   270   253.8
4   234   303
5   357   306.6
6   267   320.4
7   387   322.4
8   288   362
9   303   364.2
10   367   
11   465   
12   398   

The linear regression equation for the smooth data is
y=16.607x+219.27

Sub in 12 and you get 418.554 ~ 419.

I've attached the graph if you want to look at it.
Bachelor of Engineering (Chemical & Biomolecular) @ The University of Melbourne

Planck's constant

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Re: maths quest excercise 4C
« Reply #2 on: January 27, 2012, 01:04:10 am »
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The OP is correct.
The answer 364.2, because the question asks to use the last smoothed value and NOT the linear regression equation

Edit : In fact you are both correct, except the OP is using the method required by the question :)

« Last Edit: January 27, 2012, 01:05:45 am by argonaut »

linle

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Re: maths quest excercise 4C
« Reply #3 on: January 27, 2012, 01:09:33 am »
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But it did say predict so we would need to use the last value and sub into the equation
Bachelor of Engineering (Chemical & Biomolecular) @ The University of Melbourne

Planck's constant

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Re: maths quest excercise 4C
« Reply #4 on: January 27, 2012, 01:19:48 am »
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But it did say predict so we would need to use the last value and sub into the equation


I know, I know ... but the 'prediction' IS the last smoothed value.

Look up the the section 'prediction using moving averages' within 4C
There is Method 1 and Method 2.
OP is correctly using Method 1 (as required by the problem) and you are using Method 2


(EDIT : Here is the relevant section, to save you the trouble of looking it up

Prediction using moving averages

Because the moving average does not generate a single linear equation, there are limited
possibilities for using the resultant smoothed data for prediction. However, there are two things
that can be done.
1. Predict the next value — use the last smoothed value to predict the next time point. In
the previous example, our prediction for t = 18 would be temperature = 37.50. This is not
necessarily an accurate prediction but it is the best we can do without a linear trend equation.
2. Fit a single straight line to the smoothed data — using either the 3-median or least-squares
techniques, one could find a single equation for the smoothed data points. This is often the
preferred technique.
« Last Edit: January 27, 2012, 01:25:13 am by argonaut »