Problem Scenario Every few years the National Assessment of
Problem Scenario:
Every few years, the National Assessment of Educational Progress asks a national sample of eighth-graders to perform the same math tasks. The goal is to get an honest picture of progress in math. Here are national mean scores, on a scale of 0 to 500.
Obs
1
2
3
4
5
6
7
8
Year
1990
1992
1996
2000
2003
2005
2008
2011
Score
263
268
272
273
278
279
281
283
Homework Exercises:
1.Manually calculate the following, with Year as the independent variable (x) and Score as the dependent variable (y). (See pages 15-17 of the Module 2 lecture notes for a manual example.) Your calculations must be hand-written, not typed. (10 pts)
Statistic
Results from Manual Calculation
sx
sy
r
R2
b0
b1
Regression model
{form from the b0 and b1 computed above}
Rounding to the 2nd decimal place in your calculations is sufficient.
Provide the value for each statistic in a clearly marked table, such as that above.
Also insert the images (e.g., scanned photos) or attach a document showing your manual (hand-written) calculations for each of these statistics.
2. Input the data values for x and y into SPSS, and verify your above calculations with SPSS output. Provide SPSS output for: (3 pts)
Descriptive statistics for x and y
Correlation
Regression “coefficients”
Note that your manual calculations may not exactly equal the values provided in the SPSS output due to rounding errors.
Paste these 3 tables into your homework document.
Also attach the entire SPSS output to your homework submission.
3.Type the regression equation in the space provided below, based on values from your SPSS output: _________________________________ (1 pt)
4.In your own words, interpret the results of the correlation and the R2. (3 pts)
______________________________________________________________________________________
______________________________________________________________________________________
5.Based on the regression equation from your SPSS output, what is the predicted value of y if x=2012?
_________________________________ (2 pts)
6.Would it be appropriate to use this model to predict the value of y if x = 2025? Why or Why not? (1 pt)
______________________________________________________________________________________
| Obs | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 
| Year | 1990 | 1992 | 1996 | 2000 | 2003 | 2005 | 2008 | 2011 | 
| Score | 263 | 268 | 272 | 273 | 278 | 279 | 281 | 283 | 
Solution
Statistic
Results from Manual Calculation
2000.6
274.6
sx
7.0345
sy
6.4214
r
0.982
R2
0.9643
b0
-1518.80
b1
0.896
Regression model
y = -1518.80+0.896x
SPSS output
y =-1518.799+0.8958x
Correlation is the associatin between x and y. If r is positive then x and y are positively associated.
r lies between -1 and +1. If |r| lies near 1, strong correlation and if |r| lies near 0 weak correlation.
If r=0 x and y are not correlated at all.
R2, the coefficient of determination being positive, gives the relationship between x and y.
It is the square of correlation coefficient.
----------------------------------
If x=2012, y = 283.5506
(by subsituting 2012 for x)
---------------------------------------------
It may not be appropriate to use the model for 2025 as the trend may change when 2025 is very far from these years.
| Statistic | Results from Manual Calculation | 
|---|---|
| x bar | 2000.6 | 
| y bar | 274.6 | 
| sx | 7.0345 | 
| sy | 6.4214 | 
| r | 0.982 | 
| R2 | 0.9643 | 
| b0 | -1518.80 | 
| b1 | 0.896 | 
| Regression model | y = -1518.80+0.896x | 




