1. Section 9-3
  We already know how to calculate the correlation coefficient, r. The square of this coefficient is called the coefficient of determination.
The coefficient of determination is equal to the ratio of the unexplained variation to the total variation.
In other words, if 𝑟 2 = .81, then 81% of the variation between x and y can be explained by the relationship between x and y.
The other 19% of the variation is unexplained and is due to other factors or to sampling error.

2. Section 9-3
How to find the Standard Error of Estimate:
1) Go to STAT – Edit – Put x values into L1 and y values into L2
2) STAT – TEST – F
The Standard Error of Estimate is the standard deviation of the residuals (s).
Scroll down the list of values given as the results of the test, and find s.

3. Section 9-3
Construct a Prediction Interval for a Specific x-value (x0).
1) Determine degrees of freedom (n – 2)
2) Use regression equation and given x (x0) to find 𝑦 .
3) Find the critical t value (tc) that corresponds to the level of confidence (c ) by using the calculator (InvT( 1+𝑐 2 ), with degrees of freedom being found at n – 2).
Stat-Test-F will also give you the degrees of freedom, if you remember to look for it.
4) Use the tc value and the Se value, to calculate the margin of error (E).

4. Section 9-3
Construct a Prediction Interval for a Specific x-value (x0).
𝐸= 𝑡 𝑐 𝑆 𝑒 𝑛 + 𝑛 𝑥 0 − 𝑥 2 𝑛 𝑥 2 − 𝑥 2
𝑛= sample size
𝑥 0 is the x value that you used to find 𝑦
𝑥 is the sample mean.
𝑥 2 is total of all the squared x’s. Square first, then add them up.
𝑥 2 is the total of x’s squared. Add first, then square the answer.
The values for 𝑥 , 𝑥 2 , and 𝑥 can all be found by going to STAT– Calc–1 (1-Var Stats)
5) Find the left and right endpoints by subtracting E from 𝑦 and then adding E to 𝑦 .
These answers are your interval.

5. Example 1 (Page 526)
The correlation coefficient for the advertising expenses and company sales data as calculated in Example 4 of Section 9-1 is 𝑟≈ Find the coefficient of determination. What does this tell you about the explained variation of the data about the regression line? About the unexplained variation?
𝑟 2 = ≈0.834
About 83.4% of the variation in the company sales can be explained by the variation in the advertising expenditures.
About 16.6% (the rest) of the variation is unexplained and is due to chance or other variables.

6. Put x values into L1 and y values into L2 2) STAT–TEST–F
Example 2 (Page 528)
The regression equation for the advertising expenses and company sales data as calculated in Example 1 of Section 9-2 is 𝑦 =50.729𝑥 Find the standard error of estimate.
1) Go to STAT–Edit–
Put x values into L1 and y values into L2
2) STAT–TEST–F
Find s on the list of values given from this test.
3) The standard error of the estimate is 10.29 x
2.4
1.6
2.0
2.6
1.4
2.2 y
225
184
220
240
180
186
215

7. Example 3 (Page 530)
Using the results of Example 2, construct a 95% prediction interval for the company sales when the advertising expenses are $2100. What can you conclude?
We were told that 𝑦 =50.729𝑥 , so we plug in 2.1 for x to find 𝑦 .
𝑦 =
From here, we need to be able to use the formula for the margin of error. It’s not pretty, but it works.
𝐸= 𝑡 𝑐 𝑠 𝑒 𝑛 + 𝑛( 𝑥 0 − 𝑥 ) 2 𝑛( 𝑥 2 )−( 𝑥 ) 2
𝑡 𝑐 =2.447 (2nd VARS 4, ( )/2, with 6 degrees of freedom).
𝑠 𝑒 =10.29, from last example.
𝑛=8
𝑥 0 =2.1 (this is the x value we used to find 𝑦 .
𝑥 =1.975, ( 𝑥 2 )=32.44, ( 𝑥) =15.8
These values are from STAT- Calc–1 (1-Var Stats).

8. Example 3 (Page 530)
Using the results of Example 2, construct a 95% prediction interval for the company sales when the advertising expenses are $2100. What can you conclude?
𝐸= ( 2.1−1.975) −
≈26.857
We subtract from to get the low end of the estimate.
We add to to get the upper end of the estimate.
< y <
We can be 95% confident that when advertising expenses are $2100, the company sales will be between $183,733 and $237,447.

9. Use the data given below to construct a 99% confidence interval for y when x = 2950, if possible. If not possible, tell why. Use 0.01 for α x y
1467
606
1728
659
2150
716
2651
774
2629
797
2619
860
2533
894
3080
959
3475
1024
Enter x-values into L1 and y-values into L2
STAT–Test–F
Use L1 for Xlist and L2 for Ylist
Leave Freq: at 1
Select ≠ for a two-tailed test
For RegEQ, press Alpha Trace and then Enter
This will automatically store the linear regression equation into the Y= screen for you.
Calculate.
The standardized test statistic (t) = 8.262
The p-value is 7.416E-5 (.00007)
The Standard Error (s) is
The correlation coefficient (r) is .952
This indicates a strong positive correlation.
The coefficient of determination (r2) is .907
This means that 90.7% of the variation between x and y is due to their relationship.

10. Use the data given below to construct a 99% confidence interval for y when x = 2950, if possible. If not possible, tell why. Use 0.01 for α x y
1467
606
1728
659
2150
716
2651
774
2629
797
2619
860
2533
894
3080
959
3475
1024
t = p = 7.416E-5 (.00007) s =
r = r2 = n = 9
Because p < α, we reject the null and conclude that we have a significant relationship.
We may proceed and use the equation to make predictions.
y = .210x OR x
Setting our Graph to start at 2950 (2nd Window) gives us a y-value of (this works because the equation is already in y=)
is y-hat.
STAT-Calc-1 on L1 (the x-values) gives us:
x-bar = Σx = Σx2 =
2nd VARS 4 on (1+.99)/2 with 7 (n – 2) degrees of freedom gives us a tc value of

11. Use the data given below to construct a 99% confidence interval for y when x = 2950, if possible. If not possible, tell why. Use 0.01 for α x y
1467
606
1728
659
2150
716
2651
774
2629
797
2619
860
2533
894
3080
959
3475
1024
t = p = 7.416E-5 s = r = r2 = n = 9 tc = 𝑦 = 𝑥 = 𝑥= 𝑥 2 =
Let’s plug the numbers into the formula to find our interval.
𝐸= 𝑡 𝑐 𝑠 𝑒 𝑛 + 𝑛( 𝑥 0 − 𝑥 ) 2 𝑛( 𝑥 2 )−( 𝑥 ) becomes
𝐸=3.499(44.981) ( 2950− ) 2 9( )− 𝐸=
Now, add and subtract E from 𝑦 to get our interval.
– < y <
< y <
We can be 99% confident that when x is 2950, the actual value of y will be between and

12. Assignments:
Classwork: Page 531 #4-8 All
Homework: Pages , # 9-24 All