Eastern Conference Team (NBA) Free Throw Percentage Win Percentage Indiana .779
ID: 3173487 • Letter: E
Question
Eastern Conference Team (NBA)
Free Throw Percentage
Win Percentage
Indiana
.779
.683
Miami
.760
.659
Toronto
.782
.585
Chicago
.779
.585
Washington
.731
.537
Brooklyn
.753
.537
Charlotte
.737
.524
Atlanta
.781
.463
New York
.761
.451
Cleveland
.751
.402
Detroit
.670
.354
Boston
.777
.305
Orlando
.763
.280
Philadelphia
.710
.232
Milwaukee
.747
.183
Use the above chart to answer the following questions.
1) Calculate the sample correlation coefficient, r.
2) At a=.01 can you conclude there is a significant linear correlation between free throw percentage and percent of games won?
3) Use the calculator to produce the line of best fit.
4) Using the line of best fit, estimate the percent of games an Eastern Conference NBA team would win if they had a free through percentage of 60%. Is this a reasonable estimate? Why or why not?
5) Given the analysis you did above, would you advise an NBA coach to drill your players on free throws to improve your win percentage? Why or why not?
Eastern Conference Team (NBA)
Free Throw Percentage
Win Percentage
Indiana
.779
.683
Miami
.760
.659
Toronto
.782
.585
Chicago
.779
.585
Washington
.731
.537
Brooklyn
.753
.537
Charlotte
.737
.524
Atlanta
.781
.463
New York
.761
.451
Cleveland
.751
.402
Detroit
.670
.354
Boston
.777
.305
Orlando
.763
.280
Philadelphia
.710
.232
Milwaukee
.747
.183
Explanation / Answer
Let,
X = Free Throw Percentage
Y = Win Percentage
We put the data in excel in two columns. Then we go to Data and then Data Analysis. We select the y and x values and then we click on OK.
We get the Regression Output where we get the required details to answer these questions.
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.3913
R Square
0.1531
Adjusted R Square
0.0880
Standard Error
0.1478
Observations
15
ANOVA
df
SS
MS
F
Significance F
Regression
1
0.051362451
0.051362
2.350771
0.149190596
Residual
13
0.284039549
0.021849
Total
14
0.335402
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Intercept
-1.0275
0.965745599
-1.06399
0.306697
-3.113912534
1.058820504
X Variable 1
1.9673
1.283119181
1.533222
0.149191
-0.804703459
4.739317453
( 1. ) The value of correlation coefficient is 0.3913
( 2. ) The p-value for t-test is 0.1491 which is greater than .01 (level of significance); the null hypothesis cannot be rejected and we can conclude that there is no significant linear correlation between free throws and percentage of game won.
( 3. ) y^ = -1.0275 + 1.9673x
( 4. ) x = 0.60
y^ = -1.0275 + (1.9673*0.6) = 0.1529
The predicted value is 0.1529
Here as the regression model is not significant so it cannot be a reasonable estimate.
( 5.)
No, as there is no correlation between these two given variables so this is not reasonable to advise an NBA coach to drill the players on free throws to improve win percentage.Let,
X = Free Throw Percentage
Y = Win Percentage
We put the data in excel in two columns. Then we go to Data and then Data Analysis. We select the y and x values and then we click on OK.
We get the Regression Output where we get the required details to answer these questions.
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.3913
R Square
0.1531
Adjusted R Square
0.0880
Standard Error
0.1478
Observations
15
ANOVA
df
SS
MS
F
Significance F
Regression
1
0.051362451
0.051362
2.350771
0.149190596
Residual
13
0.284039549
0.021849
Total
14
0.335402
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Intercept
-1.0275
0.965745599
-1.06399
0.306697
-3.113912534
1.058820504
X Variable 1
1.9673
1.283119181
1.533222
0.149191
-0.804703459
4.739317453
( 1. ) The value of correlation coefficient is 0.3913
( 2. ) The p-value for t-test is 0.1491 which is greater than .01 (level of significance); the null hypothesis cannot be rejected and we can conclude that there is no significant linear correlation between free throws and percentage of game won.
( 3. ) y^ = -1.0275 + 1.9673x
( 4. ) x = 0.60
y^ = -1.0275 + (1.9673*0.6) = 0.1529
The predicted value is 0.1529
Here as the regression model is not significant so it cannot be a reasonable estimate.
( 5.)
No, as there is no correlation between these two given variables so this is not reasonable to advise an NBA coach to drill the players on free throws to improve win percentage.
SUMMARY OUTPUT
Regression Statistics
Multiple R
0.3913
R Square
0.1531
Adjusted R Square
0.0880
Standard Error
0.1478
Observations
15
ANOVA
df
SS
MS
F
Significance F
Regression
1
0.051362451
0.051362
2.350771
0.149190596
Residual
13
0.284039549
0.021849
Total
14
0.335402
Coefficients
Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Intercept
-1.0275
0.965745599
-1.06399
0.306697
-3.113912534
1.058820504
X Variable 1
1.9673
1.283119181
1.533222
0.149191
-0.804703459
4.739317453
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