e How a regression line summarizes a scatterplot . How the regression equation i
ID: 3060386 • Letter: E
Question
e How a regression line summarizes a scatterplot . How the regression equation is used to predict the Y scores at a given How the standard error of the estimate measures the errors in prediction Jesse's friend, Marty, thinks that another good predictor of time taken to reach the finish line for a 1- mile race on dirt would be the difference between the jockey (rider) and horse's weight. Therefore Marty tracks 5 horses over 5 1-mile races on dirt, recording the average difference between the weigh of the horse and Jockey (horse weight-Jockey weight) and the time (in seconds) it took the horse to reach the finish line. His data is summarized below Horse Average weight difference in lbs. (x) Average finish time in seconds (Y) 891.69 889.92 891.32 892.04 884.2 104.33 102.32 97.04 95.63 104 48 What is the strength and direction of the relationship between the average weight difference and time taken to reach the finish line? 6. 7. Using the appropriate regression equation, what is the predicted finish time for a horse that weighs 893 Ibs. more than its jockey? 8. What is the standard error of the estimate of Marty's data? 9 How much of the variance in the average time taken to reach the finish line for a 1-mile race on dirt is accounted for by variance in weight difference?Explanation / Answer
Result:
6). The correlation between average weight difference and time taken to reach the finish line is
-0.574.
The relation is negative and strength s moderate.
7).
The regression line is
Average finish time = 754.2741-0.7344*average weight difference.
When average weight difference = 893,
Predicted Average finish time = 754.2741-0.7344*893
=98.4549
8). standard error = 3.932
Regression Analysis
r²
0.330
n
5
r
-0.574
k
1
Std. Error
3.932
Dep. Var.
y
ANOVA table
Source
SS
df
MS
F
p-value
Regression
22.7988
1
22.7988
1.47
.3114
Residual
46.3734
3
15.4578
Total
69.1722
4
Regression output
confidence interval
variables
coefficients
std. error
t (df=3)
p-value
95% lower
95% upper
Intercept
754.2741
538.1159
1.402
.2556
-958.2509
2,466.7991
x
-0.7344
0.6047
-1.214
.3114
-2.6590
1.1901
Regression Analysis
r²
0.330
n
5
r
-0.574
k
1
Std. Error
3.932
Dep. Var.
y
ANOVA table
Source
SS
df
MS
F
p-value
Regression
22.7988
1
22.7988
1.47
.3114
Residual
46.3734
3
15.4578
Total
69.1722
4
Regression output
confidence interval
variables
coefficients
std. error
t (df=3)
p-value
95% lower
95% upper
Intercept
754.2741
538.1159
1.402
.2556
-958.2509
2,466.7991
x
-0.7344
0.6047
-1.214
.3114
-2.6590
1.1901
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