week 7 homework questions V2
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School
American Military University *
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Course
302
Subject
Mathematics
Date
Feb 20, 2024
Type
docx
Pages
24
Uploaded by bradymcdede
Attempt Score
17 / 20 - 85 %
Overall Grade (Highest Attempt)
17 / 20 - 85 %
stion 1
1 / 1 p
Which of the following equations are linear?
4y=8
y
2
=6x
3
+8
3y=6x+5y
2
y-x=8x
2
Hide question 1 feedback
A linear equation is a linear line. If a problem has a squared or a cubed term, it isn't linear. It is a quadratic equation.
n 2
1 Which of the following equations are linear?
y+7=3x
y=6x
2
+8
3y=6x+5y
2
y-x=8x
2
Hide question 2 feedback
A linear equation is a linear line. If a problem has a squared or a cubed term, it isn't linear. It is a quadratic equation.
n 3
1 You decided to join a fantasy Baseball league and you think the best way to pick your players is to look at their Batting Averages.
You want to use data from the previous season to help predict Batting Averages to know which players to pick for the upcoming season. You want to use Runs Score, Doubles, Triples, Home Runs and Strike Outs to determine if there is a significant linear relationship for Batting Averages.
You collect data to, to help estimate Batting Average, to see which players you should choose. You collect data on 45 players to help make your decision.
x1 = Runs Score/Times at Bat
x2 = Doubles/Times at Bat
x3 = Triples/Times at Bat
x4 = Home Runs/Times at Bat
x5= Strike Outs/Times at Bat
Is there a significant linear relationship between these 5 variables and the Batting Average?
If so, what is/are the significant predictor(s) for determining the Batting Average?
See Attached Excel for Data.
Baseball data.xlsx
Yes,
Runs Score/Times at Bat
, p-value = 0.000219186 < .05, Yes, Runs Score is a significant predictor for Batting Average.
Doubles/Times at Bat
, p-value = 0.00300543 < .05, Yes, Doubles are a significant predictor for Batting Average.
Strike Outs/Times at Bat
, p-value = 0.00000258892 < .05, Yes, Strikes Outs are a significant predictor for Batting Average.
No,
Triples/Times at Bat
, p-value = 0.291004037 > .05, No, Triples, are not a significant predictor for Batting Average.
Home Runs/Times at Bat,
p-value = 0.114060301 > .05, No, Home Runs are not a significant predictor for Batting Average.
No,
Runs Score/Times at Bat
, p-value = 0.000219186 < .05, Yes, Runs Score is a significant
predictor for Batting Average.
Doubles/Times at Bat
, p-value = 0.00300543 < .05, Yes, Doubles are a significant predictor for Batting Average.
Strike Outs/Times at Bat
, p-value = 0.00000258892 < .05, Yes, Strikes Outs are a significant predictor for Batting Average.
Yes,
Triples/Times at Bat
, p-value = 0.291004037 > .05, No, Triples, are not a significant predictor for Batting Average.
Home Runs/Times at Bat,
p-value = 0.114060301 > .05, No, Home Runs are not a significant predictor for Batting Average.
Hide question 3 feedback
You can run a Multiple Linear Regression in Excel using the Data Analysis ToolPak.
Data -> Data Analysis -> Scroll to Regression
Highlight Baseball Average for the Y Input:
Highlight all 5 columns from RS/Times at Bat to SO/Times at Bat for the X Input:
Make sure you click on Labels and Click OK.
If done correctly then,
The overall
Significance F
or p-value = 0.0000000000000012607
Because the p-value < .05, Reject Ho. Yes, there is a significant relationship, but which variables
are significant?
In the ANOVA under the p-value column we see,
Runs Score/Times at Bat
, p-value = 0.000219186 < .05, Yes, Runs Score is a significant predictor for Batting Average.
Doubles/Times at Bat
, p-value = 0.00300543 < .05, Yes, Doubles are a significant predictor for Batting Average.
Triples/Times at Bat
, p-value = 0.291004037 > .05, No, Triples, are not a significant predictor for Batting Average.
Home Runs/Times at Bat,
p-value = 0.114060301 > .05, No, Home Runs are not a significant predictor for Batting Average.
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Strike Outs/Times at Bat
, p-value = 0.00000258892 < .05, Yes, Strikes Outs are a significant predictor for Batting Average.
n 4
1 With Obesity on the rise, a Doctor wants to see if there is a linear relationship between the Age and Weight and estimating a person's Systolic Blood Pressure. Is there a significant linear relationship between Age and Weight and a person's Systolic Blood Pressure?
If so, what is/are the significant predictor(s) for Systolic Blood Pressure?
See Attached Excel for Data.
BP data
No,
Age, p-value = 0.001303023 < .05, No, Age is not a significant predictor for Systolic BP
Weight, p-value = 0.023799395 < .05, No, Weight is not a significant predictor for Systolic
BP
Yes,
Age, p-value = 0.001303023 < .05, Yes, Age is a significant predictor for Systolic BP
Weight, p-value = 0.023799395 < .05, Yes Weight is a significant predictor for Systolic BP
Yes,
Age, p-value = 0.9388 > .05, Yes, Age is a significant predictor for Systolic BP
Weight, p-value = 0 .3092 > .05, Yes Weight is a significant predictor for Systolic BP
No,
Age, p-value = 0.9388 > .05, No, Age is not a significant predictor for Systolic BP
Weight, p-value = 0 .3092 > .05, No, Weight is not a significant predictor for Systolic BP
Hide question 4 feedback
You can run a Multiple Linear Regression in Excel using the Data Analysis ToolPak.
Data -> Data Analysis -> Scroll to Regression
Highlight Systolic BP for the Y Input:
Highlight Both Age and Weight columns for the X Input:
Make sure you click on Labels and Click OK
If done correctly then,
The overall
Significance F
or p-value = .00000000013112
Because the p-value < .05, Reject Ho. Yes, there is a significant relationship, but which variables
are significant?
In the ANOVA under the p-value column we see,
Age, p-value = 0.001303023 < .05, Yes, Age is a significant predictor for Systolic BP
Weight, p-value = 0.023799395 < .05, Yes Weight is a significant predictor for Systolic BP
n 5
1 You move out into the country and you notice every Spring there are more and more Deer Fawns that appear. You decide to try and predict how many Fawns there will be for the up coming Spring.
You collect data to, to help estimate Fawn Count for the upcoming Spring season. You collect data on over the past 10 years.
x1 = Adult Deer Count
x2 = Annual Rain in Inches
x3 = Winter Severity
Where Winter Severity Index:
o
1 = Warm
o
2 = Mild
o
3 = Cold
o
4 = Freeze
o
5 = Severe
Approximately what percentage of the variation for Fawn Count is accounted for by these 3 variables in this model?
See Attached Excel for Data.
Deer data.xlsx
11.05% of variation in Fawn Count is accounted for by Adult Count, Annual Rain in inches
and Winter Severity in this model.
98.86% of variation in Fawn Count is accounted for by Adult Count, Annual Rain in inches
and Winter Severity in this model.
98.25% of variation in Fawn Count is accounted for by Adult Count, Annual Rain in inches
and Winter Severity in this model.
97.74% of variation in Fawn Count is accounted for by Adult Count, Annual Rain in inches
and Winter Severity in this model.
Hide question 5 feedback
You can run a Multiple Linear Regression Analysis using the Data Analysis ToolPak in Excel.
Data -> Data Analysis -> Scroll to Regression
Highlight Fawn Count for the Y Input:
Highlight columns Adult Count to Winter Severity for the X Input:
Make sure you click on Labels and Click OK
If done correctly then
R Square
0.977400423
Adjusted R Square
0.966100634
R-squared:
97.74% of variation in Fawn Count is accounted for by Adult Count, Annual Rain in inches and Winter Severity in this model.
If you want to give a more conservative estimate, you can use the Adjusted R-squared. This can make sure you don't over promise on what the model can do. But the interpretations are the same.
Adjusted R-squared:
96.61% of variation in Fawn Count is accounted for by Adult Count, Annual Rain in inches and Winter Severity in this model.
Note:
Correlation is a value between -1 and 1. This STAYS a decimal.
R-square gets converted from a decimal to a percentage. The Correlation IS NOT a percent,
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leave it as a decimal.
Note:
Correlation is a value between -1 and 1. This STAYS a decimal.
R-square gets converted from a decimal to a percentage. The Correlation IS NOT a percent, leave it as a decimal.
n 6
1 You decided to join a fantasy Baseball league and you think the best way to pick your players is to look at their Batting Averages.
You want to use data from the previous season to help predict Batting Averages to know which players to pick for the upcoming season. You want to use Runs Score, Doubles, Triples, Home Runs and Strike Outs to determine if there is a significant linear relationship for Batting Averages.
You collect data to, to help estimate Batting Average, to see which players you should choose. You collect data on 45 players to help make your decision.
x1 = Runs Score/Times at Bat
x2 = Doubles/Times at Bat
x3 = Triples/Times at Bat
x4 = Home Runs/Times at Bat
x5= Strike Outs/Times at Bat
Interpret the slope(s) of the significant predictors for Batting Average (if there are any).
See Attached Excel for Data.
Baseball data.xlsx
When you hold Double, Triples, Home Runs and Strike Outs constant, as Runs Score increases by 1, your Batting Average will increase by .225/Times at Bat.
When you hold Runs Score, Triples, Home Runs and Strike Outs constant, as Doubles
increases by 1, your Batting Average will increase by .358/Times at Bat.
When you hold Runs Score, Doubles, Triples, and Home Runs constant, as Strike Outs increase by 1, your Batting Average will decrease by .389/Times at Bat.
When you hold Double, Triples, Home Runs and Strike Outs constant, as Runs Score increases by 1, your Batting Average will increase by .447/Times at Bat.
When you hold Runs Score, Triples, Home Runs and Strike Outs constant, as Doubles increases by 1, your Batting Average will increase by .991/Times at Bat.
When you hold Runs Score, Doubles, Triples, and Home Runs constant, as Strike Outs increase by 1, your Batting Average will decrease by .285/Times at Bat.
There are no significant predictors.
When you hold Double, Triples, Home Runs and Strike Outs constant, as Runs Score increases by 1, your Batting Average will increase by .109/Times at Bat.
When you hold Runs Score, Triples, Home Runs and Strike Outs constant, as Doubles increases by 1, your Batting Average will increase by .313/Times at Bat.
When you hold Runs Score, Doubles, Triples, and Home Runs constant, as Strike Outs increase by 1, your Batting Average will decrease by .052/Times at Bat.
Hide question 6 feedback
You can run a Multiple Linear Regression in Excel using the Data Analysis ToolPak.
Data -> Data Analysis -> Scroll to Regression
Highlight Baseball Average for the Y Input:
Highlight all 5 columns from RS/Times at Bat to SO/Times at Bat for the X Input:
Make sure you click on Labels and Click OK
If done correctly then you look under the Coefficients for the values to write out the Regression Equation
Coefficients
Intercept
0.183162857
RS/Times at Bat
0.44668017
Doubles/Times at Bat
0.990904141
Triples/times at bat
0.621603199
HR/Times at Bat
0.27373766
SO/Times at Bat
-0.284559939
Batting Average = 0.1832 + 0.4467(RS/Times at Bat) + 0.9909(Doubles/Times at Bat) +
0.6216(Triples/Times at Bat) + 0.2737(HR/Times at Bat) -0.2846(SO/Times at Bat)
You need to interpret the slope coefficients for all significant predictors. Look at the p-
values for the coefficients to find the significant predictors.
When you hold Double, Triples, Home Runs and Strike Outs constant, as
Runs Score
increases by 1, your Batting Average will increase by .447/Times at Bat.
When you hold Runs Score, Triples, Home Runs and Strike Outs constant, as
Doubles
increases by 1, your Batting Average will increase by .991/Times at Bat.
When you hold Runs Score, Doubles, Triples, and Home Runs constant, as
Strike Outs
increase
by 1, your Batting Average will decrease by .285/Times at Bat.
n 7
0 Which residual plot has the best linear regression model?
a
b
c
d
e
f
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Hide question 7 feedback
A good residual plot has an even distribution of data points above and below the line x = 0.
n 8
0 The least squares regression line for a data set is yˆ=2.3−0.1x
and the standard deviation of the residuals is 0.13.
Does a case with the values x = 4.1, y = 2.34 qualify as an outlier?
Yes
No
Cannot be determined with the given information
Hide question 8 feedback
Plug in 4.1 for x.
y = 2.3 - .1(4.1)
y = 1.89
Residual is y-given - y-predicted.
2.34 - 1.89 = .45 -> this is the residual value.
To see if it is an outlier take 2 and multiply it by .13
2*.13 = .26
.45 is greater than .26, Yes, it is an outlier.
n 9
1 The following data represent the weight of a child riding a bike and the rolling distance achieved after going down a hill without pedaling.
Weight (lbs.)
Rolling Distance (m.)
59
26
84
43
97
48
56
20
103
59
87
44
88
48
92
46
53
28
66
32
71
39
100
49
Using the regression line for this problem, the approximate rolling distance for a child on a bike that weighs 110 lbs. is:
59.2347
58.7213
60.1846
78.4555
Hide question 9 feedback
Copy and paste the data into Excel. Then use the Data Analysis Toolpak and run a Regression.
The y-variable is the distance and the x-variable is the weight. How far the bike will travel will depend on the weight of the child. You want to predict the distance of the bike. Once you get the
Regression output, look under the
Coefficients
to find the values to use for the regression equation.
y = -8.564718163 + 0.611691023 (x)
Plug 110 in for x and solve.
y = -8.564718163 + 0.611691023 (110)
y = 58.7213
n 10
1 The following data represent the weight of a child riding a bike and the rolling distance achieved after going down a hill without pedaling.
Weight (lbs.)
Rolling Distance (m.)
59
26
84
43
97
48
56
20
103
59
87
44
88
48
92
46
53
28
66
32
71
39
100
49
Can it be concluded at a 0.05 level of significance that there is a linear correlation between the
two variables?
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yes
no
Cannot be determined
Hide question 10 feedback
Copy and paste the data into Excel. Then use the Data Analysis Toolpak and run a Regression.
The y-variable is the distance and the x-variable is the weight. How far the bike will travel will depend on the weight of the child. You want to predict the distance of the bike. Once you get the
Regression output, look under the
Significance F
value for the correct p-value to use to make your decision.
Yes, there is a significant relationship p-value = 0.000001
n 11
1 The closer the correlation coefficient is to 1, the stronger the indication of a negative linear relationship.
True
False
Hide question 11 feedback
The closer it is to -1.
n 12
1 Which of the following describes how the scatter plot appears?
Select all that apply.
positive
negative
neither positive or negative
Question 13
1 / 1
point
A new fad diet called Trim-to-the-MAX is running some tests that they can use in advertisements. They sample 25 of their users and record the number of days each has been on
the diet along with how much weight they have lost in pounds. The data are below.
Days on Diet
Weight Lost
7
5
12
7
16
12
19
15
25
20
34
25
39
24
43
29
44
33
49
35
Regression Statistics
Multiple R
0.9851
R Square
0.9705
Adjusted R Square
0.9668
Standard Error
1.9173
Observations
10
ANOVA
df
SS
MS
F
Significance F
Regression
1
967.0912
967.0912
263.0757
2.09917E-07
Residual
8
29.4088
3.6761
Total
9
996.5
Coefficients Standard Error
t Stat
P-value
Lower 95%
Upper 95%
Intercept
0.4912
1.3746
0.3574
0.7301
-2.6785
3.6610
Days on Diet
0.6947
0.0428
16.2196
0.0000
0.5960
0.7935
A strong linear correlation was found between the two variables. Find the standard error of estimate. Round answer to 4 decimal places.
___
Answer:
1.9173
Hide question 13 feedback
This is given to you in the output
Standard Error
1.9173
n 14
1 Body frame size is determined by a person's wrist circumference in relation to height. A researcher measures the wrist circumference and height of a random sample of individuals.
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Model Summary
b
Model
R
R Square
Adjusted R Square
Std. Error of the Estimate
1
.734
a
.539
.525
4.01409
a. Predictors: (Constant), Wrist Circumference
b. Dependent Variable: Height
ANOVA
a
Model
Sum of Squares
df
Mean Square
F
Sig.
1
Regression
621.793
1
621.793
38.590
.000
b
Residual
531.726
33
16.113
Total
1153.519
34
a. Dependent Variable: Height
b. Predictors: (Constant), Wrist Circumference
Model
Unstandardized Coefficients
Standardized Coefficients
t
Sig.
B
Std. Error
Beta
1
(Constant)
38.177
5.089
7.502
.000
Wrist Circumference
4.436
.714
.734
6.212
.000
What is the correct conclusion for testing that there is a significant correlation?
There is a significant correlation.
There is not a significant correlation.
Hide question 14 feedback
The p-value for the slope is 0.000, this is less than .05. Yes, it is significant. This is also the same
p-value for the overall ANOVA Sig. value.
n 15
1 To test the significance of the correlation coefficient, we use the t-distribution with how many degrees of freedom?
n – 2
n + 1
n
1
+ n
2
– 2
n
n – 1
Question 16
0 / 1
point
Choose the correlation coefficient that is represented in the scatterplot.
0.15
-0.82
0.83
Hide question 16 feedback
This has a positive direction with a moderate to strong correlation.
n 17
1 The correlation coefficient, r, is a number between:
0 and ∞
-10 and 10
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-∞ and ∞
0 and 10
0 and 1
- 1 and 1
Question 18
1 / 1
point
Bone mineral density and cola consumption has been recorded for a sample of patients. Let x represent the number of colas consumed per week and y the bone mineral density in grams per
cubic centimeter. Assume the data is normally distributed. Calculate the correlation coefficient using technology (you can copy and paste the data into Excel). Round answer to 4 decimal places.
Make sure you put the 0 in front of the decimal.
x
y
1
0.883
2
0.8734
3
0.8898
4
0.8852
5
0.8816
6
0.863
7
0.8634
8
0.8648
9
0.8552
10
0.8546
11
0.862
Answer:___
Answer:
-0.8241
Hide question 18 feedback
Use =CORREL function in Excel.
n 19
1 A teacher believes that the third homework assignment is a key predictor in how well students
will do on the midterm. Let x represent the third homework score and y the midterm exam score. A random sample of last terms students were selected and their grades are shown below. Assume scores are normally distributed.
HW3
Midterm
13.1
59.811
21.9
87.539
8.8
53.728
24.3
95.283
5.4
39.174
13.2
66.092
20.9
89.729
18.5
78.985
20
86.2
15.4
73.274
25
93.25
9.7
52.257
6.4
43.984
20.2
79.762
21.8
84.258
23.1
92.911
22
87.82
11.4
54.034
14.9
71.869
18.4
76.704
15.1
70.431
15
65.15
16.8
77.208
Find the predicted midterm score when the homework 3 score is 15. Do not round until the end, then round answer to 3 decimal places.
___
Answer:
68.534
Hide question 19 feedback
Copy and Paste the Data into Excel. Data -> Data Analysis -> Scroll to Regression
Highlight Midterm Score for the Y Input:
Highlight HW3 for the X Input:
Make sure you click on Labels and Click OK
If done correctly then you look under the Coefficients for the values to write out the Regression Equation
Coefficients
Intercept
25.90467802
HW3
2.841975886
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y = 25.905 + 2.842x
y = 25.905 + 2.842(15)
Or
Midterm = 25.90467802 + 2.841975886(HW3)
Midterm = 25.90467802462 + 2.841975886(15)
n 20
1 A teacher believes that the third homework assignment is a key predictor in how well students
will do on the midterm. Let x represent the third homework score and y the midterm exam score. A random sample of last terms students were selected and their grades are shown below. Assume scores are normally distributed.
HW3
Midterm
13.1
59.811
21.9
87.539
8.8
53.728
24.3
95.283
5.4
39.174
13.2
66.092
20.9
89.729
18.5
78.985
20
86.2
15.4
73.274
25
93.25
9.7
52.257
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6.4
43.984
20.2
79.762
21.8
84.258
23.1
92.911
22
87.82
11.4
54.034
14.9
71.869
18.4
76.704
15.1
70.431
15
65.15
16.8
77.208
Find the y-intercept and slope for the regression equation using technology (you can copy and
paste the data into Excel). Round answer to 3 decimal places.
ŷ=___+___x
Answer for blank # 1:
25.905
(50 %)
Answer for blank # 2:
2.842
(50 %)
Hide question 20 feedback
Copy and Paste the Data into Excel. Data -> Data Analysis -> Scroll to Regression
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Highlight Midterm Score for the Y Input:
Highlight HW3 for the X Input:
Make sure you click on Labels and Click OK
If done correctly then you look under the Coefficients for the values to write out the Regression Equation
Coefficients
Intercept
25.90467802
HW3
2.841975886
y = 25.905 + 2.842x
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