MATH302 Week 7 Knowledge Check
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American Military University *
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Course
302
Subject
Statistics
Date
Apr 3, 2024
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Pages
25
Uploaded by alexwhite101
Week 7 Knowledge Check Homework Practice Questio…
Attempt 1 of 4
Attempt Score
20 / 20 - 100 %
Overall Grade (Highest Attempt)
20 / 20 - 100 %
Question 1
1 / 1 point
Which of the following equations are linear?
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Question 2
1 / 1 point
y+7=3x
y=6x
2
+8
3y=6x+5y
2
y-x=8x
2
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.
Due to erosion, a river shoreline is losing several thousand pounds of soil each
year. A linear equation that expresses the total amount of soil lost per year is y
= 12,000x.
___
12000___
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Question 3
1 / 1 point
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:
1 = Warm
2 = Mild
3 = Cold
4 = Freeze
5 = Severe
Find the estimated regression equation which can be used to estimate Fawn
Count when using these 3 variables are predictor variables.
How many pounds of soil does the shoreline lose in a year? Round to a whole
number. Don't use any commas or decimals.
Answer:
The slope is 12,000. The change in the slope impacts the linear equation. Since the slope is 12,000 and this is the change in the line, the shoreline will
lose 12,000 pounds of soil each year.
See Attached Excel for Data.
Deer data.xlsx
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Fawn Count = -6.4320 + 3.5626(Adult Count) + 4.3813(Annual Rain) +
4.3878(Winter Severity)
Fawn Count = -5.5591 + 0.3071(Adult Count) + 0.3978(Annual Rain) +
0.2493(Winter Severity)
Fawn Count = 0.9661 + 0.9886(Adult Count) + 0.9774(Annual Rain) +
0.1105(Winter Severity)
Fawn Count = 0.8643 + 0.0853(Adult Count) + 0.0908(Annual Rain) +
0.0568(Winter Severity)
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 you look under the Coefficients for the values to write
out the Regression Equation
Coefficients
Intercept
-5.559106707
Adult Count
0.303715877
Annual Rain in
Inches
0.397827379
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Question 4
1 / 1 point
You are thinking about opening up a Starbucks in your area but what to know
if it is a good investment. How
much money do Starbucks actually make in a year? You collect data to, to help
estimate Annual Net Sales, in
thousands, of dollars to know how much money you will be making.
You collect data on 27 stores to help make your decision.
x1 = Rent in Thousand per month
x2 = Amount spent on Inventory in Thousand per month
x3 = Amount spent on Advertising in Thousand per month
x4 = Sales in Thousand per month
x5= How many Competitors stores are in the Area
Estimate the Annual Net Sales of a Starbucks when Rent = 2.5, Inventory =
430, Advertising =7.75,
Sales = 9.89 and Number of Competitors = 8.
See Attached Excel for Data.
Starbuck Sales data.xlsx
Winter
Severity
0.249286765
Fawn Count = -5.5591 + 0.3071(Adult Count) + 0.3978(Annual Rain) +
0.2493(Winter Severity)
274.65
275.79
Hide ques±on 4 feedback
280.81
270.77
You can run a Multiple Linear Regression in Excel using the Data Analysis
ToolPak.
Data -> Data Analysis -> Scroll to Regression
Highlight Annual Net Sales for the Y Input:
Highlight all 5 columns from Rent to Competitor 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
-49.04488287
Rent/$1000
15.15714777
Inventory/$1000
0.1743754
Advertising/$1000
12.17790597
Sales per month
/$1000
14.29717285
# Competior
Stores in Area
-2.976567478
Annual Net Sales = -49.0449 + 15.1571(Rent) + 0.1744(Inventory) +
12.1779(Advertising) + 14.2972(Sales) - 2.9766(Competitor)
Plug in, Rent = 2.5, Inventory = 430, Advertising = 7.75, Sales = 9.89 and
Number of Competitors = 8
Annual Net Sales = -49.0449 + 15.1571(2.5) + 0.1744(430) + 12.1779(7.75)
+ 14.2972(9.89) - 2.9766(8)
Question 5
1 / 1 point
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
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 = 0.001303023 < .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
No,
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Question 6
1 / 1 point
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
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
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:
1 = Warm
2 = Mild
3 = Cold
4 = Freeze
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
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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.
11.05% 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.
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, leave it as a decimal.
Note:
Correlation is a value between -1 and 1. This STAYS a decimal.
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Question 7
1 / 1 point
Which residual plot has the best linear regression model?
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R-square gets converted from a decimal to a percentage. The Correlation IS
NOT a percent, leave it as a decimal.
a
b
c
d
e
f
A good residual plot has an even distribution of data points above and below
the line x = 0.
Question 8
1 / 1 point
In the context of regression analysis, what is the definition of an influential
point?
Question 9
1 / 1 point
When r is close to ____, there is either a weak linear relationship between x
and y or no linear relationship between x and y.
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Question 10
1 / 1 point
The closer the correlation coefficient is to 1, the stronger the indication of a
negative linear relationship.
Observed data points that are far from the least squares line
Observed data points that are far from the other observed data points
in the horizontal direction
Observed data points that are close to the least squares line
Observed data points that are close to the other observed data points
in the horizontal direction
-1
1
0
±∞
Correlation is a value from -1 to 1. The closer r is to 0, the worst the
correlation is. When r = 0, this means there is no correlation (zero) between
the two values.
True
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Question 11
1 / 1 point
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
False
The closer it is to -1.
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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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Question 12
1 / 1 point
Which of the following describes how the scatter plot appears? Select all that
apply.
yes
no
Cannot be determined
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
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
weak
strong
nonlinear
negative
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
CoefficientsStandard 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
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This is given to you in the output
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Question 14
1 / 1 point
What are the hypotheses for testing to see if a correlation is statistically
significant?
Question 15
1 / 1 point
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.
Model Summary
b
Model
R
R
Square
Adjusted
R
Std.
Error of
Standard Error1.9173
H
0
: r
= ±1
; H
1
: r
≠ ±
1
H
0
: ρ
= ±1 ; H
1
:
ρ
≠ ±1
H
0
: r
= 0
; H
1
: r
≠ 0
H
0
: ρ
= 0
; H
1
:
ρ
=1
H
0
: ρ
= 0
; H
1
:
ρ
≠ 0
Square
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 value of the test statistic to see if the correlation is statistically significant?
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0.539
7.502
4.436
5.089
6.212
0.734
Question 16
1 / 1 point
Select the correlation coefficient that is represented in the following
scatterplot.
You need to test the slope. The T-stat that corresponds with the slope (or
Wrist Circumference) is 6.212
-0.50
-0.82
-0.15
0.83
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Question 17
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 ques±on 17 feedback
This has a negative direction with a moderate to strong correlation.
Question 18
1 / 1 point
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. 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.
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
Use =CORREL function in Excel.
14.9
71.869
18.4
76.704
15.1
70.431
15
65.15
16.8
77.208
Answer:___
___
Answer: 0.9846
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Question 19
1 / 1 point
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. The data is
displayed below.
Regression Statistics
Use =CORREL function n in Excel.
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Multiple R
0.7938
R Square
0.6301
Adjusted R Square 0.6182
Standard Error
3.8648
Observations
33
Coefficients Standard Error t Stat
P-value
Intercept 31.6304
5.2538
6.0205 1.16E-06
x
5.4496
0.7499
7.2673 3.55E-08
What is the predicted height (in inches) for a person with a wrist
circumference of 7 inches? Round answer to 4 decimal places.
Answer:___
Answer: 69.7776
Hide ques±on 19 feedback
Use the values from the Coefficients to write out the regression equation.
Coefficients
Intercept
31.6304
x
5.4496
y = 31.6304 + 5.4496(7)
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Question 20
1 / 1 point
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
y = 69.776
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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 %)
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Copy and Paste the Data into Excel. Data -> Data Analysis -> Scroll to
Regression
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Done
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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