Run a regression analysis on the following data set, where yyis the final grade in a math class and xxis the average number of hours the student spent working on math each week. hours/weekxGradey445.6548661.4107710771284.81376.21699.41910020100State the regression equation y=m⋅x+by=m⋅x+b, with constants accurate to two decimal places. What is the predicted value for the final grade when a student spends an average of 15 hours each week on math? Grade = Round to 1 decimal place.

Question
Asked Dec 2, 2019
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Run a regression analysis on the following data set, where
y
y
is the final grade in a math class and
x
x
is the average number of hours the student spent working on math each week.
hours/week
x
Grade
y
4
45.6
5
48
6
61.4
10
77
10
77
12
84.8
13
76.2
16
99.4
19
100
20
100


State the regression equation
y
=
m

x
+
b
y=m⋅x+b
, with constants accurate to two decimal places.

What is the predicted value for the final grade when a student spends an average of 15 hours each week on math?
Grade = Round to 1 decimal place.

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Expert Answer

Step 1

According to the provided information, the final grade in a math class and the number of hours the student spent working on math each week are given.

Here, Number of hours is independent variable (x) and Grade is dependent variable (y)

The regression line can be expressed as:

у %3D тх + с
c intercept
C
m slope
x hour/week
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Image Transcriptionclose

у %3D тх + с c intercept C m slope x hour/week

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Step 2

The regression coefficient can be calculated as:

Σν-Σ
C
η
(Σ)- (Σ(Σ)
Σr) - (Στ)
11ו
help_outline

Image Transcriptionclose

Σν-Σ C η (Σ)- (Σ(Σ) Σr) - (Στ) 11ו

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Step 3

The required data can b...

х^2
yA2
х
y
ху
4
45.6
16 2079.36
182.4
5
48
25
2304
240
36 3769.96
6
61.4
368.4
10
77
100
5929
770
84.8
12
144 7191.04
1017.6
76.2
169 5806.44
990.6
13
99.4
256 9880.36
16
1590.4
19
100
361
10000
1900
20
100
400
10000
2000
Sum
2
105
692.4
1507 56960.16
9059.4
11025 479417.8
9(9059.4)-(105)(692.4)
9(1507)-11025
m=
= 3.480
692.4-(3.480x 105)
с:
9
36.33
Therefore, the regression line will be
y3.48x+36.33
n
help_outline

Image Transcriptionclose

х^2 yA2 х y ху 4 45.6 16 2079.36 182.4 5 48 25 2304 240 36 3769.96 6 61.4 368.4 10 77 100 5929 770 84.8 12 144 7191.04 1017.6 76.2 169 5806.44 990.6 13 99.4 256 9880.36 16 1590.4 19 100 361 10000 1900 20 100 400 10000 2000 Sum 2 105 692.4 1507 56960.16 9059.4 11025 479417.8 9(9059.4)-(105)(692.4) 9(1507)-11025 m= = 3.480 692.4-(3.480x 105) с: 9 36.33 Therefore, the regression line will be y3.48x+36.33 n

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