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2 answers:
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Answer:
Step-by-step explanation:
a) to create a scatter plot for the data, let data for clinics be x and that one for Bowling alleys be y then plot x values with corresponding y values on a graph tool.
<u>x y</u>
14 12
1 3
6 6
5 7
10 7
6 8
11 9
8 15
6 7
16 12
The scatter plot is attached below
b) a cluster is a group of data that is close together. From the plot, the ordered pairs forming a cluster are;
(6,6), (6,7), (5,7), and (6,8)
c) An outlier is a point in a plot which lies very far from the other values. In this case, point its point (8,15)
d)The linear regression equation is one written in the form of
Y=a+bX where ;
Y=the dependent variable
X=the explanatory variable
a=the intercept, which is the value of y when x=o
b=slope of the line
The formula to apply here is ;
a=(∑y)(∑x²)-(∑x)(∑xy) / n(∑x²)-(∑x)² where n is number of samples
b= n(∑xy)-(∑x)(∑y) / n(∑x²)-(∑x)²
Form a table like below
<u>x y xy x² y²</u>
14 12 168 196 144
1 3 3 1 9
6 6 36 36 36
5 7 35 25 49
10 7 70 100 49
6 8 48 36 64
11 9 99 121 81
8 15 120 64 225
6 7 42 36 49
16 12 192 256 144
83 86 813 871 850------------sum
∑x=83 ∑y=86 ∑xy=813 ∑x²=871 ∑y²=850 and n=10
Applying the formula
a=(∑y)(∑x²)-(∑x)(∑xy) / n(∑x²)-(∑x)²
a=[(86 * 871) - (83*813) ]/ 10 (871-83²)
a=[74906 - 67479] / 10(-6018)
a=[7427] / -60180
a= - 0.1234
b= n(∑xy)-(∑x)(∑y) / n(∑x²)-(∑x)²
b=10(813)-(83*86) / 10 (871)-(83²)
b=[8130-7138] / 8710 -6889
b= 992/1821
b=0.5448
The liner regression equation will be;
Y=a+bX
Y= -0.1234 + 0.5448 X
e) correlation coefficient r is given by;
r= n∑xy - (∑x)(∑y) / √n(∑x²)-(∑x)² × √n(∑y²)-(∑y)²
r=10*813 - (83*86) / √10 *871 -83² ×√10*850 - 86²
r=8130-7138 / √8710-6889 × √8500-7396
r= 992 / (√1821 × √1104)
r=992/ 42.67×33.23
r=992/1417.9
r=0.6996
f) r² is important because it shows you the proportion of the variance of one variable that is predictable from the other variable.
r=0.6996
r²=0.6996² =0.4894
Answer:
(b)
Step-by-step explanation:
Given
Required
The graph
First, make y the subject
Divide through by 3
Let x = 3
Let x = 6
So, we plot the graph through
and
See attachment for graph
