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Liner Regression
The calculations for our sample size n = 10 are described below. The linear regression model is y = a + bx
Table:
Distance x miles
Time y mins
xy
x2
y2
3.5
16
56.0
12.25
256
2.4
13
31.0
5.76
169
4.9
19
93.1
24.01
361
4.2
18
75.6
17.64
324
3.0
12
36.0
9.0
144
1.3
11
14.3
1.69
121
1.0
8
8.0
64
14
42.0
196
1.5
9
13.5
2.25
81
4.1
65.6
16.81
Σx = 28.9
Σy = 136
Σxy = 435.3
Σx2 = 99.41
Σy2= 1972
The Slope b = {(10 * 435.3) - (28.9 * 136)}/ {(10 * 99.41) - (28.9)2} = 2.66
And the intercept a = {136 - (2.66 * 28.9)}/10 = 5.91
Now we insert these values in the linear model describing as
y = 5.91 + 2.66x Or
Delivery time (mins) = 5.91 + 2.66 delivery distance in miles
The slope of the regression line is the estimated number of minutes per mile required for a delivery. The intercept is the estimated time to prepare for the journey and to deliver the goods that is the time required for each journey other than the actual traveling time.
Tangents with Polar Coordinates Here we now require to discuss some calculus topics in terms of polar coordinates. We will begin with finding tangent lines to polar curves.
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