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ŷ= 1.795x +2.195 is the equation for the line of best fit for the data
<h3>How to use regression to find the equation for the line of best fit?</h3>Consider the table in the image attached:
∑x = 29, ∑y = 74, ∑x²= 125, ∑xy = 288, n = 10 (number data points)
The linear regression equation is of the form:
ŷ = ax + b
where a and b are the slope and y-intercept respectively
a = ( n∑xy -(∑x)(∑y) ) / ( n∑x² - (∑x)² )
a = (10×288 - 29×74) / ( 10×125-29² )
= 2880-2146 / 1250-841
= 734/409
= 1.795
x' = ∑x/n
x' = 29/10 = 2.9
y' = ∑y/n
y' = 74/10 = 7.4
b = y' - ax'
b = 7.4 - 1.795×2.9
= 7.4 - 5.2055
= 2.195
ŷ = ax + b
ŷ= 1.795x +2.195
Therefore, the equation for the line of best fit for the data is ŷ= 1.795x +2.195
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Find the annual earnings if a worker is paid $743.75 semimonthly.
Answer: A worker is paid $743.75 semimonthly.
Semimonthly is twice a month.
Therefore, a worker is paid $ monthly
There are 12 months in a year.
Therefore a worker is paid $ annually
Hence, the option b. $17,850.00 is correct