The data show the chest size and weight of several bears. Find the regression eq
ID: 3262779 • Letter: T
Question
The data show the chest size and weight of several bears. Find the regression equation, letting chest size be the independent (x) variable. Then find the best predicted weight of a bear with a chest size of 51 inches. Is the result close to the actual weight of 442 ounds? Use a significance level of 0.05.
Chest size (inches): 45 50 43 43 52 52
Weight (pounds): 352 374 275 314 440 367
a. What is the regresion equation?
^y = __ + __x (Round to two decimal places as needed.)
b. What is the best predicted value?
^y is about __ pounds (Round to one decimal place as needed.
Explanation / Answer
According to the given question
Chest size (inches): 45 50 43 43 52 52
Weight (pounds): 352 374 275 314 440 367
We will find an equation of the regression line in 4 steps.
Step 1: Find XY and X2 as it was done in the table below.
X Y XY XX
45 352 15840 2025
50 374 18700 2500
43 275 11825 1849
43 314 13502 1849
52 440 22880 2704
52 367 19084 2704
Step 2: Find the sum of every column:
X=285 , Y=2122 , XY=101831 , X2=13631
Step 3: Use the following equations to find a and b:
a =YX2XXY / nX2(X)2
= 212213631285101831 / 6136312852
172.643
b =nXYXY / nX2(X)2
=61018312852122613631(285)211.08
Step 4: Substitute a and b in regression equation formula
y = a + bx
y = 172.64 + 11.08x
To find the best predicted value of a bear we have to find correlation coefficient
We know that
X=285 , Y=2122 , XY=101831 , X2=13631 , Y2=766290
Now Use the following formula to work out the correlation coefficient.
r = nXYXY / [nX2(X)2][nY2(Y)2]
= 61018312852122 / [6136312852][676629021222]
0.8521
Now to predict the best value of x putting in the equation we get y = 392.44 with a significance level of 0.05 value will be close to 400.
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