Question

Find the regression​ equation, letting overhead width be the predictor​ (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is 1.5 cm. Can the prediction be​ correct? What is wrong with predicting the weight in this​ case? Use a significance level of 0.05.

Overhead Width​ (cm)	8.2	8.1	9.7	8.6	9.8	8.9
Weight​ (kg)	156	182	241	171	241	210

The regression equation is

​(Round to one decimal place as​ needed.)

The best predicted weight for an overhead width of 1.5 cm is nothing kg.

​(Round to one decimal place as​ needed.)

Can the prediction be​ correct? What is wrong with predicting the weight in this​ case?

A.

The prediction cannot be correct because a negative weight does not make sense. The width in this case is beyond the scope of the available sample data.

B.

The prediction cannot be correct because there is not sufficient evidence of a linear correlation. The width in this case is beyond the scope of the available sample data.

C.

The prediction cannot be correct because a negative weight does not make sense and because there is not sufficient evidence of a linear correlation.

D.

The prediction can be correct. There is nothing wrong with predicting the weight in this case.

Answer 1

Sum of X = 53.3
Sum of Y = 1201
Mean X = 8.8833
Mean Y = 200.1667
Sum of squares (SS_X) = 2.6683
Sum of products (SP) = 123.6167

Regression Equation = ŷ = bX + a

b = SP/SS_X = 123.62/2.67 = 46.3

a = M_Y - bM_X = 200.17 - (46.33*8.88) = -211.4

ŷ = 46.3X - 211.4

Now

X Values
∑ = 53.3
Mean = 8.883
∑(X - M_x)² = SS_x = 2.668

Y Values
∑ = 1201
Mean = 200.167
∑(Y - M_y)² = SS_y = 6562.833

X and Y Combined
N = 6
∑(X - M_x)(Y - M_y) = 123.617

R Calculation
r = ∑((X - M_y)(Y - M_x)) / √((SS_x)(SS_y))

r = 123.617 / √((2.668)(6562.833)) = 0.9341

The sample size is n=6, so then the number of degrees of freedom is ddf=n−2=6−2=4

The corresponding critical correlation value rc​ for a significance level of α=0.05, for a two-tailed test is:

rc​=0.811

Observe that in this case, the null hypothesis is rejected if ∣r∣>rc​=0.811.

Here r>rc, so test is significant

Now for x=1.5,

ŷ = (46.3*1.5) - 211.4=-141.95

A.The prediction cannot be correct because a negative weight does not make sense. The width in this case is beyond the scope of the available sample data.