Question

The data show the time intervals after an eruption​ (to the next​ eruption) of a certain geyser. Find the regression​ equation, letting the first variable be the independent​ (x) variable. Find the best predicted time of the interval after an eruption given that the current eruption has a height of 149 feet. Use a significance level of 0.05. Height​ (ft) 136 140 134 144 102 109 104 116 Interval after​ (min) 83 84 94 92 67 67 84 84.

What is the regression equation?

Answer 1

Sum of X = 985
Sum of Y = 655
Mean X = 123.125
Mean Y = 81.875
Sum of squares (SS_X) = 2066.875
Sum of products (SP) = 862.125

Regression Equation = ŷ = bX + a

b = SP/SS_X = 862.13/2066.88 = 0.4171

a = M_Y - bM_X = 81.88 - (0.42*123.13) = 30.5177

ŷ = 0.4171X + 30.5177

X Values
∑ = 985
Mean = 123.125
∑(X - M_x)² = SS_x = 2066.875

Y Values
∑ = 655
Mean = 81.875
∑(Y - M_y)² = SS_y = 706.875

X and Y Combined
N = 8
∑(X - M_x)(Y - M_y) = 862.125

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

r = 862.125 / √((2066.875)(706.875)) = 0.7133

The sample size is n=8, so then the number of degrees of freedom is df=n−2=8−2=6

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

rc​=0.707

Observe that in this case, the null hypothesis is rejected if ∣r∣>rc​=0.707.

As r>rc, so test is significant

Now for x=149,

ŷ = (0.4171*149) + 30.5177=92.6656