COMPLETELY CONFUSED WITH STATISTICS

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 94 feet. Use a significance level of 0.05.

Height (ft) 84 122 78 108 73 105 122 78

Interval after (min) 76 77 67 87 61 77 87 70

What is the regression equation?

Answer 1

We are given the data:

 

X = Height in feet: 84, 122, 78, 108, 73, 105, 122, 78
Y = Interval in minutes:  76, 77, 67, 87, 61, 77, 87, 70

 

First calculate the statistical parameters that will be needed to determine the equation for the regression line:

 

N = 8

Standard deviation of x, sx = 20.3452

Standard deviation of y, sy = 9.6622

r = SUM(xy)/SQRT[(SUM(x^2)SUM(y^2)) ] = 0.8239

b = r sy/sx = 0.3913

 

Mean of x, Mx = 96.25

Mean of y, My = 75.25
A = My – bMx = 37.5878

 

Use calculated statistical parameters to determine the regression line, Y = bX + A .

 

Y(x) = 0.3913x + 37.5878

Y(94) = 74 minutes

 

Now, calculate the standard error of our result by calculating the standard error of the slope, Sb and the standard error of the estimate, S est.

 

Sum of squared deviations of x, SSX = 2897.5

Sum of squared deviations of y, SSY = 653.5

S est. = SQRT((1 – r^2)SSY/(N – 2)) = 5.9141

Sb = S est/SSX^1/2 = 0.1099

 

The confidence interval for the slope at the 0.05 significance level:

 

lower limit:  B - (t .95)(Sb) = 0.1188

substituting this estimate of b into the regression equation gives:

Y(94) = 49 minutes

 

upper limit:  B + (t .95)(Sb) = 0.6638
substituting this estimate of b into the regression equation gives:
Y(94) = 100 minutes

 

Where t .95 is the 95 % confidence value
from a t distribution table = 2.48