Share
Explore BrainMass

# Regression : Forecasting and Confidence Intervals

The data below present the results of a hydrological investigation of the Snake River watershed. The main purpose of the investigation was to forecast the water yield (y inches) from April to July using the weighted water content of snow (x), estimated on April 1.

Year X Y Year X Y
1919 23.1 10.5 1928 37.9 22.9
1920 32.8 16.7 1929 30.5 14.1
1921 31.8 18.2 1930 25.1 12.9
1922 32.0 17.0 1931 12.4 8.8
1923 30.4 16.3 1932 35.1 17.4
1924 24.0 10.5 1933 31.5 14.9
1925 39.5 23.1 1934 21.1 10.5
1926 24.2 12.4 1935 27.6 16.1
1927 52.2 24.9

Sum x = 511.20
Sum y = 267.20
Sum x^2 = 16597
Sum y^2 = 4554
Sum x*y = 8649.8

From mini tab I get the regression equation to be

C2=0.521C1.

Assuming the relationship between x and y to be approximately linear, use the method of least squares estimation to obtain an appropriate equation for forecasting y which passes through the origin.

Find point estimates and 95% confidence intervals for:

1) the slope of the true regression line of y on x;
2) the standard deviation of the 'error' about this line;
3) the true expected value of y when x=30.0

also obtain a 95% prediction interval for y when x=30.0

#### Solution Preview

Find point estimates and 95% confidence intervals for:
1) the slope of the true regression line of y on x;
Slope is
B1 = Sum x*y / Sum x^2 = 0.5019
Intercept is
Bo = Ym - B1*Xm = 0.6251

Let e = Y-(Bo +B1*X)
Compute SUM (e^2) = 45.62
Then variance is &#963;2= SUM (e^2) /(n-2)= 45.62/(17-2)=3.04
variance of B1 ...

#### Solution Summary

Forecasting and Confidence Intervals are investigated. The solution is detailed and well presented.

\$2.19