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    Regression : Forecasting and Confidence Intervals

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    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

    © BrainMass Inc. brainmass.com October 9, 2019, 4:29 pm ad1c9bdddf
    https://brainmass.com/math/interpolation-extrapolation-and-regression/regression-forecasting-confidence-intervals-33761

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    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 σ2= 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.

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