STAT 350: 95-3

Assignment 6

  1. Consider a design with 5 data points tex2html_wrap_inline52 , tex2html_wrap_inline54 , tex2html_wrap_inline56 , tex2html_wrap_inline58 and tex2html_wrap_inline60 so that the design matrix is

    displaymath62

    If we fit a simple linear regression of the form

    displaymath64

    for which the design matrix is as above, evaluate the non-centrality parameter of the t test of tex2html_wrap_inline68 when in fact tex2html_wrap_inline70 and tex2html_wrap_inline72 . What would the power of a two sided 1% level t test of this null hypothesis be? How many times would we have to replicate this design to get a power of 0.9 for a 1% level test?

    Solution: We have

    displaymath76

    and

    displaymath78

    The variance of tex2html_wrap_inline80 is tex2html_wrap_inline82 and the non-centrality parameter is tex2html_wrap_inline84 . From Table B 5 p 1347 I get a power of 0.31 (3 df for error). If we replicate the design m times we would have a non-centrality parameter of tex2html_wrap_inline88 . The infinity line in the table would suggest an m of 1 but because the degrees of freedom is so low we have to do some trial and error. An m of 2 would give a non-centrality parameter of 5.66. There would be 8 degrees of freedom and the power would be around 0.95 so m=2 would be adequate.

  2. For the design points tex2html_wrap_inline96 as in the previous question evaluate the non-centrality parameter of the F test of the hypothesis tex2html_wrap_inline100 in the model

    displaymath102

    Assume that in fact tex2html_wrap_inline104 and tex2html_wrap_inline106 .

    Solution: The non-centrality parameter is

    displaymath108

    Here tex2html_wrap_inline110 , tex2html_wrap_inline112 is a tex2html_wrap_inline114 matrix with all entries equal to 1/5 and

    displaymath116

    We find tex2html_wrap_inline118 and finally that the non-centrality parameter is 66/2.5=26.4.

  3. Question 10.7 parts a, c, d, e and f.

    The data set in the text has one number different than the data set on the computer disk. The results below are for the data off the disk. Students who took the data from the book got somewhat different results but I marked them right, since they were right for the data they used.

    Solution: The fitted line is

    displaymath120

    a) The residual plot is

    This plot suggests that the residuals are bigger when Speed is bigger; this is evidence of heteroscedasticity.

    c) The desired plot is

    This plot suggests a linear fit of tex2html_wrap_inline122 against Speed might work.

    d) The regression equation is

    displaymath124

    The required weights are 

     0.0146 0.0032 0.0052 0.0032 0.0146 0.0052 
 0.0052 0.0032 0.0146 0.0032 0.0146 0.0052

    e) The weighted least squares regression line is

    displaymath126

    These estimates are little changed compared to the standard errors.

    OLS WLS OLS WLS
    Parameter Estimate Estimate SE SE
    Intercept -5.75 -6.23 16.73 13.17
    Slope 0.1875 0.1891 0.0538 0.0506

    f) The standard errors from weighted least squares are rather smaller though more so for the less important parameter, the intercept.

  4. Suppose

    displaymath128

    where the errors tex2html_wrap_inline130 have variances tex2html_wrap_inline132 and the tex2html_wrap_inline134 are known quantities. Find an explicit algebraic formula for the weighted least squares estimate of tex2html_wrap_inline136 .

    Solution: The matrix X is a single column of n ones. The matrix W is diagonal with tex2html_wrap_inline134 down the diagonal. The vector WX has entries tex2html_wrap_inline148 and finally tex2html_wrap_inline150 . Next tex2html_wrap_inline152 is the inner product between the vector WX with ith entry tex2html_wrap_inline134 and Y with ith entry tex2html_wrap_inline164 so tex2html_wrap_inline166 . Finally tex2html_wrap_inline168 which is what is called a weighted average.


Richard Lockhart
Wed Apr 2 23:57:05 PST 1997