STAT 330: 98-1

Assignment 5 Solutions

  1. Chapter 6 Q 13: Let tex2html_wrap_inline76 be the population mean. Then tex2html_wrap_inline78 so we must evaluate tex2html_wrap_inline76 . The definition is


    so that tex2html_wrap_inline82 .

  2. Chapter 6 Q 15: The moment of order r is tex2html_wrap_inline86 which is given by


    Make the substitution tex2html_wrap_inline90 for which tex2html_wrap_inline92 and get


    For the special case r=2 we get the formula in the book (using tex2html_wrap_inline96 ).

    1. According to the fact tex2html_wrap_inline98 so that tex2html_wrap_inline100 . Thus tex2html_wrap_inline102 is unbiased.
    2. You square these numbers, add then divide by 20 and get 74.505 as in the appendix.

  3. Chapter 6 Q 28:

    1. The log of the density is tex2html_wrap_inline104 so that the log likelihood is


      To maximize this we take the derivative with respect to tex2html_wrap_inline108 and set the result equal to 0 getting


      which has root tex2html_wrap_inline112 which is the same as the unbiased estimate above.

    2. The median, m, of the distribution satisfies


      The integral may be done by substituting tex2html_wrap_inline90 to get


      Solve this to get tex2html_wrap_inline122 and then the mle of m is


  4. Chapter 8 Q 26b: We are given tex2html_wrap_inline128 so that the test rejects when t exceeds the t critical value on 7 degrees of freedom. The nearest curve in Table A.13 is for 6 degrees of freedom and the quantity d is (4.00-3.5)/1.25= .4 For 6 df this appears to correspond to a tex2html_wrap_inline138 of around 0.76 while for 9 df we would get about 0.65. My estimate is about a third of the way between them (because 7 is a third of the way between 6 and 9) or roughly .72.
  5. Chapter 8 Q 30b: See the formula on page 319 and plug in to get


    so that n=24 would be needed. Notice that the formula for the sample size for a two sided level tex2html_wrap_inline144 test is just the same as that\ for a one sided level tex2html_wrap_inline146 test.

    It is also acceptable to do this using the t-test graphs as in the previous question. In fact, I think this is what the text intended really. If so you get a sample size of 19.

  6. Chapter 9 Q 6 b,c: The rejection region of the test is


    which can be rewritten as


    The power function at tex2html_wrap_inline152 is then the area to the right of the right hand side of this last formula under a standard normal curve. Plugging in numbers shows that we want the area to the left of -0.5 which is .31. That is tex2html_wrap_inline156 approximately. To get a probability of type II error equal to 0.1 we need


    Putting m=40 and plugging all the other numbers we get


    which leads to n of roughly 37 (actually a bit over so rounding up to 38 would be normal).

  7. Chapter 9 Q 12: This is just the formula at the foot of page 351. We get


    which we round up to 50 for safety's sake.

  8. Chapter 9 Q 73: This is a paired comparison problem. You take the differences, getting 3.2, -3.4, 0.4, 0.5, 0.3, -1.4, -0.3, 0, 3.5, -3.7, 3.7, 3.9, 1.6, 3.2. The one sample t statistic is 1.22 on 13 degrees of freedom leading to a 1 sided P value of 12% which is only weak evidence of higher TSI for the treatment than for the placebo.

Richard Lockhart
Fri Feb 13 15:37:53 PST 1998