STAT 330: 95-3

Assignment 3 Solutions

  1. Q 59, Chapter 9. Let tex2html_wrap_inline89 and tex2html_wrap_inline91 be the mean dry densities for soil samples gathered by the two methods (in practical terms I wonder what the population of soil was). This is a straightforward application of the two independent samples confidence interval methodology. We have tex2html_wrap_inline93 , tex2html_wrap_inline95 , tex2html_wrap_inline97 and tex2html_wrap_inline99 . I count tex2html_wrap_inline101 and tex2html_wrap_inline103 . The pooled estimate of the standard deviation is then tex2html_wrap_inline105 which is 3.70, leading to the pooled estimate of the standard error of tex2html_wrap_inline107 being


    Thus use of the two sample t based confidence interval (on 33 degrees of freedom) gives the interval tex2html_wrap_inline113 or -0.19 to 5.29. Using the interval which does not pool seems inadvisable here since one of the two sample sizes is small. However, the unpooled estimated standard error of the difference in means is


    Thus this interval, using the normal multiplier 1.96 is -0.1 to 5.2. (Aside: there is an approximation due to Welch or Satterthwaite to the distribution of the t-statistic which leads to using the large sample statistic but with a t multiplier with degrees of freedom being




    which comes to 20.16 degrees of freedom leading to the multiplier 2.085 and the interval tex2html_wrap_inline125 which runs from -0.22 to 5.32.)

  2. Chapter 9, Q 60: Another 2 sample unpaired t problem with no real evidence that the two population standard deviations are unequal so we use the two sample unpaired t-test (two tailed). We get tex2html_wrap_inline131 and use 16 degrees of freedom leading to the test statistic t = 6.38. The P value from this statistic is minute - tex2html_wrap_inline137 .
  3. Chapter 9, Q 66: Let tex2html_wrap_inline139 be the probability that an egg survives at 11 degrees and tex2html_wrap_inline141 the probability that an egg survives at 30 degrees. The null hypothesis is that these two probabilities are the same while the alternative appears to be that they are not the same. Under the null hypothesis we estimate the common value by tex2html_wrap_inline143 . The test statistic is


    leading to a P-value of 0.85% which is strong evidence against equal probabilities.

  4. Chapter 9, Q 70: This is another problem with two independent samples. It might have been paired if there were 8 hospitals involved with one room of each type in each hospital. I don't think that is what is intended here. However: if you did use a paired test you get valid conclusions but only 7 degrees of freedom rather than 14 as in the unpaired case. I get a P-value of 35% and conclude that there is little solid evidence of a difference. If I found out that the two hospitals were of different kinds then I would not know whether the difference in bacteria counts was due to carpeting or to the different natures of the hospitals - this is called confounding. In general confounding can be avoided only by randomization which requires a controlled experiment (and not an observational study) in which the experimenter controls which rooms are carpeted and chooses those rooms at random.
  5. Chapter 9, Q 74: The interval is


    or tex2html_wrap_inline153 . The subscript 1 refers to foreign drivers.

  6. Chapter 9, Q 75:
    1. A standard two sample problem with 2 independent samples. The z statistic is


      Since the sample sizes are equal the pooled and unpooled versions are identical. The statistic works out to -6.4 which is overwhelming evidence that the two filling operations produce different mean weights.

    2. This is a one sample problem testing tex2html_wrap_inline159 against the alternative that it is more than 1400. The statistic is


      yielding a one sided P-value of 13.6% from a t distribution on 29 degrees of freedom or of 13.1% from the normal curve showing that it doesn't really matter which you use. This is only very weak evidence against the null hypothesis which wold be accepted.

  7. In 1879, over the period from June 5 to July 2, Michelson carried out a number of measurements of the speed of light. The first 20 measurements and last 20 measurements (minus 299000 km/sec) and several summary statistics are recorded below.

    First 20 Second 20 Difference
    850 890 -40
    740 840 -100
    900 780 120
    1070 810 260
    930 760 170
    850 810 40
    950 790 160
    980 810 170
    980 820 160
    880 850 30
    1000 870 130
    980 870 110
    930 810 120
    650 740 -90
    760 810 -50
    810 940 -130
    1000 950 50
    1000 800 200
    960 810 150
    960 870 90
    Average=909 Average=831.5 Average=77.5
    SD=104.9 SD=54.2 SD=109.8

    Has the bias of the measurements changed between the first 20 and the last 20?


    This is not a paired data problem; I just calculated the differences to confuse. There is no natural pairing between the first measurement and the first of the last 20. Taking the two independent samples model the question is whether the mean tex2html_wrap_inline167 of the first 20 is the same as the mean tex2html_wrap_inline169 of the last 20. The idea is that each mean is the true speed of light plus a bias and so the means are equal if and only if the biases are the same.

    It is not obvious that the standard deviations are unchanged (and the statistical evidence is that the standard deviation has indeed changed). We thus do not pool the estimates of the standard deviations and our test statistic is


    (Numerically, though, since the two sample sizes are the same the statistic has the same value as the two sample t statistic using a pooled variance estimate.) Looking in normal tables and carrying out a 2 tailed test we get a P value of 0.3% and conclude that the bias has indeed changed.

    Although the data are not paired the paired comparisons t-test is still valid and the resulting statistic is 3.16 on 19 degrees of freedom yielding a P-value of 0.5% which leads to the same conclusion. The previous calculation based on the unpaired method would have had 38 degrees of freedom if the assumption of equal variances were valid and an indistinguishable P-value.

  8. Annual records were kept in the Prussian army for the number of deaths by horsekick.

    Year Number of Year Number of
    deaths deaths
    1875 3 1885 5
    1876 5 1886 11
    1877 7 1887 15
    1070 9 1888 6
    1879 10 1889 11
    1880 18 1890 17
    1881 6 1891 12
    1882 14 1892 15
    1883 11 1893 8
    1884 9 1894 4
    Total 92 104

    1. Use a Poisson model and obtain a 95% confidence interval for the long run average mean number of deaths per year.

      I did this in class and obtained two intervals: the first, for the parameter tex2html_wrap_inline183 (which is the annual rate) is


      and the second is the interval for tex2html_wrap_inline187 based on solving a quadratic


      This last interval translates to one for tex2html_wrap_inline183 by dividing by n=20 to get




      which is about the same width but centred slightly differently. The difference is negligible.

    2. Develop a test of the hypothesis that there has been no change in this underlying death rate over the time period in question as follows. Let N be the number of deaths in the first 10 years and M the number in the second 10 years. If the Poisson model with constant death rate is credible then M and N have the same mean. What is the standard error of N-M in terms of the Poisson parameter? How can you estimate this standard error? How can you use this to test the hypothesis that there is no change in mean?

      If tex2html_wrap_inline183 is the per year rate and this is the same for all 20 years then both N and M have Poisson distributions with parameters tex2html_wrap_inline205 . Then tex2html_wrap_inline207 and the natural test statistic is


      leading to a (two sided) P-value of 39% which is not significant.

Richard Lockhart
Tue Feb 3 10:59:11 PST 1998