Corrections

Corrections to notes

If I find any errors in the notes I have handed out, a corrected version and an explanation of what has changed will appear here. The files of printed notes that can be downloaded here have already been changed to take account of these corrections.

  • Page 1. 5/20=0.25, not 0.2.
  • Page 4. In (a) it should be k = 0,...,n, not 1,...,n.
  • Page 4. In (b) there is a E(X) which should be printed with the fancier version of E.
  • Page 11. In the displayed equation at the bottom of the page, it should be e^{-lambda}, not e^{-n lambda}, within the first product.
  • Page 13. In the first paragraph of this page, there twice appears (\theta-\hat\theta^2) which should be (\theta-\hat\theta)^2.
  • Page 15. In examples (a), (b) and (c) it should start X_1,...X_n, not X_1,...,X_k.
  • Page 15. In (c) it should say "In Example 2.3 (c)", not "In Example 2.3 (b)".
  • Page 27. Add a subscript to f in the definition of likelihood for composite hypothesis, i.e., f_X(x | theta).
  • Page 28. In line 2, it should be mu_1>m_0, not >mu_2.
  • Page 28. In the penultimate line of the second paragraph it should really be P(bar{X}>c | H_0), not P(bar{x}>c | H_0).
  • Page 28. Delete "the" from "the k" at the end of the second paragraph.
  • Page 28. At the end of the first and second paragraphs of Example 6.4, we should have mu=5 and mu=6 (not mu=0 and mu=1).
  • Page 30. In the derivation of the power function W(mu) in the top half of the page, there are three mu_1 that should just be mu.
  • Page 30. Towards the bottom, where the likelihood ratio is being calculated one equation should start with (sigma_0/sigma_1)^n not (sigma_1/sigma_0)^n.
  • Page 32. The calculation of the first p-value is wrong. This should be P(T=4 or 5)=5(0.5^4)(0.5) + 0.5^5=6/32=0.1875. The difference between the p-values is even greater.
  • Page 38. There is a spurious ) after `say' in line 7 of 9.2.
  • Page 39. In example 9.2, it gives the same answer, but is more consistent with H_0 to write e.g., e_{11}=(293/22071)11037=146.52.
  • Page 42. The penultimate line is missing a superscript 2 on (x-\mu).
  • Page 44. At the end of the page 44, we should have $S_{XX} \sim \sigma^2 \chi^2_{n-1}$, not $S_{XX}/(n-1) \sim$.
  • Page 45. In 11.1 the l.h.s. of the second displayed equation should be 1-alpha, not alpha.
  • Page 46. In the top line, 100 should be 50.
  • Page 51. Five lines above Example 12.2 there is a dot missing in the subscript of \bar{x}_{i dot}.
  • Page 53. In the first sentence of 13.2 it should say that epsilon_i is N(0,\sigma^2).
  • Page 56. 8 lines from the bottom it should be =beta, not =0.
  • Page 61. The y axis of the graphs should start with -3, not -2.
  • Page 63. In the middle of the page alpha_q should be alpha_m.
  • Page 63. Three lines from the bottom, `square' should be `squared'.

    Corrections to examples sheets

    If I find any errors in the examples sheets I have handed out, a corrected version and an explanation of what has changed will appear here.

  • Sheet 1, Q1. Remove n>1.
  • Sheet 1, Q5. (a) It should be (1-T/n)^2, not (1-T/2)^2.
  • Sheet 1, Q8. For R it should be 2.236, not 2.239, since what we want is Phi(2.236)-Phi(-2.236)=sqrt{.95}, i.e., Phi(2.236)=(1+sqrt{.95})/2.

  • Sheet 2, Q7. Some things were typed wrong here. It should be: A total of $n$ observations is made, of which $n_i$ fall into category $C_i$ $(i=1,2,\dots ,k)$. Show carefully how to use a $\chi^2$ distribution in testing the above hypothesis, and describe how you would carry out the test, using a statistic of the form \sum_{i=1}^k (n_i-e_i)^2/e_i, where $e_i$ is the expected number of observations in category $C_i$ under (an appropriate case of) the null hypothesis.

  • Sheet 3, Q4. It should be

    (a) Find a 95\% confidence interval for $\mu$ and show that its width is about 3.16.

    (d) Consider the event that the 95\% confidence interval for $\mu$ is narrower when $\s^2$ is known than it is when $\s^2$ is unknown, given that both intervals are computed from the same data, $X_1,\ldots,X_n$, where these are IID samples from $N(\mu,\s^2)$.

    Show that if $n=5$ then this event has probability $1-F_4^{-1}(a)$, where $a=4(1.96/2.78)^2=1.988$, and $F_4$ is the cdf of the $\chi_4^2$ distribution. Verify from tables that this probability is a bit less than $0.75$ (actually, $0.748$). To what do you think this probability tends as $n\rightarrow\infty$?

    Corrections to supervisors' examples sheets cribs

  • Sheet 1, Q5. Same error as in the question. 1/2 should be 1/n.
  • Sheet 1, Q8. There is an R that should be a C.

    Last modified: Tue Mar 11 10:26:12 1997