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Goodness of fit 7.43. Use the chisquare test to determine the goodness of fit of the data in Problem 7.30. x2 (38 7.45 Since the number of parameters used in estimating the expected frequencies is m 1 (namely, the parameter p of the binomial distribution), n k 1 m 6 1 1 4. For n 4, x2 9.49. Hence the fit of the data is good. 0.95 For n 4, x2 0.711. Therefore, since x2 7.54 0.711, the fit is not so good as to be incredible. 0.05 The P value is P(x2 7.45) 0.11. 33.2)2 33.2 (144 161.9)2 161.9 (342 316.2)2 316.2 (287 308.7)2 308.7 (164 150.7)2 150.7 (25 29.4)2 29.4 7.44. Determine the goodness of fit of the data in Problem 7.32.
4.13)2 (18 20.68)2 (42 38.92)2 (27 27.71)2 (8 7.43)2 0.959 4.13 20.68 38.92 27.71 7.43 Since the number of parameters used in estimating the expected frequencies is m 2 (namely, the mean m and the standard deviation s of the normal distribution), n k 1 m 5 1 2 2. For n 2, x2 5.99. Therefore, we conclude that the fit of the data is very good. 0.95 For n 2, x2 0.103. Then, since x2 0.959 0.103, the fit is not too good. 0.05 2 The P value is P(x 0.959) 0.62. x2 (5 Contingency tables 7.45. Work Problem 7.13 by using the chisquare test.
The conditions of the problem are presented in Table 713. Under the null hypothesis H0 that the serum has no effect, we would expect 70 people in each of the groups to recover and 30 in each group not to recover, as indicated in Table 714. Note that H0 is equivalent to the statement that recovery is independent of the use of the serum, i.e., the classifications are independent. Table 713 Frequencies Observed Recover Group A (using serum) Group B (not using serum) TOTAL 75 65 140 Do Not Recover 25 35 60 TOTAL 100 100 200 Table 714 Frequencies Expected under H0 Recover Group A (using serum) Group B (not using serum) TOTAL (75 70 70)2 (65 70 70 70 140 70)2 (25 30 Do Not Recover 30 30 60 30)2 TOTAL 100 100 200 (35 30 30)2 CHAPTER 7 Tests of Hypotheses and Significance
To determine the number of degrees of freedom, consider Table 715, which is the same as Tables 713 and 714 except that only totals are shown. It is clear that we have the freedom of placing only one number in any of the four empty cells, since once this is done the numbers in the remaining cells are uniquely determined from the indicated totals. Therefore, there is 1 degree of freedom. Table 715 Recover Group A Group B TOTAL 140 60 Do Not Recover TOTAL 100 100 200 Since x2 3.84 for 1 degree of freedom, and since x2 2.38 3.84, we conclude that the results are 0.95 not significant at a 0.05 level. We are therefore unable to reject H0 at this level, and we conclude either that the serum is not effective or else withhold decision pending further tests. The P value of the observed frequencies is P(x2 2.38) 0.12. Note that x2 2.38 is the square of the z score, z 1.54, obtained in Problem 7.13. In general, the chisquare test involving sample proportions in a 2 2 contingency table is equivalent to a test of significance of differences in proportions using the normal approximation as on page 217. Note also that the P value 0.12 here is twice the P value 0.0618 in Problem 7.13. In general a onetailed test using x2 is equivalent to a twotailed test using x since, for example, x2 x2 corresponds to x x0.95 or 0.95 x x0.95. Since for 2 2 tables x2 is the square of the z score, x is the same as z for this case. Therefore, a rejection of a hypothesis at the 0.05 level using x2 is equivalent to a rejection in a twotailed test at the 0.10 level using z.

