11-1 Goodness-of-Fit 581 SOLUTION REQUIREMENT CHECK (1) The data come from randomly selected subjects. (2) The data do consist of frequency counts, as shown in Table 11-2. (3) If the 2784 last digits are in 10 categories that are equally likely, each expected frequency is 2784>10 = 278.4, so each expected frequency does satisfy the requirement of being a value that is at least 5. All of the requirements are satisfied. The claim that the digits do not occur with the same frequency is equivalent to the claim that the relative frequencies or probabilities of the 10 cells 1p0, p1, c, p92 are not all equal. (This is equivalent to testing the claim that the distribution of digits is not a uniform distribution.) Step 1: The original claim is that the digits do not occur with the same frequency. That is, at least one of the probabilities, p0, p1, c, p9, is different from the others. Step 2: If the original claim is false, then all of the probabilities are the same. That is, p0 = p1 = p2 = p3 = p4 = p5 = p6 = p7 = p8 = p9. Step 3: The null hypothesis must contain the condition of equality, so we have H0: p0 = p1 = p2 = p3 = p4 = p5 = p6 = p7 = p8 = p9 H1: At least one of the probabilities is different from the others. Step 4: No significance level was specified, so we select the common choice of a = 0.05. Step 5: Because we are testing a claim about the distribution of the last digits being a uniform distribution (with all of the digits having the same probability), we use the goodness-of-fit test described in this section. The x2 distribution is used with the test statistic given in the preceding Key Elements box. Step 6: The observed frequencies O are listed in Table 11-2. Each corresponding expected frequency E is equal to 2784>10 = 278.4 (because the 2784 last digits would be uniformly distributed among the 10 categories). The Excel add-in XLSTAT is used to obtain the results shown in the accompanying screen display, and Table 11-3 shows the manual computation of the x2 test statistic. The test statistic is x2 = 4490.174, the P-value is less than 0.0001, the critical value is x2 = 16.919 (found in Table A-4 with a = 0.05 in the right tail and degrees of freedom equal to k - 1 = 10 - 1 = 9). The test statistic and critical value are shown in Figure 11-2. XLSTAT TABLE 11-3 Calculating the x2 Test Statistic for the Last Digits of Weights Last Digit Observed Frequency O Expected Frequency E O − E 1O − E22 1O − E22 E 0 1175 278.4 896.6 803,891.56 2887.5415 1 44 278.4 -234.4 54,943.36 197.3540 2 169 278.4 -109.4 11,968.36 42.9898 3 111 278.4 -167.4 28,022.76 100.6565 4 112 278.4 -166.4 27,688.96 99.4575 5 731 278.4 452.6 204,846.76 735.8001 6 96 278.4 -182.4 33,269.76 119.5034 7 110 278.4 -168.4 28,358.56 101.8626 8 171 278.4 -107.4 11,534.76 41.4323 9 65 278.4 -213.4 45,539.56 163.5760 x2 = a1O - E22 E = 4490.174 continued
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