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CHAPTER 10 Nonparametric Tests
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10.2. Work Problem 10.1 by using a normal approximation to the binomial distribution.
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For a normal approximation to the binomial distribution, we use the fact that the z score corresponding to the number of heads is Z X s m X Np !Npq .
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Because the variable X for the binomial distribution is discrete while that for a normal distribution is continuous, we make a correction for continuity (for example, 3 heads are really a value between 2.5 and 3.5 heads). This amounts to decreasing X by 0.5 if X Np and to increasing X by 0.5 if X Np. Now N 12, m Np (12)(0.5) 6, and s !(12)(0.5)(0.5) 1.73, so that !Npq z (3 0.5) 1.73 6 1.45
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Since this is greater than 1.96 (the value of z for which the area in the left-hand tail is 0.025), we arrive at the same conclusion in Problem 10.1. Note that Pr5Z 1.456 0.0735, which agrees very well with the Pr5X 3 heads6 0.07299 of Problem 10.1.
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10.3. The PQR Company claims that the lifetime of a type of battery that it manufactures is more than 250 hours. A consumer advocate wishing to determine whether the claim is justified measures the lifetimes of 24 of the company s batteries; the results are listed in Table 10-3. Assuming the sample to be random, determine whether the company s claim is justified at the 0.05 significance level. Table 10-3 271 253 264 230 216 295 198 262 211 275 288 252 282 236 294 225 291 243 284 253 272 219 224 268
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Let H0 be the hypothesis that the company s batteries have a lifetime equal to 250 hours, and let H1 be the hypothesis that they have a lifetime greater than 250 hours. To test H0 against H1, we can use the sign test. To do this, we subtract 250 from each entry in Table 10-3 and record the signs of the differences, as shown in Table 10-4. We see that there are 15 plus signs and 9 minus signs. Table 10-4
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Fig. 10-2
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Using a one-tailed test at the 0.05 significance level, we would reject H0 if the z score were greater than 1.645 (Fig. 10-2). Since the z score, using a correction for continuity, is z (15 0.5) (24)(0.5) 1.02
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!(24)(0.5)(0.5)
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the company s claim cannot be justified at the 0.05 level.
10.4. A sample of 40 grades from a statewide examination is shown in Table 10-5. Test the hypothesis at the 0.05 significance level that the median grade for all participants is (a) 66, (b) 75.
CHAPTER 10 Nonparametric Tests
Table 10-5 71 78 67 73 67 46 95 40 55 84 70 78 64 93 43 70 82 72 70 64 66 54 73 86 74 78 57 76 58 86 64 62 79 48 60 95 61 52 83 66
(a) Subtracting 66 from all the entries of Table 10-5 and retaining only the associated signs gives us Table 10-6, in which we see that there are 23 pluses, 15 minuses, and 2 zeros. Discarding the 2 zeros, our sample consists of 38 signs: 23 pluses and 15 minuses. Using a two-tailed test of the normal distribution with probabilities 1 0.025 in each tail (Fig. 10-3), we adopt the following decision rule: 2 (0.05) Accept the hypothesis if 1.96 Reject the hypothesis otherwise. Table 10-6 0 z 1.96.
Fig. 10-3
Since
0.5)
(38)(0.5)
!Npq
!(38)(0.5)(0.5)
we accept the hypothesis that the median is 66 at the 0.05 level. Note that we could also have used 15, the number of minus signs. In this case z with the same conclusion. (b) Subtracting 75 from all the entries in Table 10-5 gives us Table 10-7, in which there are 13 pluses and 27 minuses. Since z (13 0.5) (40)(0.5) !(40)(0.5)(0.5) 2.06 (15 0.5) (38)(0.5) !(38)(0.5)(0.5) 1.14
we reject the hypothesis that the median is 75 at the 0.05 level. Table 10-7
Using this method, we can arrive at a 95% confidence interval for the median grade on the examination. (see Problem 10.30.)
The Mann Whitney U test 10.5. Referring to Table 10-2, determine whether there is a difference at the 0.05 significance level between cables made of alloy I and alloy II.
We organize the work in accordance with steps 1, 2, and 3 (described earlier in this chapter):