SECTION 10.1 Goodness-of-Fit Test 531 O E O − E 1O − E22 1O − E22 E 80 83.333 -3.333 11.108889 0.133307201 95 83.333 11.667 136.118889 1.633433202 8883.333 4.667 21.780889 0.261371713 83 83.333 -0.333 0.110889 0.001330673 76 83.333 -7.333 53.772889 0.645277249 78 83.333 -5.333 28.440889 0.341292033 x 2 = Σ1O - E22 E ≈ 3.016 The figure shows the location of the rejection region and the chi-square test statistic. Because x 2 is not in the rejection region, you fail to reject the null hypothesis. 5 15 20 25 0 2χ = 9.236 2χ 2χ ≈ 3.016 α= 0.10 Rejection region Interpretation There is not enough evidence at the 10% level of significance to reject the claim that the distribution of the different-colored candies in bags of dark chocolate M&M’s® is uniform. TRY IT YOURSELF 3 A researcher claims that the number of different-colored candies in bags of peanut M&M’s® is uniformly distributed. To test this claim, you randomly select a bag that contains 180 peanut M&M’s® and determine the frequency of each color. The results are shown in the table below. Using a = 0.05, test the researcher’s claim. (Adapted from Mars, Incorporated) Color Frequency, f Brown 22 Yellow 27 Red 22 Blue 41 Orange 41 Green 27 Answer: Page A42 You can use technology and a P-value to perform a chi-square goodness-of-fit test. For instance, using a TI-84 Plus and the data in Example 3, you obtain P = 0.6975171071, as shown at the left. Because P 7 a, you fail to reject the null hypothesis. TI-84 PLUS χ2=3.016012072 p=0.6975171071 df=5 CNTRB={0.133307... χ2GOF-Test
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