Question

A randomly selected dice (2 die) is been tested to be fair (all sides are equally likely to come up on a roll). Use α=0.05 to test the claim if samples of rolls came up with same frequency.

3-A randomly selected dice (2 die) is been tested to be fair (all sides are equally likely to come up on a roll). Use a=0.05 to test the claim if samples of rolls came up with same frequency. 4 6 5 88 7 104 8 90 9 70 B2 96

Answer 1

Categories Observed Expected (fo-fe)</fe 4 82 530*0.1667=88.351 (82-88.351)288.351 = 0.457 (88-88.351)2/88.351 = 0.001 5 88 530*0.1667=88.351 6 96 530*0.1667=88.351 (96-88.351)2/88.351 = 0.662 7 104 530*0.1667=88.351 (104-88.351)288.351 = 2.772 8 90 530*0.1667=88.351 (90-88.351)2/88.351 = 0.031 (70-88.351)2/88.351 = 3.812 9 70 530*0.1667=88.351 Sum = 530 530 7.734

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H. : P1 = 0.1667, P2 = 0.1667, p = 0.1667, P4 = 0.1667, p5 = 0.1667, p6 = 0.1667 H,: Some of the population proportions differ from the values stated in the null hypothesis This corresponds to a Chi-Square test for Goodness of Fit. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, the number of degrees of freedom is df = 6 - 1=5, so then the rejection region for this test is R = {x? : x² > 11.07}. (3) Test Statistics The Chi-Squared statistic is computed as follows: 72 X2 (O; - E;) E; 0.457 +0.001 +0.662 +2.772 +0.031 +3.812 = 7.734 i=1 (4) Decision about the null hypothesis Since it is observed that x2 = 7.734 < x2 = 11.07, it is then concluded that the null hypothesis is not rejected.

Hence, we can say that samples of rolls came up with same frequency.

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