Question

please show all work and explain

here is the data and explanation of 12.4

here is appendix D

12.12. * In Problem 12.4, find the value of x2. If the dice really are true, what is the probability of getting a value of X2this large or larger? Explain whether the evidence suggests the dice are loaded. (See Appendix D for the necessary probabili- ties.) 12.4.I throw three dice together a total of 400 times, record the number of sixes in each throw, and obtain the results shown in Table 12.12. Assuming the dice Table 12.12. Number of occurrences for each result for three dice thrown together 400 times; for Problem 12.4. Result Bin number k Occurrences Ok No sixes One six Two or three sixes 217 148 35 are true, use the binomial distribution to find the expected number Ek for each of the three bins and then calculate χ2. Do I have reason to suspect the dice are loaded? Appendix D Probabiles for ON Squared Table D. The pescentage peobability Prob, (Blanks indicate probabilities less than 005%) obeaining a value of in an experiment with d degrees of freedom, as a function of d andx Appendix D d005 1.0 1.5 2.0 2.5 3.0 3.5 4,0 45 5,055 60 80 ว0.0 Probabilities for Chi Squared 1100 48 32 22 683 6 46 34 25 19 14 05 02 2100 61 37 22 14 82 50 30 18 1.1 07 04 02 3 100 1 58 29 15 07 04 02 0 4 100 74 41 5 100 8 42 19 75 29 1 04 0 92 40 1.7 07 03 01 01 0 02 04 06 08 10 12 14 16 18 20 22 24 26 28 3.0 1 100 65 53 44 37 32 27 24 21 18 16 14 11 94 83 2 100 82 67 55 45 7 30 25 20 17 14 1 1 74 6I s0 3100 756931 24 94 1 6 S0 38 29 4 100 94 81 66 52 41 3 23 17 13 92 66 48 34 24 1.7 5 100 96 85 TO 55 42 31 22 16 75 5135 23 16 10 If a series of measurements is grouped into bins-1,, n, we denote by o the number of measurements observed in the bin k. The number expected (on the basis of some assumed or expected distribution) in the bin is denoted by E The extent to which the observations fit the assumed distribution is indicated by the reduced chi squared, defined as 6 00 98 88 73 57 4 30 21 14 95 62 40 25 16 LO 06 7 10076 59 43 30 20 13 82 51 31 19 11 7 04 8 1009 92 78 60 43 29 19 12 2 42 24 14 08 04 02 9 300 99 94 80 6 44 29 18 3 15 19 10 05 3 0.1 10 100 100 95 82 3 29 17 10 55 29 15 08 04 02 0.1 11100D00 %8364 4428 16 914.8 24 1.2 0.6 0.3 0.1 0.1 12100100 %8465 45 28 16 84422.009 04020.1 3 100 100 97 86 66 45 27 1577 3717 07 03 01 0 14 100 100 8 7 67 45 27 14 7 33 14 06 02 0.1 15 100 1008 68 5 26 14 65 29 12 05 02 0.1 where d is the number of degrees of freedom, d-潍-c,and c is the number of constraints (see Section 12.3). The expected average value of 2 is 1. If 2 1, the observed results do not fit the assumed distribution; it1, the agreement is satisfactory 16 100 100 98 89 69 45 26 13 60 25 10 04 0. 17 100 100 99 90 TO 45 25 12 55 22 08 03 0.1 18 100 1009 90 TO 6 25 12 51 20 07 02 0 19 100 1009 91 71 6 25 717 06 02 20 100 10092 72 46 24 43 15 05 01 This test is made quantitative with the probabilities shown in Table D. Let·2 denote the value ofR actually obtained in an experiment with d degrees of free- dom The number Probį2>X-1) is the probability of obtaining a value of 2 as large as the observed, if the measurements seally did follow the assumed disiribution. Thus, if Prob is large, the observed and expected distribu tions are consistent, if it is small, they probably disagree. In particular, if ProbAR > -') is less than 5%, we say the disagreement is significant and reject the assumed distribution a the 5% level. If it is less than 1%, the disagreement is called highly signifiant, and we reject the assumed distribution at the 1% level 22 100 10099 93 73 6 3 0 3.7 12 04 0 24 100 100 100 94 7446 23 92 32 09 03 26 100 100 100 95 5 46 22 85 27 7 02 28 100 100 100 95 76 46 21 78 23 06 01 30100100 100%77 47 21 722.00.30.1 For example, suppose we obtain a reduced chi squared of 2.6 (that is, 2.6) in an experiment with six degrees of freedom (d 6) Acconding t Table D, the probability of getting 2.6 is 1.6%, if the measurements were governed by thc assumed distribution. Thus, at the 5% level (but not quite at the 1% level), we would reject the assumed distributio. For further discussion, see The values in Table D were calculated from the integral See, for example, E. M. Pugh and G. H. Winslow, The Analysis of Physical Mea- rments (Addison-Wesley, 1966), Section 12-5 Chapter 12

Answer 1

probability of getting 0 sixes when we throw a dice 3 times = (5/6)^3 = 125/216

probability of getting 1 six = 3 * (1/6) * (5/6)^2 =75/216

probability of 2 or 3 six = 1 - 125/216 - 75/216 = 16/216

The following table is obtained: Categories Observed Expected (fo-fe)2/fe Category1 217 400 125/216 231.481 (217-231.481)/231.481- 0.906 (148-138.889)2/138.889- 0.598 Category 2 148 400 75/216-138.889 Category3 35 400*16/216-29.63 (35-29.63)2/29.63 0.973 Sum 400 400 2.477 The following null and alternative hypotheses need to be tested Ho : p -125/216,p2-75/216,p3 16/216 Ha: 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 3-1-2, so then the rejection region for this test is R h2 x2 > 5.991)

(2) Rejection Region Based on the information provided, the significance level is a 0.05, the number of degrees of freedom is df 3-1 2, so then the rejection region for this test is Rx2x5.991) (3) Test Statistics The Chi-Squared statistic is computed as follows: X2 = > (01-50 = 0.906 + 0.598 + 0.973 = 2.477 Ei i=1 (4) Decision about the null hypothesis Since it is observed that X2-2.477 < χ,: 5.991, it is then concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the α-_ 0.05 significance level.