D ICh. 12 Homework! [1] A game machine was designed to yield live equally probable outcomes (A, B, C, D, E) for each trials For checking of any bias it might have, 500 trials were made as follows Fill in the blanks. By calculating Chi-square, determine whether the machine has hias or not with a 0.05 AC H. 85 observed expected (100) (oneyle DE 107 70597 306 L (100)(100) (100) (100) [2] Go back to the gender dependence problem. The null hypothesis H, is stated as; the opinions on an environment legislation are not dependent on genders. Given observed and expected frequencies (expected frequencies in parentheses) as the following table, fill in the blanks. Calculate Chi-square and then make your judgment with a = 0.05. support 35(38) oppuse 10(8) How total 50 so opinion 514) 314) male Famil female (8) 50 41(38) 76 (column total) 16 100 JE 300 male to Semale (o-ele [Ch. 13 Homework] [1] Suppose the result of different productions by five workers (in 13.1) were acquired as follows. A B C 52 58 63 64 test production worker 54 worker2 62 worker3 57 workers 61 workers 62 60 69 63 67 62 65 sample mean sample variance sample st.d Calculate the between-treatment estimate and within-treatment estimate for variance a State if the three production methods A, B and C can be regarded as 'equivalent in efficiency based on the above data. 18:549|| 21.026 23.337317 22 307 24.996 27438 3578 27.204 30.144 32.852 36.191 33.396 36415 39.364 47980 11.2 VARIANCE TEST Test statistic about population variance x = (-1) Consider the following issue. London Bus Company hopes its drivers running on schedule within arrival time variance less than 3.0 min. From randomly selected 16 arrivals at City Hall station, the variance was found as s = 3.7 min. Is it excessive? Bus Company wants judgment with significance level of a =0.05 The corresponding Chi-square is -08-12 15x12= 18.5 13.0 which is smaller than x 25.0 (for d.f-15). Therefore, from this testing, it is concluded that London Bus drivers are running on schedule. ICh. 11 Homework [] Calculate the 95% confidence interval estimate of a population variance o from the sample of randomly selected 25 bottles, of which net weight variance is s* = 5.65(gram) [2] Check again the running on schedule problem above. After secondary examination on 11 arrivals at London Museum, the arrival time variance was found as s = 5.5 min. What is your opinion? (hint compare the acquired x value with x 05 and oxford.f-10) 3.229 6.362 2261 8.547 7.633 350? 10.117 1 10.560 6 12.401 15848 15.659
