<p> Homework 7 due Mar. 21</p><p>8-61 (9-62) 8-68 (9-69) 9-4 (10-4) 9-20 (10-22) </p><p>8-61. Mean = np = 6(0.25) = 1.5</p><p>Value 0 1 2 3 4 Observed 4 21 10 13 2 Expected 8.8989 17.7979 14.8315 6.5918 1.6479</p><p>The expected frequency for value 4 is less than 3. Combine this cell with value 3:</p><p>Value 0 1 2 3-4 Observed 4 21 10 15 Expected 8.8989 17.7979 14.8315 8.2397</p><p>The degrees of freedom are k  p  1 = 4  0  1 = 3</p><p> a) 1) The variable of interest is the form of the distribution for the random variable X.</p><p>2) H0: The form of the distribution is binomial with n = 6 and p = 0.25 3) H1: The form of the distribution is not binomial with n = 6 and p = 0.25 4)  = 0.05 5) The test statistic is k 2 2 Oi  Ei  0   i1 Ei 2 2 6) Reject H0 if o  0.05,3  7.81 7) 4  8.89892 15  8.23972 2  ⋯  10.39 0 8.8989 8.2397</p><p>8) Since 10.39 > 7.81 reject H0. We can conclude that the distribution is not binomial with n = 6 and p = 0.25 at  = 0.05. b) P-value = 0.0155 (found using Minitab) 8-68. 1. The variable of interest is failures of an electronic component. 2. H0: Type of failure is independent of mounting position.</p><p>3. H1: Type of failure is not independent of mounting position. 4. a = 0.01 5. The test statistic is: r c 2 2 (oij  Eij) 0    i1j1 Eij 2 6. The critical value is .01,3  11.344 7. The calculated test statistic is 2 0  10.71 2 2 8. 0  0.01,3 , do not reject H0 and conclude evidence is not sufficient to claim that the type of failure is not independent of the mounting position at a = 0.01. P-value = 0.013 9-4. a) 1) The parameter of interest is the difference in mean burning rate, 1  2</p><p>2) H0 : 1  2  0 or 1  2</p><p>3) H1 : 1  2  0 or 1  2 4)  = 0.05 5) The test statistic is (x1  x2 )  0 z0  2 2 1  2 n1 n2</p><p>6) Reject H0 if z0 < z/2 = 1.96 or z0 > z/2 = 1.96 7) x1  18 x2  24  = 0 1  3 2  3</p><p> n1 = 20 n2 = 20 (18  24)  0 z0   6.32 (3)2 (3)2  20 20 8) Since 6.32 < 1.96 reject the null hypothesis and conclude the mean burning rates do not differ significantly at  = 0.05.</p><p> b) P-value = 2(1 (6.32))  2(1 1)  0 ‘               0   0 c)    z/2    z/2    2 2   2 2   1  2   1  2   n1 n2   n1 n2           2.5   2.5  = 1.96    1.96    (3)2 (3)2   (3)2 (3)2         20 20   20 20  = 1.96  2.64  1.96  2.64  0.68  4.6 = 0.24825  0 = 0.24825</p><p>2 2 2 2 1 2 1 2 d) x1  x2   z/2   1  2  x1  x2   z/2  n1 n2 n1 n2</p><p>(3)2 (3)2 (3)2 (3)2 18  24  1.96       18  24  1.96  20 20 1 2 20 20</p><p>7.86  1  2  4.14</p><p>We are 95% confident that the mean burning rate for solid fuel propellant 2 exceeds that of propellant 1 by between 4.14 and 7.86 cm/s.</p><p>9-20. a) 1) The parameter of interest is the difference in mean impact strength, 1  2</p><p>2) H0 : 1  2  0 or 1  2</p><p>3) H1 : 1  2  0 or 1  2 4)  = 0.05 5) The test statistic is (x1  x2 )  0 t0  s2 s2 1  2 n1 n2</p><p>6) Reject the null hypothesis if t0 < t, where t0.05,25 = 1.708 since 2  2 2  s1 s2     n1 n2     2  27.43  2  25.43 2  2   2  s1 s2      n1   n2   n1  1 n2  1   25 (truncated) 7) x1  290 x2  321</p><p> s1  12 s2  22</p><p> n1 = 10 n2 = 16 (290  321)  0 t0   4.64 (12)2 (22)2  10 16 8) Since 4.64 < 1.708 reject the null hypothesis and conclude that supplier 2 provides gears with higher mean impact strength at the 0.05 level of significance. b) P-value = P(t < 4.64): P-value < 0.0005 c) 1) The parameter of interest is the difference in mean impact strength, 2  1</p><p>2) H0 : 2  1  25</p><p>3) H1 : 2  1  25 or 2  1  25 4)  = 0.05 5) The test statistic is (x2  x1)   t0  s2 s2 1  2 n1 n2</p><p>6) Reject the null hypothesis if t0 > t, = 1.708 where 2  2 2  s1 s2     n1 n2     2  27.43  2  25.43 2  2   2  s1 s2      n1   n2   n1  1 n2  1   25 x  x   7) 1 290 2 321 0 =25 s1  12 s2  22 n1 = 10 n2 = 16 (321 290)  25 t0   0.898 (12)2 (22)2  10 16 8) Since 0.898 < 1.708, do not reject the null hypothesis and conclude that the mean impact strength from supplier 2 is not at least 25 ft-lb higher that supplier 1 using  = 0.05.</p>