Homework 7 due Mar. 21
8-61 (9-62) 8-68 (9-69) 9-4 (10-4) 9-20 (10-22)
8-61. Mean = np = 6(0.25) = 1.5
Value 0 1 2 3 4 Observed 4 21 10 13 2 Expected 8.8989 17.7979 14.8315 6.5918 1.6479
The expected frequency for value 4 is less than 3. Combine this cell with value 3:
Value 0 1 2 3-4 Observed 4 21 10 15 Expected 8.8989 17.7979 14.8315 8.2397
The degrees of freedom are k p 1 = 4 0 1 = 3
a) 1) The variable of interest is the form of the distribution for the random variable X.
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 i1 Ei 2 2 6) Reject H0 if o 0.05,3 7.81 7) 4 8.89892 15 8.23972 2 ⋯ 10.39 0 8.8989 8.2397
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.
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 i1j1 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
2) H0 : 1 2 0 or 1 2
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
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
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.
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
2 2 2 2 1 2 1 2 d) x1 x2 z/2 1 2 x1 x2 z/2 n1 n2 n1 n2
(3)2 (3)2 (3)2 (3)2 18 24 1.96 18 24 1.96 20 20 1 2 20 20
7.86 1 2 4.14
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.
9-20. a) 1) The parameter of interest is the difference in mean impact strength, 1 2
2) H0 : 1 2 0 or 1 2
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
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
s1 12 s2 22
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
2) H0 : 2 1 25
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
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.