Section 8.6 (p.491)

 

1.         U = [3 (22-20)] / (72/8)1/2 = 2

(a) Reject H0 if U ³ 1.860 (df 8, a0=0.05)  Þ  Reject H0.

(b) Reject H0 if U £ -2.306 or U ³ 2.306  Þ  Accept H0.

(c) –2.306 £ U £ 2.306  Þ  -2.306 £ [3 (22-m)] / (72/8)1/2  £ 2.306  Þ  19.694 £ m £ 24.306.

 

2.         Assume: The miles an automobile will travel per gallon of gasoline are independent and identically distributed, and that each has a normal distribution. Test the hypothesis: H0: m ³ 20 vs. H1: m < 20. Reject H0 if U £ -1.86. The data gives U = -1.809, therefore do not reject the claim (H0).

 

3.         Reject H0 if U £ -1.895 or U ³ 1.895. Here we have =-11.2/8 = -1.4 and Sn2 = 43.7 – 8(1.4)2 = 28.02, which implies U = -1.979. Therefore we reject H0.

 

6.         Let T = (X-m)/s and it has a standard normal distribution. Also let V = S Yi2/s2 and it has a c2 distribution with n degrees of freedom. Note that T and V are independent. Therefore when m=m0, let

Then U has a t distribution with n degrees of freedom, and we reject H0 when U ³ c.

 

7.         Given p(m, s2|d) = a0 when s2 = s02, we have Pr(Sn2/s02 ³ c) = a0 and Sn2/s02 has a c2 distribution with n-1 degrees of freedom. Therefore c can be obtained from the c2 table. In case of s2 < s02, Sn2/s2 has a c2 distribution with n-1 degrees of freedom. And p(m, s2|d) = Pr(Sn2/s02 ³ c) = Pr(Sn2/s2 ³ c s02/s2) < Pr(Sn2/s2 ³ c) = a0. Similarly we can show that p(m, s2|d) > a0 if s2 > s02.

 

8.         We reject H0 if Sn2/4 ³ 16.92 (c2 with 9 df, a0=0.05). Here we have Sn2/4 = 15, therefore we do not reject H0.

 

9.         When null hypothesis is correct (s2=4), Sn2/4 has a c2 distribution with 9 df. From c2 table we have Pr (Sn2/4 £ 2.700) = Pr (Sn2/4 ³ 19.02) = 0.025. Therefore c1=4(2.700) = 10.8 and c2 = 4(19.02) = 76.08.