Section 8.9 (p.511)

 

1.         We want to test H0: mA ³ mB vs. H1: mA < mB, and reject H0 if U £ c where U is defined in Equation (5) (note the difference between this hypothesis and the one used in Equation (1)). Here we have c = -1.356 (t with 12 df, a = 0.10) and U = -1.692, therefore reject H0.

 

2.         Pr (U < c1) = Pr(U > c2) = 0.05  Þ  c1=-1.782 and c2=1.782 (t with 12 df). Since U = -1.692, H0 is not rejected.

 

3.         The random variable  has a normal distribution with mean 0 and variance (s12/m)+(ks12/n). Therefore, has a standard normal distribution. Also, SX2/s12 has a c2 distribution with m-1 df and Sy2/(ks12) has a c2 distribution with n-1 df. Since these two random variables are independent, we have U2 = (1/s12)(SX2+SY2/k) has a c2 distribution with m+n-2 df. From Section 7.3 we know that U1 and U2 are independent, therefore U defined by Equation (10) has the form U = U1/U2 and it has a t distribution with m+n-2 df.

 

4.         According to Exercise 10, U has a t distribution with m+n-2 df. Therefore we reject H0 if U < -1.356 (df=12, a=0.1). Data give U = -1.672, so we reject H0.

 

5.         Similarly to Exercise 3, we can show that the following statistic U will have a t distribution with m+n-2 df when H0 is true:

H0 should be rejected if U < c1 or U > c2.

 

6.         According the results of Exercise 5 and its hypothesis, we would accept H0 if c1<U<c2. Since U has a t distribution with 12 df, we have c1=-1.782 and c2=1.782. In other words,

which implies that –0.320 < m1-m2 (=l) < 0.008