Question

Chapter 10 Section 1 Additional Problem 1 Consider the hypothesis test Ho: Mi = u2 against H:Mi < u2 with known variances ay = 10 and 62 = 5. Suppose that sample sizes nj = 10 and 12 = 15 and that j = 14.2 and I2 = 19.7. Use a = 0.05. (a) Test the hypothesis and find the P-value. (b) What is the power of the test in part (a) if ili is 4 units less than 12? (c) Assuming equal sample sizes, what sample size should be used to obtain B = 0.05 if H1 is 4 units less than 112? Assume that a = 0.05. (a) The null hypothesis rejected. The P-value is Round your answer to four decimal places (e.g. 98.7654). (b) The power is Round your answer to two decimal places (e.g. 98.76). (c)n = n2 = Round your answer up to the nearest integer.

Answer 1

a)

Ho :   µ1 - µ2 =   0  
Ha :   µ1-µ2 <   0  
          
Level of Significance ,    α =    0.05  
          
sample #1   ------->      
mean of sample 1,    x̅1=   14.2000  
population std dev of sample 1,   σ1 =    10  
size of sample 1,    n1=   10  
          
sample #2   --------->      
mean of sample 2,    x̅2=   19.7000  
population std dev of sample 2,   σ2 =    5  
size of sample 2,    n2=   15  
          
difference in sample means = x̅1 - x̅2 =    14.2-19.7=   -5.5  
std error , SE =    √(σ1²/n1+σ2²/n2) =    3.4157  
Z-statistic = ((x̅1 - x̅2)-µd)/SE =    (-5.5-0)/3.4157=   -1.610  
          
Z-critical value , Z* =        -1.6449   [excel function =NORMSINV(α)]
          
p-value =        0.0537   [excel function =NORMSDIST(z)]

Null hypotheiss is rejected, p value = 0.0537

b)

true mean ,    µ =    -4
      
hypothesis mean,   µo =    0
significance level,   α =    0.05
      
δ=   µ - µo =    -4
      
std error of mean=σx = 3.4157

Zα =   -1.6449   (left tail test)
P(type II error) , ß =   P(Z > Zα - δ/σx)  
= P(Z >    -1.6449-(-4)/3.4157)  
=P(Z>    -0.474   ) =
type II error, ß =   0.6822  
power =    1 - ß =   0.3178

c)

true mean= µ =   -4  
hypothesized mean=µo =    0  
α=   0.05  
σ = √(σ1²+σ2²) = 11.1803  
power= 1- ß =   0.95  
ß =    0.05  
δ=µ - µo =    4  
Zα=   1.6449  
Z (ß ) =    1.6449  
n = ( ( Z(ß)+Z(α/2) )*σ / δ )² =    ((1.645+1.645)*11.1803/4)^2=   84.55
so, sample size=   85