Question

2. The article "Uncertainty Estimation in Railway Track Life-Cycle Cost"+ presented the following data on time to repair (min) a rail break in the high rail on a curved track of a certain railway line. 159 120 480 149 270 547 340 43 228 202 240 218 A normal probability plot of the data shows a reasonably linear pattern, so it is plausible that the population distribution of repair time is at least approximately normal. The sample mean and standard deviation are 249.7 and 145.1, respectively. (a) Is there compelling evidence for concluding that true average repair time exceeds 200 min? Carry out a test of hypotheses using a significance level of 0.05. State the appropriate hypotheses. Ho: # = 200 Haix + 200 HoiH - 200 Hou < 200 O HOM = 200 H: > 200 O Ho: > 200 Ho: u = 200 Hou < 200 H: - 200 Calculate the test statistic and determine the p-value. (Round your test statistic to two decimal places and your p-value to four decimal places.) P-value = What can you conclude? There is compelling evidence that the true average repair time exceeds 200 min. There is not compelling evidence that the true average repair time exceeds 200 min. (b) Using = 150, what is the type II error probability of the test used in (a) when true average repair time is actually 300 min? That is, what is (300)? (Round your answer to two decimal pla B(300) =

Answer 1

Ho :   µ =   200  
Ha :   µ >   200   (Right tail test)
          
Level of Significance ,    α =    0.050  
sample std dev ,    s =    145.1000  
Sample Size ,   n =    12  
Sample Mean,    x̅ =   249.7000  
          
degree of freedom=   DF=n-1=   11  
          
Standard Error , SE = s/√n =   145.1/√12=   41.8868  
t-test statistic= (x̅ - µ )/SE =    (249.7-200)/41.8868=   1.19  
          

          
p-Value   =   0.1302   [Excel formula =t.dist(t-stat,df) ]
Decision:   p-value>α, Do not reject null hypothesis       

There is not compelling evidence that the true average repair time exceeds 200 min.

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B

true mean ,    µ =    300
      
hypothesis mean,   µo =    200
significance level,   α =    0.05
sample size,   n =   12
std dev,   σ =    150.0000
      
δ=   µ - µo =    100
      
std error of mean=σx = σ/√n =    150/√12=   43.3013

Zα =   1.6449   (right tailed test)
      
We will fail to reject the null (commit a Type II error) if we get a Z statistic <   1.645  
this Z-critical value corresponds to X critical value( X critical), such that      
      
(x̄ - µo)/σx ≤ Zα      
x̄ ≤ Zα*σx + µo      
x̄ ≤    1.6449*43.3013+200  
x̄ ≤    271.2243   (acceptance region)
      
now, type II error is ,ß =    P( x̄ ≤    271.2243 given that µ is 300
      
   = P ( Z < (x̄-true mean)/σx )  
=P( Z < (   271.2243-300)/43.3013)  
      
= P ( Z <    -0.665   )
ß =    0.2532

ß = 0.25

..............

Please let me know in case of any doubt.

Thanks in advance!

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