Question

Application Exercise:
Under specified driving conditions, an automobile manufacturer hypothesizes that its new truck will get more miles per gallon (mpg) than other automobiles in its class. For automobiles of the same class, the average is 23 with a standard deviation of 4.5 mpg. To investigate, the manufacturer tested 13 of its new truck in which the average was 20.8 mpg. What can be concluded with α = 0.01?

a) What is the appropriate test statistic?
---Select--- na z-test one-sample t-test independent-samples t-test related-samples t-test

b)
Population:
---Select--- automobile manufacturer tested trucks mpg trucks in same class specified conditions
Sample:
---Select--- automobile manufacturer tested trucks mpg trucks in same class specified conditions

c) Obtain/compute the appropriate values to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value =  ; test statistic =  
Decision:  ---Select--- Reject H0 Fail to reject H0

d) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[  ,  ]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and select "na" below.
d =  ;   ---Select--- na trivial effect small effect medium effect large effect
r² =  ;   ---Select--- na trivial effect small effect medium effect large effect

f) Make an interpretation based on the results.

Under the specified conditions, the automobiles in its class get significantly less mpg than the new truck.

Under the specified conditions, the automobiles in its class get significantly more mpg than the new truck.     

Under the specified conditions, there was no signfiicant mpg difference between the new truck and other automobiles in its class.

Answer 1

a)

Z test

b)

Population:
---Select--- automobile manufacturer tested trucks mpg trucks in same class
Sample:
---Select--- automobile manufacturer tested trucks mpg trucks in specified conditions

c)

Ho :   µ =   23                  
Ha :   µ >   23       (Right tail test)          
                          
Level of Significance ,    α =    0.01                  
population std dev ,    σ =    4.5000                  
Sample Size ,   n =    13                  
Sample Mean,    x̅ =   20.8000                  
                          
'   '   '                  
                          
Standard Error , SE = σ/√n =   4.5000   / √    13   =   1.2481      
Z-test statistic= (x̅ - µ )/SE = (   20.800   -   23   ) /    1.2481   =   -1.76
                          
  
                          
p-Value   =   0.9610   [ Excel formula =NORMSDIST(z) ]              
Decision:   p-value>α, Do not reject null hypothesis

d)

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.5758   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   4.5000   / √   13   =   1.248075
margin of error, E=Z*SE =   2.5758   *   1.24808   =   3.214829
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    20.80   -   3.214829   =   17.585171
Interval Upper Limit = x̅ + E =    20.80   -   3.214829   =   24.014829
99%   confidence interval is (   17.59   < µ <   24.01   )

e)

Cohen's d=|(mean - µ )/std dev|=   -0.49 ----Medium

r² = d²/(d² + 4) =    0.056 ---- Small

f)

Under the specified conditions, there was no signfiicant mpg difference between the new truck and other automobiles in its class.