Question

1. A company advertises that its electric vehicles provide an efficiency that is at least 20% higher than the industry norm. A consumer interest group ran an experiment with 30 test drives and found that the average efficient improvement over industry norm is 17.8%. And the sample standard deviation in those test drives is 5%.

a) Construct a 95%-confidence interval for the percentage improvement of efficiency over the industry norm.

b) Use part (a) to test the plausibility of the company’s statement?

c) Suppose there is a competitor vehicle brand. The consumer interest group also performed tests on that brand. They did 40 test drives with that brand and found that the average efficient improvement over industry norm is 19.3% with sample standard deviation being 4.4%. Use a pooled variance approach to test the hypothesis that the two brands have the same efficiency improvement.

Answer 1

a)

sample std dev ,    s =    5
Sample Size ,   n =    30
Sample Mean,    x̅ =   17.8

Level of Significance ,    α =    0.05          
degree of freedom=   DF=n-1=   29          
't value='   tα/2=   2.0452   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   5.000   / √   30   =   0.9129
margin of error , E=t*SE =   2.0452   *   0.9129   =   1.867
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    17.8   -   1.867   =   15.933
Interval Upper Limit = x̅ + E =    17.8   -   1.867   =   19.667
95% confidence interval is (   15.93% < µ <   19.67% )  

b)

Ho :   µ =   20%
Ha :   µ <   20%

since, confidence interval do not contain null hypothesis ,20% , so reject Ho

hence, there is enough evidence to reject the claim at α=0.05

c)

Ho :   µ1 - µ2 =   0  
Ha :   µ1-µ2 ╪   0  
          
Level of Significance ,    α =    0.05  
          
mean of sample 1,    x̅1=   17.8  
standard deviation of sample 1,   s1 =    5
size of sample 1,    n1=   30  

          
mean of sample 2,    x̅2=   19.3  
standard deviation of sample 2,   s2 =    4.4   
size of sample 2,    n2=   40  

          
difference in sample means =    x̅1-x̅2 =    -1.50   
          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    4.6653  
          
std error , SE =    Sp*√(1/n1+1/n2) =    1.1268  
          
t-statistic =    ((x̅1-x̅2)-µd)/SE = -1.50/1.1268 =   -1.3312  
          
Degree of freedom, DF=   n1+n2-2 =    68  
t-critical value , t* = ±1.9955 (excel formula =t.inv(α/2,df)
Decision rule : reject Ho , if | t-stat | > |t-critical|,otherwise not          

since, | t stat | =1.3312 < critical value = 1.9955, do not reject Ho

there is not enough evidence to conclude that two brands have the same efficiency improvement at α=0.05