Question

Allstate Batteries advertises and sells a 4 year (48 months) battery. You work at Consumer Digest, a consumer advocacy organization, and have been receiving complaints from consumers who felt they were not receiving their full 48 months. Test Allstate’s claim that these are 48 month batteries at the 5% level of significance. You have tested a sample of 36 cars and found a mean of 45months and a standard deviation of 1 month. State the appropriate hypotheses from the consumers’ point of view, show drawing and appropriate critical value, compute the appropriate test statistic, accept(fail to reject)/reject the null hypothesis and draw conclusion. Do you believe the batteries are as advertised? Construct a 99% confidence interval for the unknown mean.

H :

H :

Answer 1

Ho :   µ =   48
Ha :   µ ╪   48
Level of Significance ,    α =    0.05          
sample std dev ,    s =    1.0
Sample Size ,   n =    36          
Sample Mean,    x̅ =   45.0   
                  
degree of freedom=   DF=n-1=   35          
                  
Standard Error , SE = s/√n =   1.0 / √    36   =   0.1667
t-test statistic= (x̅ - µ )/SE = (   45.0 -   48   ) /    0.167 = -18
                  
critical t value, t* =    ±   2.0301   [Excel formula =t.inv(α/no. of tails,df) ]      
                  

Decision: Test stat<-2.03, Reject null hypothesis               
conclusion: there is enough evidence to conclude that batteries are not as advertise.

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Level of Significance ,    α =    0.01          
degree of freedom=   DF=n-1=   35          
't value='   tα/2=   2.7238   [Excel formula =t.inv(α/2,df) ]      
                  
Standard Error , SE = s/√n =   1.000   / √   36   =   0.1667
margin of error , E=t*SE =   2.7238   *   0.167   =   0.454
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    45.00   -   0.454   =   44.546
Interval Upper Limit = x̅ + E =    45.00   -   0.454   =   45.454
99%   confidence interval is (   44.55   < µ <   45.5   )