Question

The manufacturer of an MP3 player wanted to know whether a 10% reduction in price is enough to increase the sales of its product. To investigate, the owner randomly selected eight outlets and sold the MP3 player at the reduced price. At seven randomly selected outlets, the MP3 player was sold at the regular price. Reported below is the number of units sold last month at the regular and reduced prices at the randomly selected outlets. Regular price 138 124 89 112 116 123 98 Reduced price 124 134 154 135 118 126 133 132 Click here for the Excel Data File . At the 0.025 significance level, can the manufacturer conclude that the price reduction resulted in an increase in sales? Hint: For the calculations, assume reduced price as the first sample.

Compute the pooled estimate of the variance. (Round your answer to 3 decimal places.)

Compute the test statistic. (Round your answer to 2 decimal places.)

State your decision about the null hypothesis. Reject H0 Fail to reject H0

Answer 1

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 >   0                  
                          
Level of Significance ,    α =    0.025                  
                          
Sample #1   ---->   Reduced price
mean of sample 1,    x̅1=   132.00                  
standard deviation of sample 1,   s1 =    10.65                  
size of sample 1,    n1=   8                  
                          
Sample #2   ---->   Regular price
mean of sample 2,    x̅2=   114.29                  
standard deviation of sample 2,   s2 =    16.56                  
size of sample 2,    n2=   7                  
                          
difference in sample means =    x̅1-x̅2 =    132.0000   -   114.3   =   17.71  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    13.6985      

pooled variance = 187.649

           
std error , SE =    Sp*√(1/n1+1/n2) =    7.0896                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   17.7143   -   0   ) /    7.09   =   2.50
                          
Degree of freedom, DF=   n1+n2-2 =    13                  

  
p-value =        0.013329   [excel function: =T.DIST.RT(t stat,df) ]              
Conclusion:     p-value <α , Reject null hypothesis  

Please revert in case of any doubt.

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