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The manufacturer of an MP3 player wanted to know whether a 10 percent reduction

ID: 3151322 • Letter: T

Question

The manufacturer of an MP3 player wanted to know whether a 10 percent 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 sampled outlets. Regular price 137 123 88 118 141 126 96 Reduced price 128 137 152 137 117 101 113 118 At the .010 significance level, can the manufacturer conclude that the price reduction resulted in an increase in sales? Hint: For the calculations, assume the "Reduced price" as the first sample. Regular price 137 123 88 118 141 126 96 Reduced price 128 137 152 137 117 101 113 118 At the .010 significance level, can the manufacturer conclude that the price reduction resulted in an increase in sales? Hint: For the calculations, assume the "Reduced price" as the first sample.


The pooled variance is . (Round your answer to 2 decimal places.)

Explanation / Answer

Formulating the null and alternative hypotheses,              
              
Ho:   u1 - u2   <=   0  
Ha:   u1 - u2   >   0  
At level of significance =    0.01          
As we can see, this is a    right   tailed test.      
Calculating the means of each group,              
              
X1 =    125.375          
X2 =    118.4285714          
              
Calculating the standard deviations of each group,              
              
s1 =    16.2914307          
s2 =    19.83983487          
              
Thus, the pooled standard deviation is given by              
              
S = sqrt[((n1 - 1)s1^2 + (n2 - 1)(s2^2))/(n1 + n2 - 2)]               
              
As n1 =    8   , n2 =    7  
              
Then              
              
S =    18.01620912

So pooled variance is

S^2 =    18.01620912^2 = 324.5837911 [ANSWER, POOLED VARIANCE]

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Thus, the standard error of the difference is              
              
Sd = S sqrt (1/n1 + 1/n2) =    9.32427407          
              
As ud = the hypothesized difference between means =    0   , then      
              
t = [X1 - X2 - ud]/Sd =    0.744983311   [ANSWER, TEST STATISTIC]

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Getting the critical value using table/technology,              
df = n1 + n2 - 2 =    13          
tcrit =    +   2.650308838      
              
As t < 2.65, we   FAIL TO REJECT THE NULL HYPOTHESIS.