A manufacturer produces both a deluxe and a standard model of an automatic sande

Question

A manufacturer produces both a deluxe and a standard model of an automatic sander designed for home use. Selling prices obtained from a sample of retail outlets follow. The manufacturer's suggested retail prices for the two models show a $10 price differential. Use a.05 level of significance and test that the mean difference between the prices of the two models is $10. Develop the null and alternative hypotheses. Calculate the value of the test statistic. If required enter negative values as negative numbers, (to 2 decimals). The p-value is Can you conclude that the price differential is not equal to $10? What is the 95% confidence interval for the difference between the mean prices of the two models (to 2 decimals)?

Explanation / Answer

a)
Formulating the null and alternative hypotheses,              
              
Ho:   ud   =   10  
Ha:   ud   =/   10   [ANSWER]

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The differences are

13
11
8
7
10
7
7

At level of significance =    0.05          
As we can see, this is a    two   tailed test.      
              
Calculating the standard deviation of the differences (third column):              
              
s =    4.956630087          
              
Thus, the standard error of the difference is sD = s/sqrt(n):              
              
sD =    1.873430079          
              
Calculating the mean of the differences (third column):              
              
XD =    9          
              
As t = [XD - uD]/sD, where uD = the hypothesized difference =    10   , then      
              
t =    -0.533780263   [ANSWER, TEST STATISTIC]

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As df = n - 1 =    6          
              
              
Also, using p values, as this is two tailed,              
              
p =        0.612680031   [ANSWER]

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As P > 0.05, NO, WE CANNOT CONCLUDE THAT THE DIFFERENTIAL IS NOT $10.   [ANSWER]

**********************************

b)

For the   0.95   confidence level,      
              
alpha/2 = (1 - confidence level)/2 =    0.025          
t(alpha/2) =    2.446911851          
              
lower bound = [X1 - X2] - t(alpha/2) * sD =    4.415881738          
upper bound = [X1 - X2] + t(alpha/2) * sD =    13.58411826          
              
Thus, the confidence interval is              
              
(   4.415881738   ,   13.58411826   ) [ANSWER]

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