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.
H0 = d Selectgreater than 10greater than or equal to 10equal to 10less than or equal to 10less than 10not equal to 10Item 1
Ha = d Selectgreater than 10greater than or equal to 10equal to 10less than or equal to 10less than 10not equal to 10Item 2

Calculate the value of the test statistic. If required enter negative values as negative numbers. (to 2 decimals).


The p-value is Selectless than .01between .10 and .05between .05 and .10between .10 and .20between .20 and .40greater than .40Item 4

Can you conclude that the price differential is not equal to $10?
SelectYesNoItem 5

What is the 95% confidence interval for the difference between the mean prices of the two models (to 2 decimals)?
( , )

Model Price ($) Model Price ($) Retail Outlet Deluxe Standard Retail Outlet Deluxe Standard 1 39 27 5 40 30 2 39 29 6 39 35 3 46 35 7 35 29 4 38 31

Explanation / Answer

a)

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

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b)

At level of significance =    0.05          
As we can see, this is a    two   tailed test.      
The differences are

12
10
11
7
10
4
6

              
Calculating the standard deviation of the differences (third column):              
              
s =    4.917501213          
              
Thus, the standard error of the difference is sD = s/sqrt(n):              
              
sD =    1.858640755          
              
Calculating the mean of the differences (third column):              
              
XD =    8.571428571          
              
As t = [XD - uD]/sD, where uD = the hypothesized difference =    10   , then      
              
t =    -0.768610838   [ANSWER]      

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c)

              
As df = n - 1 =    6          
              
Then the P value is, using table,

P > 0.40 [ANSWER]              

[Actual P value using technology is p = 0.471284793.]

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d)

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

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e)

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 =    4.023498482          
upper bound = [X1 - X2] + t(alpha/2) * sD =    13.11935866          
              
Thus, the confidence interval is              
              
(   4.023498482   ,   13.11935866   ) [ANSWER]

      

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