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.

	Model Price ($)				Model Price ($)
Retail Outlet	Deluxe	Standard		Retail Outlet	Deluxe	Standard
1	40	27		5	40	30
2	39	28		6	39	32
3	43	35		7	36	29
4	38	31

1. 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.
   H ₀ =  _d Selectgreater than 10greater than or equal to 10equal to 10less than or equal to 10less than 10not equal to 10Item 1  
   H _a =  _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  
   
2. What is the 95% confidence interval for the difference between the mean prices of the two models (to 2 decimals)? Use a t-table.
   ( ,  )

Answer 1

a)

null Hypothesis:μd		is equal to	10
alternate Hypothesis: μd		is not equal to	10

  value of the test statistic =-1.11

p-value is  between .20 and .40

No

b)

for 95% CI; and 6 degree of freedom, value of t=			2.447
therefore confidence interval=sample mean -/+ t*std error
margin of errror          =t*std error=		2.202
lower confidence limit                     =		6.80
upper confidence limit                    =		11.20
from above 95% confidence interval for population mean =(6.80 , 11.20)