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
   H _a =  _d  
   
   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?
   
   
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

using minitab>stat>basic stat>paired t

we have

Paired T-Test and CI: Deluxe, Standard

Paired T for Deluxe - Standard

N Mean StDev SE Mean
Deluxe 7 39.29 2.14 0.81
Standard 7 30.29 2.69 1.02
Difference 7 9.000 2.380 0.900

95% CI for mean difference: (6.798, 11.202)
T-Test of mean difference = 10 (vs ≠ 10): T-Value = -1.11 P-Value = 0.309

1 ) the null and alternative hypotheses is
H 0 = ud = 10
H a = ud   10 (two tailed )

the value of the test statistic t =-1.11
The p-value is 0.309

since p value is greater than 0.05 so we cannot conclude that  the price differential is not equal to $10

2)the 95% confidence interval for the difference between the mean prices of the two models is (6.80, 11.20)