Question

Need helping figuring out part B of the problem.

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 39 27 5 40 30 2 39 29 6 39 35 الع 46 35 7 35 29 38 31 a. 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. Ho= ud equal to 10 ✓ Ha = uld not equal to 10 Calculate the value of the test statistic. If required enter negative values as negative numbers. (to 2 decimals). -1.29 The p-value is between .20 and .40 Can you conclude that the price differential is not equal to $10? No b. What is the 95% confidence interval for the difference between the mean prices of the two models (to 2 decimals)? Use a t-table. 11.28

Answer 1

S. No Deluxe Standard diff:(d)=x1-x2 1 39 27 12 144.00 2 39 29 10 100.00 3 46 35 11 121.00 4 38 31 7 49.00 5 40 30 10 100.00 6 39 35 4 16.00 7 29 36.00 2d2=566 35 total mean dbar degree of freedom -n-1 Std deviaiton S=v(Ed?-(Ed)*/n)/(n-1) std error=Se=S/Vn- 6 Ed=60 8.5714 6 8 2.935821 1.1096

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=		2.7153
lower confidence limit                     =		5.86
upper confidence limit                    =		11.29

from above 95% confidence interval =(5.86 ,11.29)

(please try 5.85 ,11.28 if this comes wrong)