Question

a. What are the confidence limits (α=0.05) for the difference between the true mean value of Y when X₁ = 3 and the true mean value of Y when X₂ = -2?

Fe (X) Loss in MDD (Y)

 1  0.01 127.6
 2  0.48 124.0
 3  0.71 110.8
 4  0.95 103.9
 5  1.19 101.5
 6  0.01 130.1
 7  0.48 122.0
 8  1.44  92.3
 9  0.71 113.1
10  1.96  83.7
11  0.01 128.0
12  1.44  91.4
13  1.96  86.2

Thirteen specimens of 90/10 Cu-Ni alloys, each with a specific iron were tested in a corrosion-wheel setup. The wheel was rotated in salt seawater at 30 ft/s for 60 days. The corrosion was measured in weight loss in milligram/square decimeter/day, MDD. The following data were collected. Show all work.

Answer 1

a) The 95% Confidence interim for the contrast between evident mean estimations of Y for X1 = 3 and X2 = - 2 can be effectively determined by the recipe:

where Y_{1 &} Y₂ are the anticipated mean qualities of Y for X1 = 3 & X2 = - 2 respectively. here i have done some calculations you can find the values by substituting in above formula

Sample Size , n=   13, t critical value=   t α/2 =    2.20098516 ,SSE=   (Sx*Sy - S²xy)/Sx =    102.85,

Se =    √(SSE/(n-2)) =    3.0578,

X Value=   3
Confidence Level=   95%
  
Sample Size , n=   13
Degrees of Freedom,df=n-2 =   11
critical t Value=tα/2 =   2.201
X̅ =Σx/n = 0.873
Σ(x-x̅)² =Sxx   5.709
Standard Error of the Estimate,Se=   3.058
h Statistic = (1/n+(X-X̅)^2/Sxx) =    0.869
Predicted Y (YHat) = Ŷ=   129.7866 -24.0199   *3 = 57.727