X | Y | XY | X² | Y² |
0.318 | 3.1 | 0.9858 | 0.101124 | 9.61 |
0.29 | 7.6 | 2.204 | 0.0841 | 57.76 |
0.34 | 4 | 1.36 | 0.1156 | 16 |
0.248 | 8.6 | 2.1328 | 0.061504 | 73.96 |
0.367 | 3.1 | 1.1377 | 0.134689 | 9.61 |
0.269 | 11.1 | 2.9859 | 0.072361 | 123.21 |
sample size , n = 6
here, x̅ =Σx/n = 0.3053 , ȳ =
Σy/n = 6.25
SSxx = Σx² - (Σx)²/n = 0.01
SSxy= Σxy - (Σx*Σy)/n = -0.64
SSyy = Σy²-(Σy)²/n = 55.78
.a)
X | Y | XY | X² | Y² | |
total sum | 1.8320000 | 37.5000000 | 10.8062000 | 0.5693780 | 290.1500000 |
correlation coefficient , r = Sxy/√(Sx.Sy) = -0.862
b)
correlation hypothesis test
Ho: ρ = 0
Ha: ρ ╪ 0
n= 6
alpha,α = 0.05
correlation , r= -0.8617
t-test statistic = r*√(n-2)/√(1-r²) =
-3.397
DF=n-2 = 4
p-value = 0.0274
Decison: p value < α , So, Reject
Ho
critical t-value = 2.7764
t = -3.40
critical t = 2.78
c)
std error ,Se = √(SSE/(n-2)) =
1.8946
a=estimated slope , ß1 = SSxy/SSxx = -0.644
/ 0.010 = -64.3328
b = intercept, ß0 = y̅-ß1* x̄ = 25.8930
d)
Predicted Y at X= 0.3 is
Ŷ = 25.893 +
-64.333 * 0.3
= 6.59%
e)
Confidence Level= 80%
Sample Size , n= 6
Degrees of Freedom,df=n-2 = 4
critical t Value=tα/2 = 1.533 [excel
function: =t.inv.2t(α/2,df) ]
X̅ = 0.31
Σ(x-x̅)² =Sxx 0.010007
standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =
0.780
margin of error,E=t*Std error=t* S(ŷ) =
1.5332 * 0.7800 =
1.1959
Confidence Lower Limit=Ŷ +E =
6.593 - 1.1959 =
5.40
Confidence Upper Limit=Ŷ +E = 6.593
+ 1.1959 =
7.79
f)
t = -3.40
critical t = 2.78
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