In: Statistics and Probability
Given are five observations for two variables, x and y. (Round your answers to two decimal places.)
xi 1 2 3 4 5
yi 3 8 6 10 15
(a) Use sŷ* = s 1 n + (x* − x)^2 / Σ(xi − x)^2 to estimate the standard deviation of ŷ* when x = 5.
(b) Use Sŷ* ± tα/2sŷ* to develop a 95% confidence interval for the expected value of y when x = 5.
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(c) Use spred = s 1 + 1 n + (x* − x)^2 / Σ(xi − x)^2 to estimate the standard deviation of an individual value of y when x=5.
(d) Use ŷ* ± tα/2spred to develop a 95% prediction interval for y when x = 5.
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ΣX | ΣY | Σ(x-x̅)² | Σ(y-ȳ)² | Σ(x-x̅)(y-ȳ) | |
total sum | 15.00 | 42.00 | 10.00 | 81.20 | 26.00 |
mean | 3.00 | 8.40 | SSxx | SSyy | SSxy |
Sample size, n = 5
here, x̅ = Σx / n= 3.000
ȳ = Σy/n = 8.400
SSxx = Σ(x-x̅)² = 10.0000
SSxy= Σ(x-x̅)(y-ȳ) = 26.0
estimated slope , ß1 = SSxy/SSxx = 26/10=
2.6000
intercept,ß0 = y̅-ß1* x̄ = 8.4- (2.6 )*3=
0.6000
Regression line is, Ŷ= 0.600 +
( 2.600 )*x
SSE= (SSxx * SSyy - SS²xy)/SSxx =
13.6000
std error ,Se = √(SSE/(n-2)) =
2.1292
a)
X̅ = 3.00
Σ(x-x̅)² =Sxx 10.00
Standard Error of the Estimate,Se= 2.1292
Predicted Y at X= 5 is
Ŷ= 0.60000 +
2.60000 *5= 13.600
standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =
1.649
b)
Confidence Level= 95%
Sample Size , n= 5
Degrees of Freedom,df=n-2 = 3
critical t Value=tα/2 = 3.182 [excel
function: =t.inv.2t(α/2,df) ]
margin of error,E=t*Std error=t* S(ŷ) =
3.1824 * 1.649 =
5.2486
Confidence Lower Limit=Ŷ +E =
13.600 - 5.249 =
8.3514
Confidence Upper Limit=Ŷ +E = 13.600
+ 5.249 =
18.8486
c) standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =
2.6932
d)
margin of error,E=t*std error=t*S(ŷ)=
3.182 * 2.693 =
8.5710
Prediction Interval Lower Limit=Ŷ -E =
13.600 - 8.571 =
5.029
Prediction Interval Upper Limit=Ŷ +E =
13.600 + 8.571 =
22.171