Question

Video Given are five observations for two variables, x and y x1 2 3 4 5 Y3 8 6 12 14 Round your answers to two decimal places. a. Using the following equation Estimate the standard deviation of y* when x = 4 0742 b. Using the following expression: Develop a 95% confidence interval for the expected value of y when x-4 6.03 to 10.76 c. Using the following equation: I(x*- Estimate the standard deviation of an individual value of y when x4 1.818 d. Using the following expression: Develop a 95% prediction interval for y when x 4, If your answer is negative, enter minus (-) sign. 2.62 to14.18

Answer 1

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
1	3	4	31.36	11.2
2	8	1	0.36	0.6
3	6	0	6.76	0
4	12	1	11.56	3.4
5	14	4	29.16	10.8

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	15	43	10	79.2	26
mean	3	8.6	SSxx	SSyy	SSxy

sample size ,   n =   5          
here, x̅ =   3       ȳ =   8.6  
                  
SSxx =    Σ(x-x̅)² =    10          
SSxy=   Σ(x-x̅)(y-ȳ) =   26          
                  
slope ,    ß1 = SSxy/SSxx =   2.6          
                  
intercept,   ß0 = y̅-ß1* x̄ =   0.8          
                  
so, regression line is   Ŷ =   0.8000   +   2.6000   *x

std error ,Se =    √(SSE/(n-2)) =    1.9664

----------------------

a)

Sy = s√(1/n+(X-X̅)²/Sxx) =1.966√(1/5+(4-3)²/10)=1.077033

b)

Degrees of Freedom,df=n-2 =   3
critical t Value=t(0.05/2,3)=   3.182
Predicted Y (YHat) = 11.200
  
margin of error,E=t*Sy=    3.4276
Confidence Lower Limit=Ŷ -E =   7.7724
Confidence Upper Limit=Ŷ +E =   14.6276

c)

S_pred = s√(1+1/n+(X-X̅)²/Sxx) =1.966* √(1+1/5+(4-3)²/10)=2.242023

d)

prediction interval

Degrees of Freedom,df=n-2 =   3
critical t Value=tα/2 =   3.182

margin of error,E=t*S_pred=    7.1351
Prediction Interval Lower Limit=Ŷ -E =   4.0649
Prediction Interval Upper Limit=Ŷ +E =   18.3351