Question

1. Given are five observations for two variables, x and y. x i 1 2 3 4 5 y i 3 7 5 12 13 Round your answers to two decimal places. a. Using the following equation: Estimate the standard deviation of ŷ* when x = 3. b. Using the following expression: Develop a 95% confidence interval for the expected value of y when x = 3. to c. Using the following equation: Estimate the standard deviation of an individual value of y when x = 3. d. Using the following expression: Develop a 95% prediction interval for y when x = 3. If your answer is negative, enter minus (-) sign. to

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2.

he following data are the monthly salaries y and the grade point averages x for students who obtained a bachelor's degree in business administration. GPA Monthly Salary ($) 2.7 3,500 3.4 3,900 3.7 4,200 3.2 3,800 3.5 4,200 2.7 2,300 The estimated regression equation for these data is  = -859.1 + 1,409.1x and MSE =196,932. a. Develop a point estimate of the starting salary for a student with a GPA of 3.0 (to 1 decimal). $ b. Develop a 95% confidence interval for the mean starting salary for all students with a 3.0 GPA (to 2 decimals). $ (  ,  ) c. Develop a 95% prediction interval for Ryan Dailey, a student with a GPA of 3.0 (to 2 decimals). $ (  ,  )

Answer 1

1)

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	15.00	40.00	10.00	76.00	25.00
mean	3.00	8.00	SSxx	SSyy	SSxy

sample size ,   n =   5          
here, x̅ = Σx / n=   3.000   ,     ȳ = Σy/n =   8.000  
                  
SSxx =    Σ(x-x̅)² =    10.0000          
SSxy=   Σ(x-x̅)(y-ȳ) =   25.0          
                  
estimated slope , ß1 = SSxy/SSxx =   25.0   /   10.000   =   2.5000
                  
intercept,   ß0 = y̅-ß1* x̄ =   0.50000          
                  
so, regression line is   Ŷ =   0.50   +   2.500   *x

a)

Predicted Y at X=   3   is          
Ŷ =   0.5000   +   2.5000   *3=   8.000

b)

X Value=   3              
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) ]          
                  
X̅ =    3.00              
Σ(x-x̅)² =Sxx   10.00              
Standard Error of the Estimate,Se=   2.1213              
                  
Predicted Y at X=   3   is          
Ŷ =   0.5000   +   2.5000   *3=   8.000
                  
standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    0.949              
margin of error,E=t*Std error=t* S(ŷ) =   3.1824   *   0.949   =   3.0191
                  
Confidence Lower Limit=Ŷ +E =    8.000   -   3.019   =   4.981
Confidence Upper Limit=Ŷ +E =   8.000   +   3.019   =   11.019

c)

For Individual Response Y  
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   2.324

d)

margin of error,E=t*std error=t*S(ŷ)=    3.182   *   2.324   =   7.3953
                  
Prediction Interval Lower Limit=Ŷ -E =   8.000   -   7.395   =   0.605
Prediction Interval Upper Limit=Ŷ +E =   8.000   +   7.395   =   15.395

=================

2)

a)

Predicted Y at X=   3   is          
Ŷ =   -859.0909   +   1409.0909   *3=   3368.2

b)

standard error, S(ŷ)=Se*√(1/n+(X-X̅)²/Sxx) =    204.385              
margin of error,E=t*Std error=t* S(ŷ) =   2.7764   *   204.385   =   567.4650
                  
Confidence Lower Limit=Ŷ +E =    3368.182   -   567.465   =   2800.72
Confidence Upper Limit=Ŷ +E =   3368.182   +   567.465   =   3935.65

c)

For Individual Response Y                  
standard error, S(ŷ)=Se*√(1+1/n+(X-X̅)²/Sxx) =   488.575              
margin of error,E=t*std error=t*S(ŷ)=    2.776   *   488.575   =   1356.5008
                  
Prediction Interval Lower Limit=Ŷ -E =   3368.182   -   1356.501   =   2011.68
Prediction Interval Upper Limit=Ŷ +E =   3368.182   +   1356.501   =   4724.68