The following sample observations were randomly selected Do

The following sample observations were randomly selected. (Do not round the intermediate values. Round your answers to 2 decimal places.)

X:

4

5

3

6

10

Y:

10.7

5

9.9

16.3

22.5

Determine the 0.95 confidence interval for the mean predicted when
X = 8 ( , )

Determine the 0.95 prediction interval for an individual predicted when
X = 8 ( , )

Solution

For predicting y for a given x, we have to find linear regression equation between x and y.

Explanation

We will find an equation of the regression line in 4 steps.

Step 1: Find X?Y and X2 as it was done in the table below.

Step 2: Find the sum of every column:

?X=23.5 , ?Y=59.45 , ?X?Y=400.875 , ?X2=165.25

Step 3: Use the following equations to find a and b:

ab=?Y??X2??X??XYn??X2?(?X)2=59.45?165.25?23.5?400.8754?165.25?23.52?3.711=n??XY??X??Yn??X2?(?X)2=4?400.875?23.5?59.454?165.25?(23.5)2?1.898

Step 4: Substitute a and b in regression equation formula

y = a + b?x= 3.711 + 1.898?x

Use this to get y value when x =8

y = 3.711+1.898(8) =18.895

Confidence interval for y(8) = (18.895- e, 18.895+e)

For 95% we can use 1.96 sigma

Std dev of x^2 = 4(165.25)-552.25/4^2 = 6.796

std error = 1.303

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conf interval for X = (8-1.96*1.303, 8+1.96*1.303)

= (5.45,10.55)

Confidence interval for Y = {Y(5.45), Y(10.55)}