Suppose an All Greens store in Sonoma, California, wants to estimate a range of advertising costs appropriate to its store. If it spends too little on advertising, it will not reach enough customers. However, it does not want to overspend on advertising for this type and size of store. At this store, x1 = 163, x2 = 2.4, x3 = 188, x5 = 6.6, and x6 = 10. Use these data to predict x4 (advertising costs) and find an 80% confidence interval for your prediction. (Use 2 decimal places.)

The advertising cost, X₄ = 5.626 million

The 80% confidence limits for X₄ is (5.041 , 6.100)

The 80% prediction limits for X₄ is (4.048 , 7.094)

Step-by-step explanation:

Using MINITAB

The regression equation is X₄ = 4.14 + 0.0431 X₁ - 0.800 X₂ + 0.00059 X₃ - 0.661 X₅ + 0.057 X₆

Predictor Coef SE Coef T P

Constant 4.142 1.626 2.55 0.019

X₁ 0.043089 0.009466 4.55 0.000

X₂ -0.7998 0.2515 -3.18 0.005

X₃ 0.000590 0.004221 0.14 0.890

X₅ -0.6606 0.1542 -4.28 0.000

X₆ 0.0574 0.1254 0.46 0.652

S = 1.07911 R-Sq = 93.4% R-Sq(adj) = 91.8%

Analysis of Variance

Source DF SS MS F P

Regression 5 345.966 69.193 59.42 0.000

Residual Error 21 24.454 1.164

Total 26 370.420

Source DF Seq SS

X₁ 1 309.464

X₂ 1 8.699

X₃ 1 5.994

X₅ 1 21.566

X₆ 1 0.244

Unusual Observation

Obs X₁ X₄ Fit SE Fit Residual St Resid

17 398 5.500 7.714 0.641 -2.214 -2.55 R

27 400 7.000 7.366 1.025 -0.336 -1.00 X

Where R is observation with a large standardized residual.

Where X is observation whose X values give it large influence.

Predicted values for new Observations

New

Obs Fit SE Fit 80% Cl 80% Pl

1 5.571 0.400 (5.041 , 6.100) (4.048 , 7.094)

Values of Predictors for New Observations

New

Obs X₁ X₂ X₃ X₅ X₆

1 163 2.40 188 6.60 10.0

∴ The advertising cost, X₄ = 5.626 million, The 80% confidence limits for X₄ is (5.041 , 6.100), and The 80% prediction limits for X₄ is (4.048 , 7.094)

its a trick question

step-by-step explanation:

dogs love water, just turn on the hose and it will run in circles chasing the water all day : )