Cascade Manufacturing has just received an order from their largest customer for

Question

Cascade Manufacturing has just received an order from their largest customer for 80 motors.

a. Use the regression model from part 1 to provide a 90% confidence interval for the average cost of an order of 80 motors.

b. Again, using the regression model from part 1, produce a 90% prediction interval for the cost of this particular order.

I have no clue where to begin here. The regression model from part 1 just refers to a regression model made in excel. Please go step by step and use excel wherever possible. Thank you!

$        5,964

Order Size Total Cost 56 $        7,531 54 $        6,329 68 $        8,413 60 $        7,793 38 $        5,360 42 $        4,838 22 $        2,551 34 $        3,899 66 $        8,326 46 $        5,465 14 $        2,283 46 $        5,413 36 $        4,238 52 $        6,911 40 $        6,315 58 $        8,243 20 $        2,866 44 $        6,775 28 $        4,289 12 $        1,475 30 $        3,590 70 $        9,439 46 $        6,760 36 $        5,170 62 $        7,780 42 $        4,896 70 $        8,816 58 $        8,116 42 $        6,212 48 $        5,551 54 $        7,080 72 $        9,826 48 $        6,129 40 $        5,094 26 $        3,568 28 $        3,738 38 $        5,332 28 $        3,286 56 $        6,664 48 $        5,990 54 $        7,093 32 $        3,975 72 $        9,046 40 $        4,906 38 $        5,324 14 $        2,734 64 $        8,138 42 $        5,376 58 $        7,763 46

$        5,964

Explanation / Answer

Output from the excel linear regression:

We can find the model's standard error =671.546

=> standard error= standard deviation/sqrt(n)

=>standard deviation = standard error*sqrt(n)

=>SD = 671.546 * sqrt(60) = 5201.77

Now we also have the mean of y = 5801.18

90% CI for average cost of this order = (5801.18 -1.645*5201.77, 5801.18 +1.645*5201.77) = (-2755.73,14358)

For 80 motors, the predicted value of y = 10134.648501563

So 90% confidence interval = (10134.648501563 - 1.645*5201.77 ,10134.648501563 + 1.645*5201.77 ) = (1577.73,18691.56)

SUMMARY OUTPUT Regression Statistics Multiple R 0.945003956 R Square 0.893032478 Adjusted R Square 0.890803988 Standard Error 671.5460064 Observations 50 ANOVA df SS MS F Significance F Regression 1 180720781.5 180720781.5 400.734335 6.09026E-25 Residual 48 21646753.86 450974.0387 Total 49 202367535.4 Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 297.0355582 290.894665 1.021110367 0.312321219 -287.8473661 881.9184825 -287.8473661 881.9184825 Order Size 122.9701618 6.142872018 20.01834996 6.09026E-25 110.6190898 135.3212338 110.6190898 135.3212338

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