b. Construct a 95% prediction interval of the tear rating for an individual bag

Question

b. Construct a 95% prediction interval of the tear rating for an individual bag of coffee when the plate gap is

-0.3

omework: Linear Regression (3) core: 0 of 15 pts 2014 (1 compete) uestion 2, 13.8.59-T Wher suvey cata indicated that a cofee company neeced to improve ts package-sealing process, an experiment was ccnducted to detemine the actors in the bag-sealing equipment that might be affecting the liner of the bag Data were collected on 19 bags and the plate gap on the bag-sealing equipment was used to predict the tear rating of a bag Camplete parts ai th Ell Clck the icon to view the data table Data Tabe a. Construct a 90% confidence nterval estimate of the mean tear rating for al bags of coffee when the plate gap is-0.3 Bag Plate gap Tear rating (Type integers or decimals Round to three decimal places asneedec. Use ascending order) 0.3 0.00 2.10 1.80 0.00 -030 0.30 -0.30 0.30 -2.10 2.10 2.40 -2.10 0.00 -3.00 - 1.80 2.10 1.80 0.00 0.47 0.86 0.37 n 33 0.74 1.98 0.28 0.16 10 .94 13 14 15 10 17 18 19 1.56 0.03 0.46 4.14 0.01

Explanation / Answer

The statistical software output for this problem is:

Simple linear regression results:
Dependent Variable: Tear Rating (Y)
Independent Variable: Plate Gap (X)
Tear Rating (Y) = 0.78840428 + 0.41613543 Plate Gap (X)
Sample size: 19
R (correlation coefficient) = 0.5429505
R-sq = 0.29479524
Estimate of error standard deviation: 1.0697358

Parameter estimates:

Analysis of variance table for regression model:

Predicted values:

Hence,

a) 95% confidence interval: (0.138, 1.189)

b) 95% prediction interval: (-1.654, 2.981)

Parameter Estimate Std. Err. Alternative DF T-Stat P-value Intercept 0.78840428 0.24546373 0 17 3.2118973 0.0051 Slope 0.41613543 0.15610169 0 17 2.665797 0.0163

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