Question

When survey data indicated that a coffee company needed to improve its​ package-sealing process, an experiment was conducted to determine the factors in the​ bag-sealing equipment that might be affecting the ease of opening the bags without tearing the inner 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. Complete parts​ (a) through ​(c).

Bag	Plate gap (X)	Tear rating (Y)
1	0.3	0.06
2	-0.3	0.04
3	1.8	0.42
4	1.8	0.82
5	-0.3	0.33
6	-0.3	0.36
7	-0.3	0.78
8	-0.3	1.91
9	0	0.23
10	-1.8	0.11
11	-1.5	0.14
12	2.7	3.96
13	-1.5	0.09
14	-0.3	0.57
15	-2.7	0.06
16	-2.1	0.13
17	1.8	0.48
18	2.1	4.13
19	-0.3	0.03

a. Construct a 90% confidence interval estimate of the mean tear rating for all bags of coffee when the plate gap is 0.3. OsHy|x=0.3 (Type integers or decimals. Round to three decimal places as needed. Use ascending order.) b. Construct a 90% prediction interval of the tear rating for an individual bag of coffee when the plate gap is 0.3. sYx=0.3 (Type integers or decimals. Round to three decimal places as needed. Use ascending order.) c. Why is the interval in (a) narrower than the interval in (b)? O A. Since the single predicted observation is larger than the mean, the confidence interval is narrower than the prediction interval. O B. Since there is more variation in predicting an individual value than in estimating a mean value, the confidence interval is narrower than the prediction interval. O C. Since there is more variation in estimating a mean value than in predicting an individual value, the confidence interval is narrower than the prediction interval. OD. Since the mean is smaller than the single predicted observation, the confidence interval is narrower than the prediction interval.

Answer 1

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) = 0.80334399 +0.51127977 Plate gap (X) Sample size: 19 R (correlation coefficient) = 0.62855991 R-sq = 0.39508756 Estimate of error standard deviation: 0.98850712 Parameter estimates: Parameter Estimate Std. Err. DF 95% L. Limit 95% U. Limit Intercept 0.80334399 0.22698605 17 0.32444528 1.2822427 Slope 0.51127977 0.15343828 17 0.18755329 0.83500624 Analysis of variance table for regression model: Source DF SS MS F-stat P-value Model 1 10.849491 10.849491 11.103241 0.0039 Error 17 16.611488 0.97714634 Total 18 27.460979 Predicted values: X value Pred. Y 0.3 0.95672792 s.e.(Pred. y) 90% C.I. for mean 90% P.I. for new 0.23352459 (0.55048698, 1.3629689) (-0.81021939, 2.7236752)

Hence,

a) 90% confidence interval:

(0.550, 1.363)

b) 90% prediction interval:

(-0.810, 2.724)

c) Option B is correct.