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).

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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). Click the icon to view the data table. a. Construct a 90% confidence interval estimate of the mean tear rating for all bags of coffee when the plate gap is 0.3. SMYIX = 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)? ооо A. 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. 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. C. Since the single predicted observation is larger than the mean, the confidence interval is narrower than the prediction interval. D. Since the mean is smaller than the single predicted observation, the confidence interval is narrower than the prediction interval.

Bag	Plate gap (X)	Tear rating (Y)
1	0.3	0.07
2	0.3	0.06
3	1.5	0.46
4	2.1	0.89
5	0	0.37
6	0.3	0.34
7	-0.3	0.78
8	-0.3	1.91
9	0.3	0.26
10	-1.8	0.18
11	-2.1	0.14
12	3	3.81
13	-1.5	0.09
14	0.3	0.58
15	-3	0.01
16	-1.5	0.05
17	1.8	0.48
18	1.8	4.21
19	0.3	0.04

Answer 1

a) From the given data

S.NO X^2 Y^2 XY Plate gap (X) 0.3 Tear rating (Y) 0.07 1 0.09 0 0.021 2 0.3 0.09 0 0.018 0.06 0.46 3 1.5 2.25 0 0.69 1.869 4 2.1 0.89 4.41 1 0 0.37 0 0 0 5 6 0.09 0 0.102 -0.234 7 0.34 0.78 1.91 0.09 0.3 -0.3 -0.3 0.3 1 8 0.09 4 9 0.26 0.09 0 -0.573 0.078 -0.324 -0.294 10 -1.8 0.18 3.24 0 11 -2 0.14 4.41 0 12 3 9 15 3.81 0.09 13 -1.5 0 2.25 0.09 11.43 -0.135 0.174 14 0.3 0.58 0 15 -3 0.01 9 0 16 -1.5 0.05 2.25 0 -0.03 -0.075 0.864 17 1.8 0.48 0 3.24 3.24 18 1.8 4.21 18 7.578 19 0.3 0.04 0.09 0 0.012 Total: 1.5 14.73 44.01 38 21.17

The sum of squares obtained from the table above are: Τη 12 1 SSxx ΣΧ? -, Χ; i=1 =1 44.01 1 (1.5)2 19 43.8916 2 n SSYY ΣΥ? Υ; n 1=1 =1 = 38.4601 1 (14.73)? 19 27.0405 η SSXY ΣΧΥ; 8 - (ΣΑ)(Σ Υ; η 1=1 21.171 1 (1.5) (14.73) 19 20.0081

The slope and y-intercept coefficients are computed using the following formulas: B1 SSXY SSxx 20.0081 43.8916 0.4559 Bo = 7 – 81 x Ý 0.7753 – 0.4559 x 0.0789 0.7393 Therefore, the regression equation is: û = 0.7393 +0.4559 X Now that we have the regression equation, we can compute the predicted value for X = 0.3, by simply plugging in the value of X = 0.3 in the regression equation found above: Ý 0.7393 +0.4559 x 0.3 0.876

Now, we are going to compute the 90% confidence interval for the mean response value Y = 0.876. The first step consists of finding the standard error of the estimate, for which we need to deal with the regression sum of squares and with the sum of squared errors. We know that the total sum of squares is: S STotal = S Syy = 27.0405 Also, the regression sum of squares is computed as follows: SSRegression BSSXY 0.4559 x 20.0081 9.1208 Since we know that SSTotal = SSRegression + SSError, then we can compute the sum of squared errors as follows: SSError = SSTotal - SS Regression 27.0405 - 9.1208 17.9197

Hence, the means squared error is : SSError MSE n - 2 17.9197 19 - 2 1.0541 Now, taking square root, we get the standard error of the estimate: Ô = MSE V1.0541 1.0267

Since we want to compute a 90% confidence intervel for the mean response value û = 0.876, the significance level used is a = 0.1. The critical t-value for df = n - 2 = 19 – 2 = 17 degrees of freedom, and a = 0.1 is te = 1.74. Now, with all this information, we are ready to compute the margin of error of the confidence interval: E ta/2:n–21 C + (Xo - X) SSxx n 1.74 x 1.0541 (0.3 -0.0789) 43.8916 19 0.4141 Hence, the 90% confidence intervel for the predicted value Y = 0.876 is: CI (0.876 – E, 0.876 + E) (0.876 – 0.4141, 0.876 +0.4141) (0.462, 1.2901)

Answers: (0.462 , 1.290)

b) Prediction Interval:

Since we want to compute a 90% prediction intervel for the predicted value û = 0.876, the significance level used is a = 0.1. The critical t-value for df = n - 2 = 19 – 2 = 17 degrees of freedom, and a = 0.1 is te = 1.74. Now, with all this information, we are ready to compute the margin of error of the prediction interval: 1 E = ta/2:1-21 (1+ + (X, - X2 SSxx n 1.74 x 1 1.0541 1+ 19 (0.3 -0.07892 43.8916 1.8368 Hence, the 90% prediction intervel for the predicted value Y = 0.876 is: PI (0.876 – E, 0.876 – E) (0.876 – 1.8368, 0.876 +1.8368) (-0.9608, 2.7129)

Answer: (-0.961, 2.713)

c) Correct answer: Option (B)