Question

*1.21. Airfreight breakage. A substance used in biological and medical research is shipped by air freight to users in cartons of 1,000 ampules. The data below, involving 10 shipments, were collected on the number of times the carton was transferred from one aircraft to another over the shipment route (X) and the number of ampules found to be broken upon arrival (Y). Assume that first-order regression model (1.I) is appropriate. 2 5 7 8 9 10 Xi: Y: 16 9 17 12 22 13 8 15 19 11 0 2 0 2

Answer 1

The data on the number of transfers (X) and the number of broken ampules (Y) for 10 shipments is provided in the question The regression line can be obtained as follows: j = a+bx Here, a and b are the intercept and slope of the regression line. The value of a and b can be calculated using the formula a -br Σν Σχ And, nΣχο,- Σ ηΣΥ- Σx The necessary calculation is done using Excel. The screenshot is shown below. B C x^2 y2 C2^2 1 2 1 16 -В2"C2 B2 2 0 9 B3 C3 -B3^2 3С3^2 3 -B4 C4 17 =B4^2 C4^2 4 -B5 C5 5 0 12 B5 2 C5 2 22 B6*C6 B6 2 C6^2 6 1 13 -B7 C7 B7^2 -С7^2 7 -C8^2 8 0 8 B8 CS B8 2 15 =B9 C -B9^2 -C9 2 9 B10 C10 B11 .C11 |=SUM(D2D11 ) C10 2 C11 2 -SUM(F2F11 ) 19 =B10^2 10 -B11 2 -SUME2Ε11 ) 0 11 11 12 Sum =SUM(B2B11) -SUM(C2C11 ) The obtained result is shown below A C E F y^2 256 x^2 1 16 16 1 2 0 9 0 81 2. 17 34 289 4 4 0 12 144 5 3 22 66 9 484 6 1 13 13 1 169 7 0 8 0 0 64 8 1 15 15 1 225 19 38 4 361 10 11 0 0 121 11 182 Sum 10 142 20 2194 12

Therefore, the value of b is calculated as follows nΣy,- Σ: Σν ηΣΥ-Στ (10x182)-10x142) (10x20)-(10) = 4 And, the value of a is calculated as follows a y-br Σν Σχ 142 10 (4) 10 = 10 =10.2 Hence, the obtained regression equation for the number of broken ampules is y =10.2+4x (a) The predicted value of number of broken ampules for X=2 can be determined by substituting 2 in the regression equation. Hence, the required predicted value is calculated as follows j = 10.2 +4x 10.2 + (4x2) - 18.2 The predicted value of number of broken ampules for X =4 is calculated as follows:

j = 10.2 +4x 10.2 +(4x4) =26.2 Therefore, the predicted value ofnumber ofbroken ampules for X 2 and X =4 are 18.2 and 26.2, respectively The 99% confidence interval can be computed by using the formula, SSE |1 CI X SS n-2 SSE -,ttaf2 x SS n-2 The value of error sum of square (SSE) is calculated as follows ΣτΣν. Σν- Σν Σ- SSE Σ- 10 x142 182 (142) 10 2194 (10) 10 20 10 =17.6 The value of SS is calculated as follows Σ. Ss,-Σ- (10) =20 10 =10 Hence, the 99% confidence interval for X2 is calculated as follows SSE CI too,10-2 X SS (2-10/10) 17.6 1 foo10-2=3.36 from t distributi on table =18.2 3.36x 10 2 V10 10 =(15.97,20.43)

And, the 99% confidence interval for X=4 is calculated as follows: SSE CI too.10-2 SS n-2 (4-10/10) 17.6 1 =18.2 +3.36 x fo.op10-2=3.36 from t distribution table V10-2 V10 10 (21.22,31.18) The 99% confidence interval for X = 2 indicates that on an average the predicted value of number of broken ampules lies within the interval (15.97,20.43) with 99% confident The 99% confidence interval for X = 4 indicates that on an average the predicted value of number of broken ampules lies within the interval (21.22,31.18) with 99% confident (b) According to the question, the next shipment has 2 transfer. That is, x 2. So, the 99% prediction interval is calculated as follows = (x SSE PI =yttoo0-2 X. 1 X1 n-2 -X n J 17.6 V10-2 V (2-10/10) =18.2 3.36x fooe 0-3.36 from t distribution table 10 10 (12.75,23.65) The 99% prediction interval indicates that the predicted value number of broken ampules will lie between 12.75 and 23.65 with probability 0.99 Hence, the required value of prediction interval is12.75,23.65)

According to question, three shipment are made, and each shipment has two transfer. Therefore, the total number of transfers is 3x2 =6. That is, x 6 The 99% prediction interval for the number of broken ampules is calculated as follows: fooyp.10-2X o.o.10-2X (Mean number of troken ampules SSE SSE n-2 n-2 (-r [1 (х, —х X X + SS SS n 3.36 3.36 17.6 17.6 Mean number (102+4x6)x, (10.2+4x6) 10-2 of troken ampules, V10-2 (6-10/10) (6-10/10) 1 1 X. 10 10 10 10 Mean number 42.22 26.18 of broken ampules Therefore, the 99% prediction interval for mean number of broken ampules is 26.18.42.22) Hence, the 99% prediction interval for total number of broken ampules is calculated as follows: Mean number 26.18 42.22 of broken ampules (Total number of broken ampules) Total number of transfer Total number 26.18 -42.22 26.18 x6 42.22 x6 of broken ampules (Total numbo <42.22x6 26.18 x6 of broken ampules Total number 157.08 253.32 of broken ampules) Hence, the required prediction interval for total number of broken ampules is (157.08,253.32 VI