Question

Only need d

4.12. Typographical errors. Shown below are the number of galleys for a manuscript (X) and the total dollar cost of correcting typographical errors (Y) in a random sample of recent orders handled by a firm specializing in technical manuscripts. Since Y involves variable costs only, an analyst wished to determine whether regression-through-the-origin model (4.10) is appropriate for studying the relation between the two variablės. i: 1 23 ' 56 7 8910 11 12 x,: 7 12 10 10, 14 25 30 25 18 1046 Y: 128 213 191 178 250 446 540 457 324 177 75 107 a. Fit regression model (4.10) and state the estimated regression function art One Simple Linear Regression Plot the estimated regression function and the data. Does a linear regression function through the origin appear to provide a good fit here? Comment. In estimating costs of handling prospective orders, management has used a standard of $17.50 per galley for the cost of correcting typographical errors. Test whether or not this standard should be revised; use α .02. State the alternatives, decision rule, and conclusion. b. c. d. Obtain aprediction interval for thecorrectioncost on a forthcoming job involving 10 galleys. Use a confidence coefficient of 98 percent.

Answer 1

Well calculations of the prediction interval to follow:

These are the data that have been provided for the dependent and independent variables:

Obs.	X	Y
1	7	128
2	12	213
3	10	191
4	10	178
5	14	250
6	25	446
7	30	540
8	25	457
9	18	324
10	10	177
11	4	75
12	6	107

Now, with the provided sample values of the predictor and the response variable, we need to construct the following table to compute the estimated regression coefficients:

Obs.	X	Y	$X_i^2$​	$Y_i^2$​	$X_i \cdot Y_i$​
1	7	128	49	16384	896
2	12	213	144	45369	2556
3	10	191	100	36481	1910
4	10	178	100	31684	1780
5	14	250	196	62500	3500
6	25	446	625	198916	11150
7	30	540	900	291600	16200
8	25	457	625	208849	11425
9	18	324	324	104976	5832
10	10	177	100	31329	1770
11	4	75	16	5625	300
12	6	107	36	11449	642
Sum =	171	3086	3215	1045162	57961

The sum of squares obtained from the table above are:

12 2 3215 (171)2 12 778.25 1045162 - 3086) 12 -251545.6667 X,Yi - 57961 - (171) (3086) 12 13985.5

The slope and y-intercept coefficients are computed using the following formulas:

SSxy βι SSxx 13985.5 778.25 17.9704 -257.1667-17.9704 x 14.25 1.0878

Therefore, the regression equation is:

$\hat Y = 1.0878 + 17.9704 X$

Now that we have the regression equation, we can compute the predicted value for X=10, by simply plugging in the value of X=10 in the regression equation found above:

​

Y1.0878 +17.9704 x 10 180.7923

Now, we are going to compute the 98% prediction interval for the predicted value $\hat Y = 180.7923$ . 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:

SSRegression - β1SSXY 17.9704 x 13985.5 - 251325.6797

Since we know that SSTotal-SSResionSSError, then we can compute the sum of squared errors as follows: SSErrorSSrtal SS Regression 251545.6667-251325.6797 219.9869 Hence, the means squared error is MSE - SSError 2-2 219.9869 12-2 21.9987 Now, taking square root, we get the standard error of the estimate: = V21.9987 4.6903

Since we want to compute a 98% prediction intervel for the predicted value Y-180.7923, the significance level used is α = 0.02. The critical t-value for df = n-2 12-2-10 degrees of freedom, and α = 0.02 is tc 2.764. Now, with all this information, we are ready to compute the margin of error of the prediction interval: SSxx 1425 ) 2.764,V21 9967(1+1ฐ.10 778.25 13.4581 Hence, the 98% prediction intervel for the predicted value Y-180.7923 is: PI(180.7923 - E, 180.7923-E) (180.7923 - 13.4581, 180.7923+13.4581) (167.3342, 194.2504)

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