1. Text Exercise 5.20 (a) (b) 1 for man-made shade X 1 = 0 otherwise Y =  0 +  1 X 1 +  2 X 2 +  or E(Y) =  0 +  1 X 1 +  2 X 2 Do this by first defining the appropriate dummy variable(s), and then writing a regression model. You should realize that there is only one qualitative independent variable to identify. type of shade (man-made, tree, none) 1 for tree shade X 2 = 0 otherwise Check some answers brefore submitting Homework #17. Homework #17Score____________/ 10Name ______________

2. (c)Do this by writing a statement interpreting each parameter in the model of part (b).  0 =  1 =  2 = the mean of Y for no shade the amount that the mean milk production for man-made shade exceeds the mean milk production for no shade the amount that the mean milk production for tree shade exceeds the mean milk production for no shade

3. Additional HW Exercise 5.2 (a) (b) The mean length of fish is being studied for North Lake, Blue Lake, and Harvey Lake. A 0.05 significance level is chosen for a hypothesis test to see if there is any evidence that mean length of fish is not the same for the three lakes. Fish are randomly selected from each lake, and the lengths in inches are recorded as follows: North13 17 15 18 17 Blue15 12 16 11 16 Harvey14 10 12 13 11 After defining the appropriate dummy variable(s), write a regression model for the prediction of Y = “length of fish” from “lake”. 1 for North Lake X 1 = 0 otherwise 1 for Blue Lake X 2 = 0 otherwise Y =  0 +  1 X 1 +  2 X 2 +  or E(Y) =  0 +  1 X 1 +  2 X 2 With the model defined in part (a), write the formula for the population mean length of fish for each lake. mean for North =  0 +  1 mean for Blue =  0 +  2 mean for Harvey = 00

4. (c) Do each of the following calculations: y N  = y B  = y H  = y  = SSR = SST = SSE = (13 + 17 + 15 + 18 + 17) / 5 =16 (15 + 12 + 16 + 11 + 16) / 5 =14 (14 + 10 + 12 + 13 + 11) / 5 =12 (13 + 17 + … + 13 + 11) / 15 = 14 (5)(16 – 14) 2 + (5)(14 – 14) 2 + (5)(12 – 14) 2 =40 (13 – 14) 2 + (17 – 14) 2 +,,, + (13 – 14) 2 + (11 – 14) 2 = 88 88 – 40 =48

5. 1.-continued (d) (e) Complete the one-way ANOVA table displayed. SourcedfSSMSfP-value Error Total 12 14 2 40 48 88 20 4 Lake 5.000.025 < P < 0.05 Summarize the results (Step 4) of the f test to see if there is sufficient evidence that mean length of fish is not the same for the for the three lakes at the 0.05 level. have sufficient evidence to reject H 0. We conclude that mean length of fish is not the same for the for the three lakes (0.025 < P < 0.05). Since f 2,12 = 5.00 and f 2,12;0.05 =3.89, we

6. (f)Based on the results in part (e), decide whether or not a multiple comparison method is necessary. If no, then explain why not; if yes, then use Tukey’s HSD multiple comparison method. Since we have rejected H 0, we need a multiple comparison method to identify for which lakes mean length of fish is significantly different. q  (k,v) =  NB = q(k,v)q(k,v) s—2 s—2 1 1 — + — n N n B = k = s = v =3  4 = 2 12 q 0.05 (3,12) =3.77 2 ——  2 1 1 — + — 5 5 =  NH =  BH = q(k,v)q(k,v) s—2 s—2 1 1 — + — n N n H = 3.77 2 ——  2 1 1 — + — 5 5 = q(k,v)q(k,v) s—2 s—2 1 1 — + — n B n H = 3.77 2 ——  2 1 1 — + — 5 5 = 3.4 y N  =16 y B  =14 y H  =12 yN –yN –y B  = yN –yN –y H  = yB –yB – 2 4 2 With  = 0.05, we conclude that mean length of fish is larger for North Lake than for Harvey Lake.

7. Additional HW Exercise 5.3 (a) (b) A study is being conducted to see if the drying time for a brand of outdoor paint can be predicted from temperature in O F and wind velocity in miles per hour. Data from random observation has been stored in the SPSS data file drypaint. drtm =  0 +  1 (tmp) +  2 (wnd) +  Write a first-order model for the prediction of drying time from temperature and wind velocity. Use the Analyze > Regression > Linear options in SPSS to obtain SPSS output displaying the ANOVA table and coefficients in the least squares prediction equation for the first-order model in part (a). To have the mean and standard deviation displayed for the dependent and independent variables, click on the Statistics button, and select the Descriptives option. Title the output to identify the homework exercise (Additional HW Exercise 5.3 - part (b)), your name, today’s date, and the course number (Math 214). Use the File > Print Preview options to see if any editing is needed before printing the output. Attach the printed copy to this assignment before submission.

8. (c) Summarize the results (Step 4) of the f test to see if there is sufficient evidence that the prediction of drying time from temperature and wind velocity is significant at the 0.05 level. Since f 2,19 = 417.907 and f 2,19;0.05 = 3.52, we have sufficient evidence to reject H 0. We conclude that the prediction of drying time from temperature and wind velocity is significant (P < 0.01). OR (P < 0.001)

9. Additional HW Exercise 5.3 - continued (d) (e) In order to see if we can improve prediction, the complete second order model is now considered. Write a complete second-order model for the prediction of drying time using the complete second order model with independent variables temperature and wind velocity. drtm =  0 +  1 (tmp) +  2 (wnd) +  12 (tmp)(wnd) +  11 (tmp) 2 +  22 (wnd) 2 +  Use SPSS to create three new variables, one named tmp_wnd equal to the product of the variables temperature and wind velocity, one named tmp2 equal to the square of temperature, and one named wnd2 equal to the square of wind velocity. Use the Analyze > Regression > Linear options in SPSS to obtain SPSS output displaying the ANOVA table and coefficients in the least squares prediction equation for the complete second-order model in part (d). Since you only need the ANOVA table and the coefficients in the least squares prediction equation, you may choose to delete all other sections of the SPSS output. Title the output to identify the homework exercise (Additional HW Exercise 5.3- part (e)), your name, today’s date, and the course number (Math 214).

10. (f) Summarize the results (Step 4) of the f test to see if there is sufficient evidence that the prediction of drying time using the complete second order model with independent variables temperature and wind velocity is significant at the 0.05 level. Since f 5,16 = 838.869 and f 5,16;0.05 = 2.85, we have sufficient evidence to reject H 0. We conclude that the prediction of drying time using the complete second order model with independent variables temperature and wind velocity is significant (P < 0.01). OR (P < 0.001) Use the File > Print Preview options to see if any editing is needed before printing the output. Attach the printed copy to this assignment before submission.

11. Additional HW Exercise 5.3 - continued (g) Calculate the partial f statistic for the hypothesis test to see if there is sufficient evidence that adding temperature times wind velocity, temperature squared, and wind velocity squared after temperature and wind velocity, is significant at the 0.05 level. SSR(tmp, wnd, tmp_wnd, tmp2, wnd2) = SSR(tmp, wnd) = SSR(tmp_wnd, tmp2, wnd2 | tmp, wnd) = MSR(tmp_wnd, rain2, temp2 | tmp, wnd) = MSE(tmp, wnd, tmp_wnd, tmp2, wnd2) = Numerator degrees of freedom for f statistic = Denominator degrees of freedom for f statistic = 29729.365 29179.456 29729.365 – 29179.456 = 549.909 549.909 / (5 – 2) =183.303 7.088 3 16 f 3,16 = 183.303 ———– = 7.088 25.861