1.06 1.04 To check the accuracy of the prediction method, the linear model y= Bo+B1x+ε is fit. If the prediction method is accurate, the value of Bo will be 0 and the value of ẞ1 will be 1. Note: This problem has a reduced data set for ease of performing the calculations required. This differs from the data set given for this problem in the text.

Lecture(8.8): The amount of time people engage in physical  activity mat be related to health outcome. Those who report that they spend more than 15 hours are put into one while those who spend less 10 were put into another group. (Those who fall in between 10 and 15 were left out of the study); The reaearcher then ask the participants to wear a monitor for one month. The average time  in minutes is recorded and shown below . Is there any evidence that on average people who watch less than 10 hours    watching televsion spend more time on physical activity.?. Test the hypotheses at (alpha=0.05) using the 5 step procedure <10 hours       75        63      118      35           82 >15 hours      62         6        78       43            22    33

EXAMPLE 8.5 | Alloy Adhesion An article in the Journal of Materials Engineering ["Instrumented Tensile Adhesion Tests on Plasma Sprayed Thermal Barrier Coatings" (1989, Vol. 11(4), pp. 275-282)] describes the results of tensile adhesion tests on 22 U-700 alloy specimens. The load at specimen failure is as follows (in megapascals): 19.8 15.4 11.4 19.5 10.1 18.5 14.1 8.8 14.9 7.9 17.6 13.6 O 0.01 O 0.025 O 0.05 O 0.95 O 0.975 7.5 12.7 16.7 11.9 15.4 11.9 15.8 11.4 The sample mean is x = 13.71, and the sample standard deviation is s = 3.55. Figures 8.6 and 8.7 show a box plot and a normal probability plot of the tensile adhesion test data, respectively. These displays provide good support for the assumption that the population is normally distributed. We want to find a 95% CI on μ. Since n = 22, we have n - 1 = 21 degrees of freedom for t, so to.025,21 = 2.080. The resulting CI is X-1/2-1/√x+1a/2n-1³/√n 13.71-2.080 (3.55)/√/22 ≤ ≤ 13.71 +2.080 (3.55)/√22 13.711.57 ≤ ≤ 13.71 +1.57 12.14 ≤…

6 A study was conducted that measured the total brain volume (TBV) (in mm3) of patients that had schizophrenia and patients that are considered normal.  Table #1 contains the TBV of the normal patients and Table #2 contains the TBV of schizophrenia patients ("SOCR data Oct2009," 2013).  Table #1: Total Brain Volume (in mm3) of Normal Patients 1663407 1583940 1299470 1535137 1431890 1578698 1453510 1650348 1288971 1366346 1326402 1503005 1474790 1317156 1441045 1463498 1650207 1523045 1441636 1432033 1420416 1480171 1360810 1410213 1574808 1502702 1203344 1319737 1688990 1292641 1512571 1635918         Table #2: Total Brain Volume (in mm3) of Schizophrenia Patients 1331777 1487886 1066075 1297327 1499983 1861991 1368378 1476891 1443775 1337827 1658258 1588132 1690182 1569413 1177002 1387893 1483763 1688950 1563593 1317885 1420249 1363859 1238979…