no. of absences, x | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
Final exam score, y | 88.3 | 85.5 | 82.8 | 81.2 | 78.3 | 73.4 | 64.4 | 70.9 | 64.7 | 66.4 |
n | |
3 | .997 |
4 | .950 |
5 | .878 |
6 | .811 |
7 | .754 |
8 | .707 |
9 | .666 |
10 | .632 |
11 | .602 |
12 | .576 |
13 | .553 |
14 | .532 |
15 | .514 |
16 | .497 |
17 | .482 |
18 | .468 |
19 | .456 |
20 | .444 |
21 | .433 |
22 | .423 |
23 | .413 |
24 | .404 |
25 | .396 |
26 | .388 |
27 | .381 |
28 | .374 |
29 | .367 |
30 | .361 |
The accompanying data represent the number of days absent, x, and the final exam score, y, for a sample of college students in a general education course at a large state university. Complete parts (a) through (e) below.
a) y = __x + __
b) interpret the slope and the y-intercept, if appropriate
c) predict the final exam score for a student who missed four and five class periods
Ans:
a)
Regression equation:
y'=-2.7727 x+88.067
b)slope indicate that for each absence there will be decrease of on average 2.7727 scores in final exam.
y-intercept indicates that if there is no absence,average score is 88.067
c)
For x=4,y'=-2.7727*4+88.067=76.98
For x=5,y'=-2.7727*5+88.067=74.20
2 62.8 MEAN = 3 71.9 MEDIAN = 4 69.6 MODE= 5 74.1 RANGE = 6 66.6 MIN. = 7 76.5 MAX.= 8 66.4 STDEV. = 9 73.1 MY HEIGHT =67.00 10 71.6 Z SCORE FOR MY HEIGHT = 11 69.3 12 64.0 13 70.9 14 62.2 15 63.3 16 67.7 17 65.2 18 64.2 19 69.4 20 71.7 21 64.6 22 69.0 23 71.3 24 69.1 25 71.6 26 75.9 27 66.2 28 67.4 29 64.6 30 69.6 31...
Consider the graph 12 10 6, 9) y-f(x 8 (2, 7) (4, 5) (0, 3) (8, 0) 10 (a) Using the indicated subintervals, approximate the shaded area by using lower sums s (rectangles that lie below the graph of f) (b) Using the indicated subintervals, approximate the shaded area by using upper sums S (rectangles that extend above the graph of f) +-14 points SullivanCalc1 5.1.019 Approximate the area A under the graph of function f from a to b...