Question

A study was conducted to investigate for any possible correlation that may exist between the number of tardiness and the final grade (%) of the students in statistics class. Based on this data, Kuya, who is enrolling in the class next trimester, may anticipate his grade from being tardy every other week (6 times). What would his estimated grade be?

STUDENT	TARDY	GRADE
1	1	89
2	2	94
3	0	97
4	0	87
5	1	94
6	0	92
7	2	98
8	9	50
9	0	92
10	2	95
11	0	84
12	1	96
13	2	88

The appropriate test for this problem is:

a. correlation

b. regression

c. multiple regression

The obtained statistic is:

a. - .48

b. - .84

c. = .48

d. = .84

The associated p value is:

a. .0003

b. .003

c. .03

d. .3

Decision is:

a. reject the null

b. retain the null

Conclusion is (closest answer due to rounding error):

a. predicted grade is 79.62

b. predicted grade is 69.62

c. there is a significant negative correlation

d. no conclusion can be drawn

Answer 1

Ans:

STUDENT	TARDY(x)	GRADE(y)	y'=-4.3256x+95.578	(y-y')^2	(x-1.5385)^2
1	1	89	91.2524	5.073	0.290
2	2	94	86.9268	50.030	0.213
3	0	97	95.578	2.022	2.367
4	0	87	95.578	73.582	2.367
5	1	94	91.2524	7.549	0.290
6	0	92	95.578	12.802	2.367
7	2	98	86.9268	122.616	0.213
8	9	50	56.6476	44.191	55.674
9	0	92	95.578	12.802	2.367
10	2	95	86.9268	65.177	0.213
11	0	84	95.578	134.050	2.367
12	1	96	91.2524	22.540	0.290
13	2	88	86.9268	1.152	0.213
Total	20	1156	1156.002	553.586	69.231
mean=	1.538461538

a)

Correlation

b)

Sample correlation coefficient,r=-0.84

standard error for slope=SQRT(553.586/11)/SQRT(69.231)=0.8526

t-test statistic=-4.3256/0.8526=-5.073

c)df=13-2=11

p-value=tdist(5.073,11,2)=0.0003

d)Reject the null hypothesis.

e)

when x=6

y'=-4.3256*6+95.578=69.62

predicted grade is 69.62