Question

4 or 5 Total Assuming that a single student is randomly selected from the sample data: (a) What is the probability that the student skips class sometimes, often or almost always and has an A average grade? (b) What is the probability that the student has an average grade of B or less? Yo What is the probability that the student skips class rarely or never, given that she/he has a B average grade? (d) Test at the 1% significance level the claim that the frequency of classes skipped and the average grade are independent. Should H, be rejected? (Hint: use the test of independence). Can we infer that there is an association between lecture attendance and first year grade average? ( Bad on your results in wa vice should Brock academic advisors and other student Support Workers given to students

Answer 1

Fundamental definition of probability:

Probability = number of favorable outcomes total number of outcomes

Assuming that a single student is randomly selected from the sample data:
(a) What is the probability that the student skips class sometimes, often or almost always and has an A average grade?
total of favorable outcomes=students(sometimes+often+almost always)=(7+0)=7
total outcomes=total students with Grade A=69
Required probability=7/69=0.101449275

(b) What is the probability that the student has an average grade of B or less?
total of favorable outcomes=student has an average grade of B or less= student with grade (C or less,B)=(183+201)=384
total outcomes=total students graded=453
Required probability=384/453=0.847682119

(c) What is the probability that the student skips class rarely or never, given that she/he has a B average grade?
total of favorable outcomes=the student skips class rarely or never=(55+88)=143
total outcomes=total students with Grade B=201
Required probability=143/201=0.711442786

(d) Test at the 1% significance level the claim that the frequency of classes skipped and the average grade are independent. Should H₀ be rejected? (Hint: use the chi- square test of independence). Can we infer that there is an association between lecture attendance and first year grade average?

Theory:

² test of Independence of Atnbutes het hoo a tributer he A and B. Atterbut A be drided into a darres mi classes A., Ar... Am. Attribute B 'n' clases BIBL... Bn. The various sell frequences can be presented in following manifold table known as contingency table.. - mxn Contgency table A B B B, B . . . B, Total AL CAIB CABL)... (A. Bil... CABO CAL C A, B) + (AIB)- +(A, Bn Az (A261) CA2B2) ... (A₂B) . . . (A2Bn) (A2) Ai (AiBi) (Ai B2) ... (AiBJ) ... (AiBn) (AS) Am I (Ambi) (Amby) .. (AMB) ... (Ambn) (Am) Total (B) (B) (B) . (B) N I all here, (Ar Bj) denotes the cell frequency ic. number of objects possessing both the attribute Ar and By. Also, I Ai = 1 Bi =N, when N = total of all all frequencies. • Each cell frequency must be greater than 5. • N should be 350. Ho: The two altributes A and B are independent (discouated) V. Hii 11 are dependent (associated) Test statistic 72 = 5 [(A, B;) - (A: B; )] ² %N X 2 6040 (A483 ) J Reject Ho, if XCallutated 7 Xem) (4) ; & (Ai bj ) = 0 Obserwed freg. 5,- (A5) (83) - (now total) (Lolumn total) (A1B). = E = Expected frey. Ν.

Here, m=3 and n=4. Attribute A is average grades and attribute B is Attendance.

Computation in MS-Excel:

Grade	Attendance	OiB	EiB			Oi	Ei	(Oi-Ei)	(Oi-Ei)^2	(Oi-Ei)^2/Ei
C or less	1	21	(107*183)/453	43.2252		21	43.225	-22.2252	493.958	11.42756
	2	70	(189*183)/453	76.351		70	76.351	-6.35099	40.33512	0.528285
	3	72	(131*183)/453	52.9205		72	52.921	19.07947	364.0262	6.878733
	4 or 5	20	(26*183)/453	10.5033		20	10.503	9.496689	90.1871	8.58654
B	1	55	(107*201)/453	47.4768		55	47.477	7.523179	56.59822	1.192123
	2	88	(189*201)/453	83.8609		88	83.861	4.139073	17.13192	0.20429
	3	52	(131*201)/453	58.1258		52	58.126	-6.12583	37.52577	0.645595
	4 or 5	6	(26*201)/453	11.5364		6	11.536	-5.53642	30.65199	2.656975
A	1	31	(107*69)/453	16.298		31	16.298	14.70199	216.1484	13.26226
	2	31	(189*69)/453	28.7881		31	28.788	2.211921	4.892592	0.169952
	3	7	(131*69)/453	19.9536		7	23.914	-16.9139	286.0803	11.96292
	4 or 5	0	(26*69)/453	3.96026						57.5152
TOTAL

critical value		15.1	CHIINV(0.01,6-1)
degrees of freedom=(m-1)(n-1)-1=5 as 1 df lost due to pooling
(m-1)(n-1)	(3-1)*(4-1)	6
calculated value > critical value, Reject H0.

Yes, we infer that there is an association between lecture attendance and first year grade average.

(e) Based on your results in (d) what advice should Brock academic advisers and other student support workers give to students?

As there is association between attendance and grade, one can advise students to attend lecture regularly, to attend as many lectures as are taken. Not to miss any. Or atleast try to attend maximum number of lectures.