Question

solve the problem make sure to explain in words what you did with the problem and state your conclusions in terms of the problem.

Part I: Choose to do one of the following: 1) Test the claim that the mean Unit 3 Test scores of data set 7176 is greater than the mean Unit 3 Test scores of data set 7178 at the .05 significance level

Class 7176				Class 7178
Unit Test 3	Course Grade	Attendance		Unit Test 3	Course Grade	Attendance
238	63	96		291	95	100
208	55	48		301	91	83
258	89	96		261	68	87
264	84	96		0	53	91
324	98	100		0	23	44
0	62	44		284	93	96
0	56	66		307	77	78
274	87	96		0	44	70
274	83	96		208	72	57
0	0	18		0	56	78
179	71	100		0	73	91
268	86	100		0	28	39
241	60	87		231	64	91
0	8	26		307	87	100
278	84	96		301	87	96
307	89	87		228	52	74
294	87	100		255	73	70
175	76	74		304	85	83
129	66	87		0	37	44
284	82	100		0	60	48
297	90	79		0	25	35
255	74	74		0	37	48
268	88	100		0	67	100
215	77	39		301	94	87
146	71	87		284	71	57
304	88	100		321	87	100
				311	91	96
				274	83	96
				307	97	100
				278	92	91
				0	52	91

Answer 1

let us describe the problem as a testing of hypothesis problem where our null hypothesis is the means of the two samples are equal

H0: mA=mB ( mA mean of unit data of class 7176 & mB is the mean of unit data of class 7178)

and the alternative hypothesis is  

Ha: mA>mB (greater)

we can perform the test by two sample t test with unequal sample sizes

the test statistics is

here nA = sample size of first sample and nB = sample size of second sample

and

is the pooled variance of the two samples

R has a nice function to do such tests

R- Code:-

r=read.csv("Book1.csv")
t.test(x = r$Unit.Test.31,y=r$Unit.Test.32, alternative = "greater",var.equal = F) # Unit.Test.31= unit data for class 7176 similarly Unit.Test.32 is for class 7178

Output:-

Welch Two Sample t-test

data: r$Unit.Test.31 and r$Unit.Test.32

t = 1.1682, df = 54.093, p-value = 0.1239

alternative hypothesis: true difference in means is greater than 0

95 percent confidence interval:
-16.46524 Inf
sample estimates:

mean of x mean of y
210.7692 172.7097

Comment:-

Here P-value of the test is comparatively large than 0.05 which implies the mean of 1st sample is not greater than the mean of 2nd sample.