Question

The given data is the grades for people in this class. The goal here is to determine the factors that effect student's Grade in the class. 4) Find the mean and median for the men's and the women's Quizzes. Gender Men Women 5) Test the claim that the majority of students at this class are women. F M F F M F F F F M M F F F M F F F F M M F F M M F F F M F F M F M F Quizzes 61 55 59 54 86 62 49 73 66 58 70 66 73 75 83 69 66 87 74 80 49 78 91 MyStatLab 46 50 51 56 57 59 59 64 66 66 66 67 68 71 72 73 74 74 75 75 75 78 81 83 84 84 86 86 86 87 88 89 90 90 92 92 Grade 59 55 53 65 72 66 61 59 59 73 70 67 70 74 64 72 72 79 75 75 62 73 88 68 72 76 73 78 82 83 73 82 85 81 80 81 6) Test the claim that men and women have the same mean Grade. Women Men = = 55 67 67 66 89 74 77 63 81 87 78 82 84 SE SE M 81 - Phoso Ghi 81 F F F 81 92 94 95 77 90

Answer 1

(a) There are 26 F and 13 Males out of a total of 39 Students

The Mens Quizzes scores are as below

1	M	55
2	M	86
3	M	58
4	M	70
5	M	83
6	M	80
7	M	49
8	M	55
9	M	67
10	M	74
11	M	81
12	M	78
13	M	84

Average = Sum of observations / Total Observations = 920 / 13 = 70.77

The Median. Since n is odd, the median = (n + 1)/ 2 th number = (13 + 1) / 2 = 7th number = 49

For Females

1	F	61
2	F	59
3	F	54
4	F	62
5	F	49
6	F	73
7	F	66
8	F	66
9	F	73
10	F	75
11	F	69
12	F	66
13	F	87
14	F	74
15	F	78
16	F	91
17	F	67
18	F	66
19	F	89
20	F	77
21	F	63
22	F	87
23	F	82
24	F	81
25	F	81
26	F	92

Average = Sum of observations / Total Observations = 1888 / 26 = 72.62

The Median = Middle value

Since n is even, the median = Average of n/2th and the next number = Average of 26/2 = 13th and 14th numbers

= (87 +74) / 2 = 80.5

	Men	Women
Mean	70.8	72.62
Median	49	80.5

________________________________________

(a) Test For proportion

The Hypothesis:

p = 0.5

p > 0.5

This is a right tailed test

The Test Statistic: $\hat{p}$ = 26 / 39 = 0.667

$Z = \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}} = \frac{0.667-0.5}{\sqrt{\frac{0.5*0.5}{39}}} = 2.8$

The p Value:    The p value (Right tail) for Z = 2.08, is; p value = 0.0188

The Critical Value:   The critical value (Right tail) at $\alpha$ = 0.05 (default level) Z critical = +1.645

The Decision Rule: If Z observed is > Z critical Then Reject H0.

Also If the P value is < $\alpha$ , Then Reject H0

The Decision:    Since Z observed (2.08) is > Z critical (1.645), We Reject H0.

Also since P value (0.0188) is < $\alpha$ (0.05), We Reject H0.

The Conclusion:   There is sufficient evidence at the 95% significance level to conclude that the majority of students in the class are women

____________________________________________

For Grades the sample statistics are as below

	Men	Women
Total	937	1898
n	13	26
Mean	72.08	73
SD	8.391	8.841

Since s₁/s₂ = 8.391 / 8.841 = 0.949 (it lies between 0.5 and 2) we used the pooled variance.

The degrees of freedom used is n1 + n2 - 2 = 13 + 26 - 2 = 37 (since pooled variance is used)

$S_p^2 = \frac{(n_1-1)*s_1^2+(n_2-1)*s_2^2}{n_1+n_2-2} = \frac{12*8.391^2+25*8.841^2}{26+13-2} = 75.648$

The Hypothesis:

HO: Hi = 21

$Ha: \mu_1 \neq \mu_2$

This is a Two tailed test.

The Test Statistic:We use the students t test as population standard deviations are unknown.

$t = \frac{\bar{x}_1-\bar{x}_2}{\sqrt{S_p^2*(\frac{1}{n1}+\frac{1}{n2})}} = \frac{72.08-73}{\sqrt{75.648*(\frac{1}{26}+\frac{1}{13})}} = -0.32$

The p Value:   The p value (2 Tail) for t = -0.32, df = 37, is; p value = 0.7575

The Critical Value:   The critical value (2 tail) at $\alpha$ = 0.05 (default level), df = 37, t critical = - 2.03 and + 2.03

The Decision Rule: If t observed is > t critical or If   t observed is < -t critical, Then Reject H0.

Also If the P value is < $\alpha$ , Then Reject H0

The Decision: Since t lies in between -2.03 and +2.03, We Fail To Reject H0

Also since P value (0.7575) is > $\alpha$ (0.05), We Fail to Reject H0.

The Conclusion: There isn’t sufficient evidence at the 95% significance level to warrant rejection of the claim that Men and Women have the same mean grade.

___________________________________________

Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)².

Men
#	X	X - Mean	(X - Mean)2
1	55	72.08	291.7264
2	72	72.08	0.0064
3	73	72.08	0.8464
4	70	72.08	4.3264
5	64	72.08	65.2864
6	75	72.08	8.5264
7	62	72.08	101.6064
8	68	72.08	16.6464
9	72	72.08	0.0064
10	82	72.08	98.4064
11	82	72.08	98.4064
12	81	72.08	79.5664
13	81	72.08	79.5664
Total	937		844.923

Women
#	X	X - Mean	(X - Mean)2
1	59	73	196
2	63	73	100
3	65	73	64
4	66	73	49
5	61	73	144
6	59	73	196
7	59	73	196
8	67	73	36
9	70	73	9
10	74	73	1
11	72	73	1
12	72	73	1
13	79	73	36
14	75	73	4
15	73	73	0
16	88	73	225
17	76	73	9
18	73	73	0
19	78	73	25
20	83	73	100
21	73	73	0
22	85	73	144
23	80	73	49
24	81	73	64
25	77	73	16
26	90	73	289
Total	1898		1954