Question

The subjects in the data are college students. In the data, id is student ID, anxiety is student’s anxiety score via Anxiety Scale, selfest is student’s self-esteem score via Rosenberg Self-esteem Scale, GPA is student’s GPA; for gender, 0=female, 1=male; for grade, 1=freshman, 2=junior, 3=senior. We have known that population mean for Anxiety Scale is μ=60 with σ=10.

Raise relevant questions ( 2 questions is fine) about the data extensively, the questions can be either about descriptive analysis or inferential analysis, for example, “what are the mean and standard deviation of …” or “whether female students are more … than male students”. Then you should select appropriate statistical analyses for your questions and explain why you select these analyses. Finally, conduct analyses and report the results

This is the data

id	anxiety	selfest	gender	grade	GPA
1	61	25	0	1	3.31
2	71	29	0	1	2.67
3	67	28	1	1	3.89
4	52	28	0	1	4.00
5	81	20	1	1	2.00
6	81	21	1	1	2.56
7	69	28	1	1	3.12
8	60	38	0	1	3.47
9	67	29	1	1	3.29
10	58	28	0	1	3.22
11	48	28	1	1	3.68
12	66	24	1	1	3.78
13	49	37	1	1	3.84
14	75	25	1	1	3.11
15	79	27	1	1	3.05
16	84	21	0	1	3.33
17	70	22	1	1	2.21
18	52	35	0	1	2.95
19	63	33	0	1	2.44
20	62	32	1	1	3.00
21	70	30	1	1	3.91
22	63	25	1	1	2.44
23	61	37	0	1	3.68
24	71	28	0	1	3.27
25	68	29	1	1	3.18
26	62	36	0	1	2.47
27	64	28	0	1	3.45
28	62	28	1	1	3.67
29	63	33	1	1	3.78
30	60	25	0	1	3.79
31	57	29	0	1	3.10
32	57	35	0	1	3.20
33	80	23	0	1	2.10
34	68	28	0	1	2.89
35	70	26	1	1	2.71
36	58	28	0	1	3.02
37	73	24	0	1	3.09
38	65	35	0	1	3.94
39	51	36	0	1	3.95
40	72	23	1	1	3.03
41	62	29	1	1	3.06
42	76	31	1	1	2.94
43	68	30	1	1	2.86
44	70	32	1	1	2.06
45	65	30	0	1	3.14
46	70	28	1	1	3.19
47	35	34	0	1	3.29
48	54	30	0	1	3.64
49	62	30	0	1	3.49
50	66	31	0	1	2.78
51	79	24	0	2	1.67
52	86	18	0	2	3.18
53	65	33	1	2	4.00
54	64	35	1	2	3.97
55	78	32	0	2	3.25
56	71	25	1	2	3.54
57	65	27	1	2	3.75
58	49	41	0	2	2.22
59	70	26	1	2	2.68
60	74	27	0	2	3.45
61	60	35	1	2	3.54
62	60	34	0	2	3.65
63	71	24	1	2	3.62
64	62	38	0	2	3.69
65	60	32	1	2	2.99
66	85	26	0	2	3.07
67	69	29	0	2	3.91
68	51	35	1	2	3.89
69	62	35	1	2	3.23
70	69	19	1	2	3.36
71	44	38	1	2	3.61
72	67	27	1	2	3.04
73	78	32	0	2	2.91
74	53	29	0	2	2.59
75	80	22	1	2	3.03
76	62	34	1	2	3.56
77	74	27	0	2	3.67
78	64	25	0	2	3.85
79	76	20	1	2	3.88
80	44	33	1	2	3.47
81	60	27	1	2	3.22
82	69	25	1	2	3.03
83	74	25	1	2	3.09
84	95	19	1	2	2.89
85	75	27	0	2	2.67
86	77	20	1	2	3.01
87	76	21	1	2	3.01
88	58	31	0	2	3.22
89	53	33	1	2	3.24
90	56	27	0	2	3.53
91	73	31	0	2	3.64
92	64	38	0	2	3.71
93	69	30	0	2	3.89
94	57	33	1	2	3.01
95	92	23	1	2	2.10
96	75	24	0	2	2.19
97	97	24	0	2	2.52
98	74	28	0	2	3.09
99	59	31	0	2	3.03
100	62	23	0	2	3.45
101	47	38	0	3	3.99
102	60	26	0	3	4.00
103	68	23	0	3	4.00
104	67	32	1	3	3.89
105	50	31	0	3	3.78
106	76	24	1	3	3.67
107	56	35	1	3	3.56
108	59	29	1	3	3.67
109	72	27	0	3	3.88
110	42	37	1	3	3.81
111	45	33	0	3	3.17
112	67	30	1	3	3.67
113	59	34	0	3	2.22
114	55	37	1	3	3.98
115	52	28	0	3	2.89
116	84	21	1	3	3.78
117	65	27	0	3	2.87
118	50	33	1	3	3.67
119	82	27	0	3	2.99
120	78	22	1	3	3.78
121	59	27	0	3	3.01
122	75	31	1	3	3.86
123	91	23	0	3	3.57
124	58	27	0	3	3.66
125	67	32	0	3	3.89
126	59	28	0	3	2.62
127	61	31	0	3	3.05
128	44	32	1	3	3.01
129	67	16	0	3	3.44
130	62	28	0	3	3.47
131	66	31	1	3	3.89
132	68	30	0	3	3.87
133	63	34	1	3	3.78
134	79	22	1	3	2.55
135	60	29	0	3	3.03
136	67	31	1	3	3.09
137	71	32	1	3	3.95
138	59	36	1	3	3.78
139	78	24	1	3	3.59
140	56	32	1	3	3.66
141	66	31	1	3	3.70
142	68	30	1	3	3.78
143	72	37	0	3	2.99
144	75	29	0	3	3.98
145	74	27	0	3	3.27
146	78	22	1	3	3.49
147	63	31	1	3	3.50
148	90	14	1	3	3.61
149	42	32	1	3	3.88
150	62	32	1	3	3.93

Answer 1

Solution

I Descriptive Analysis of GPA

Back-up Theory

Let x_i = ith value, i = 1, 2, ……., n. Then,

Mean (Average), µ, = (1/n)Σ(i = 1, n)(x_i) ………………………………………………………………..……………………. (1)

Variance, σ² = (1/n)Σ(i = 1, n){(x_i – µ)²} or equivalently [{(1/n)Σ(i = 1, 20)(x_i)²} – µ² ] …………………... (2)

Standard deviation (SD), σ = sqrt(Variance) …………………………………………………………………………..….. (3)

Median

Let x⁽¹⁾ , x⁽²⁾ , x⁽³⁾ , ……….. , x^{(n - 1)} , x⁽ⁿ⁾ be the ordered set of the given values; i.e.,

x⁽¹⁾ < x⁽²⁾ < x⁽³⁾ < ……….. < x^{(n - 1)} < x⁽ⁿ⁾

Case 1: n is even, say n = 2k

Median = Average of two middle values in the ordered set ……………………………………………………… (4a)

= (x^(k) + x^{(k + 1)})/2

Case 2: n is odd, say n = 2k + 1

Median = (k + 1)th value in the ordered set; i.e., x^{(k + 1)} ……………………………………………………… (4b)

First Quartile Q1 is the median of the first half of the ordered set ……………………………………….. (5a)

and Third Quartile Q3 is the median of the second half of the ordered set…………………………….. (5b)

Now, to work out the solution

Vide (1), Mean = 3.3067 Answer 1 [using Excel Function Statistical AVERAGE]

Vide (3), Standard deviation = 0.5146 Answer 2 [using Excel Function Statistical STDEVP]

Vide (4a), Median = 3.345 Answer 3 [Average of 75^th and 76^th observations in the ordered set – given at end]

Vide (5a), Q1 = 3.01 Answer 4 [38^th observations in the ordered set – given at end]

Vide (5b), Q3 = 3.75 Answer 5 [113^th observations in the ordered set – given at end]

Minimum value = 1.67 Answer 6 [1^st observations in the ordered set – given at end]

Maximum value = 4.00 Answer 7 [150^th observations in the ordered set – given at end]

II Inferential Analysis of Selfest

Two sample t-test for equality of means between male and female.

Let X = selfest score of male

      Y = selfest score of female

Then, X ~ N(µ₁, σ₁²) and Y ~ N(µ₂, σ₂²), where σ₁² = σ₂² = σ², say and σ² is unknown.

Claim:

Male and female do not differ with respect to selfest score

Hypotheses:

Null: H₀: µ1 = µ2 Vs Alternative: H_A: µ1 ≠ µ2

Test Statistic:

t = (Xbar - Ybar)/[s√{(1/n₁) + (1/n₂)}] where

s² = {(n₁ – 1)s₁² + (n₂ – 1)s₂²}/(n₁ + n₂ – 2);

Xbar and Ybar are sample averages and s₁,s₂ are sample standard deviations based on n₁ observations on X and n₂ observations on Y respectively.

Calculations

Summary of Excel calculations is given below:

n1 =	77
n2 =	73
Xbar =	28.41558
Ybar =	29.28767
s1 =	5.289888
s2 =	4.820326
s =	5.066891
tcal =	-1.05361
α =	0.05
tcrit =	1.976122
p-value =	0.293779

Distribution, Significance Level, α Critical Value and p-value:

Under H₀, t ~ t_{n1 + n2 - 2}. Hence, for level of significance α%, Critical Value = upper (α/2)% point of t_{n1 + n2 - 2} and p-value = P(t_{n1 + n2 - 2} > | tcal |) = 2P(t_{n1 + n2 - 2} > tcal) if tcal > 0 and 2P(t_{n1 + n2 - 2} < tcal) if tcal < 0

Using Excel Function: Statistical TINV and TDIST, these are found to be as shown in the above table.

Decision:

Since | tcal | < tcrit, or equivalently since p-value > α, H₀ is accepted.

Conclusion:

There is sufficient evidence to suggest that the claim is valid. ANSWER

DONE

GPA ordered set

id	GPA [min to max]
1	1.67
2	2
3	2.06
4	2.1
5	2.1
6	2.19
7	2.21
8	2.22
9	2.22
10	2.44
11	2.44
12	2.47
13	2.52
14	2.55
15	2.56
16	2.59
17	2.62
18	2.67
19	2.67
20	2.68
21	2.71
22	2.78
23	2.86
24	2.87
25	2.89
26	2.89
27	2.89
28	2.91
29	2.94
30	2.95
31	2.99
32	2.99
33	2.99
34	3
35	3.01
36	3.01
37	3.01
38	3.01
39	3.01
40	3.02
41	3.03
42	3.03
43	3.03
44	3.03
45	3.03
46	3.04
47	3.05
48	3.05
49	3.06
50	3.07
51	3.09
52	3.09
53	3.09
54	3.09
55	3.1
56	3.11
57	3.12
58	3.14
59	3.17
60	3.18
61	3.18
62	3.19
63	3.2
64	3.22
65	3.22
66	3.22
67	3.23
68	3.24
69	3.25
70	3.27
71	3.27
72	3.29
73	3.29
74	3.31
75	3.33
76	3.36
77	3.44
78	3.45
79	3.45
80	3.45
81	3.47
82	3.47
83	3.47
84	3.49
85	3.49
86	3.5
87	3.53
88	3.54
89	3.54
90	3.56
91	3.56
92	3.57
93	3.59
94	3.61
95	3.61
96	3.62
97	3.64
98	3.64
99	3.65
100	3.66
101	3.66
102	3.67
103	3.67
104	3.67
105	3.67
106	3.67
107	3.67
108	3.68
109	3.68
110	3.69
111	3.7
112	3.71
113	3.75
114	3.78
115	3.78
116	3.78
117	3.78
118	3.78
119	3.78
120	3.78
121	3.78
122	3.79
123	3.8
124	3.84
125	3.85
126	3.86
127	3.87
128	3.88
129	3.88
130	3.88
131	3.89
132	3.89
133	3.89
134	3.89
135	3.89
136	3.89
137	3.91
138	3.91
139	3.93
140	3.94
141	3.95
142	3.95
143	3.97
144	3.98
145	3.98
146	3.99
147	4
148	4
149	4
150	4