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QUESTION 13 Psychology Final Question 1 A professor teaches two sections of a Psychology class. The...

QUESTION 13
Psychology Final Question 1
A professor teaches two sections of a Psychology class. The placement of the enrolled students in the two sections is done randomly. In Section 1, students are required to participate in weekly study groups. In Section 2, there is no such requirement. The data file FINAL.MWX (for Minitab 19) or FINAL.MTW (for Minitab 18) contains the final exam scores of students in these two sections. Let µ1 and µ2 denote the average score in Section 1 and Section 2, respectively.
Find a 90% confidence interval for µ2-µ1.
a. (-6.38, -0.82)   
b. (9.43, 14.52)   
c. (-14.46, -9.49)
d. (-5.61, -0.05)
e. (1.27, 7.89)

Class 1 Class 2
50 74
73 79
54 68
90 62
63 55
70 76
70 66
47 81
81 64
93 88
69 97
62 72
38 78
65 81
82 77
48 81
38 85
57 78
65 79
40 62
62 95
44 52
60 100
67 78
63 82
85 67
71 80
69 70
95 72
85 83
90 73
60 68
49 90
55 70
35 76
82 81
82 86
85 63
51 65
78 77
42 82
40 54
82 85
82 59
60 97
76 81
79 76
69 71
90 78
39 85
41 80
42 75
59 48
47 66
61 75
59 94
38 62
44 92
48 86
66 68
71 58
64 95
84 61
60 72
77 83
63 87
62 94
42 94
60 59
73 53
38 67
52 68
40 64
42 65
86 74
83 91
94 68
73 78
42 86
60 74
65 68
57 82
94 62
44 83
39 62
76 78
37 77
62 72
51 61
40 60
55 65
74 69
35 82
90 45
78 70
44 89
66 73
78 92
84 78
52 71
63 85
62 56
88 91
80 53
88 91
65 67
91 87
66 49
36 64
86 79
82 73
73 66
38 66
42 80
39 92
73 80
73 87
93 61
75 83
51 71
47 86
75 73
57 70
66 76
41 56
40 71
59 83
81 84
56 92
41 69
81 88
80 72
49 62
69 75
79 57
69 78
62 83
93 73
76 80
36 84
95 90
80 99
66 78
74 94
68 74
67 64
79 97
37 69
66 89
52 87
81 79
42 57
73 75
55 79
46 88
81 85
92 64
84 66
76 84
36 75
65 97
35 76
61 62
38 82
48 90
44 52
64 81
78 43
61 88
71 67
49 77
65 83
72 85
67 58
40 86
56
71
75
80
94
74
87
89
70
73
66
66
90
84
87
76
94
61
69
78
51
76
87
62
78

Solutions

Expert Solution

since, n>30 in both the samples, we will use a t-test for two independent sample means.

since the Population sample deviation is not available and the samples are independent.

the data is given as

class 1 class 2
50 74
73 79
54 68
90 62
63 55
70 76
70 66
47 81
81 64
93 88
69 97
62 72
38 78
65 81
82 77
48 81
38 85
57 78
65 79
40 62
62 95
44 52
60 100
67 78
63 82
85 67
71 80
69 70
95 72
85 83
90 73
60 68
49 90
55 70
35 76
82 81
82 86
85 63
51 65
78 77
42 82
40 54
82 85
82 59
60 97
76 81
79 76
69 71
90 78
39 85
41 80
42 75
59 48
47 66
61 75
59 94
38 62
44 92
48 86
66 68
71 58
64 95
84 61
60 72
77 83
63 87
62 94
42 94
60 59
73 53
38 67
52 68
40 64
42 65
86 74
83 91
94 68
73 78
42 86
60 74
65 68
57 82
94 62
44 83
39 62
76 78
37 77
62 72
51 61
40 60
55 65
74 69
35 82
90 45
78 70
44 89
66 73
78 92
84 78
52 71
63 85
62 56
88 91
80 53
88 91
65 67
91 87
66 49
36 64
86 79
82 73
73 66
38 66
42 80
39 92
73 80
73 87
93 61
75 83
51 71
47 86
75 73
57 70
66 76
41 56
40 71
59 83
81 84
56 92
41 69
81 88
80 72
49 62
69 75
79 57
69 78
62 83
93 73
76 80
36 84
95 90
80 99
66 78
74 94
68 74
67 64
79 97
37 69
66 89
52 87
81 79
42 57
73 75
55 79
46 88
81 85
92 64
84 66
76 84
36 75
65 97
35 76
61 62
38 82
48 90
44 52
64 81
78 43
61 88
71 67
49 77
65 83
72 85
67 58
40 86
56
71
75
80
94
74
87
89
70
73
66
66
90
84
87
76
94
61
69
78
51
76
87
62
78

the 90% confidence Interval for is given by

where sp is

from the Descriptive summary, we get,

class 1 class 2
Mean 63.44571429 Mean 75.41
Standard Error 1.287902972 Standard Error 0.84619129
Median 65 Median 76
Mode 73 Mode 78
Standard Deviation 17.03735488 Standard Deviation 11.96695198
Sample Variance 290.2714614 Sample Variance 143.2079397
Kurtosis -1.099133088 Kurtosis -0.444786654
Skewness -0.026939637 Skewness -0.255490363
Range 60 Range 57
Minimum 35 Minimum 43
Maximum 95 Maximum 100
Sum 11103 Sum 15082
Count 175 Count 200
Confidence Level(90.0%) 2.129751074 Confidence Level(90.0%) 1.398370662

let X be the values of class1 and Y be the Value of class 2.

alpha = 0.1

therefore, zalpha/2 = 1.645.

63.44571429

75.41

= 290.2714614

= 143.20

n1 = 175,

n2 = 200

sp2 = 211.8112984

t373,0.05= 1.648949026

therefore, the 90% confidence Interval for is (-14.44835, -9.4802181).

therefore, the 90% Confidence Interval for   is ( 9.4802181 , 14.44835).

which is close to option b

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(-1) = zarnavniet (Z - 4)^=a/2

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