The vice president of the student organization has asked you to create an estimate of the proportion of students who have attained more than 60 credit hours. You ask her what level of confidence she wants to have in the estimate and she says that she would like to be 90% confident in the proportion estimate. Create the confidence interval using the sample data. Write a sentence or two to communicate the results to her.
Student | More than 60 credit hours attained |
1 | No |
2 | No |
3 | No |
4 | No |
5 | No |
6 | No |
7 | No |
8 | No |
9 | No |
10 | No |
11 | No |
12 | No |
13 | No |
14 | No |
15 | No |
16 | No |
17 | No |
18 | No |
19 | No |
20 | No |
21 | No |
22 | No |
23 | No |
24 | No |
25 | No |
26 | No |
27 | No |
28 | No |
29 | No |
30 | No |
31 | No |
32 | No |
33 | No |
34 | No |
35 | No |
36 | No |
37 | No |
38 | No |
39 | No |
40 | No |
41 | No |
42 | No |
43 | No |
44 | No |
45 | No |
46 | No |
47 | No |
48 | No |
49 | No |
50 | No |
51 | No |
52 | No |
53 | No |
54 | Yes |
55 | Yes |
56 | Yes |
57 | Yes |
58 | Yes |
59 | Yes |
60 | Yes |
61 | Yes |
62 | Yes |
63 | Yes |
64 | Yes |
65 | Yes |
66 | Yes |
67 | Yes |
68 | Yes |
69 | Yes |
70 | Yes |
71 | Yes |
72 | Yes |
73 | Yes |
74 | Yes |
75 | Yes |
76 | Yes |
77 | Yes |
78 | Yes |
79 | Yes |
80 | Yes |
81 | Yes |
82 | Yes |
83 | Yes |
84 | Yes |
85 | Yes |
86 | Yes |
87 | Yes |
88 | Yes |
89 | Yes |
Number of students who have attained more than 60 credit hours = 36
proportion = p = 36/89 = 0.4045
confidence interval is
given by:
p = proportion = 0.4045
z = z value for desired confindence level of 90% = 1.645
n= numbe of students = 89
Thus putting the values in above equation gives
confidence interval = (0.3189 , 0.4901)
Result : Thus we have 90% confidence that proportion of students who have attained more than 60 credit hours is between 31.89% and 49.01%.
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Thanks.
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