1. An investigation is carried out to determine whether face to face or virtual learning is more effective. If 12 students took an IT course virtual and 14 took it face to face: The data of the scores made for both types of learning are shown in the table. Virtual 66 75 85 64 88 77 74 91 72 69 77 83 Face to Face 80 83 64 81 75 80 86 81 51 64 59 85 74 77 Do the following: a) explain the type of data you have here and the hypothesis test you would use to determine the difference between the two types b) determine whether you can conclude that the mean score differs between the two types of course. You need to first set up an alternate and null hypothesis c) construct a pdf for both sets of data and explain the shape of the graph d) calculate a 95% confidence interval for each set of data. e) Determine the error in the sample mean
Percentage
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Do part d
Given :
An investigation is carried out to determine whether face to face or virtual learning is more effective . If 12 student took an IT course virtual and 14 took it face to face . Data represents scores made for both types of learning .
Find mean and standard deviation for both the sample data :
For virtual :
The sample size is . The provided sample data along with the data required to compute the sample mean and sample variance are shown in the table below:
X | X2 | |
66 | 4356 | |
75 | 5625 | |
85 | 7225 | |
64 | 4096 | |
88 | 7744 | |
77 | 5929 | |
74 | 5476 | |
91 | 8281 | |
72 | 5184 | |
69 | 4761 | |
77 | 5929 | |
83 | 6889 | |
Sum = | 921 | 71495 |
The sample mean is computed as follows:
For face to face :
The sample size is . The provided sample data along with the data required to compute the sample mean and sample variance are shown in the table below:
X | X2 | |
80 | 6400 | |
83 | 6889 | |
64 | 4096 | |
81 | 6561 | |
75 | 5625 | |
80 | 6400 | |
86 | 7396 | |
81 | 6561 | |
51 | 2601 | |
64 | 4096 | |
59 | 3481 | |
85 | 7225 | |
74 | 5476 | |
77 | 5929 | |
Sum = | 1040 | 78736 |
The sample mean is computed as follows:
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