Assignment
Please conduct hypothesis tests for the following questions.
1) A research study was conducted to examine the efficacy of studying in groups. Students were randomly assigned to one of three groups: a single person group (individual studying), a group that had two study partners, and a group that had three study partners. After four weeks of studying, the students were given an exam. The raw scores are below. The data are presented below.
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Group 1
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Group 2
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Group 3
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|
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78
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82
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74
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|
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87
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79
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86
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|
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76
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82
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86
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65
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83
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85
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92
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81
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95
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78
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90
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94
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77
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89
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86
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|
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88
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79
|
71
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|
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85
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83
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88
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|
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92
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81
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85
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Mean
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81.8
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82.9
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85
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Standard deviation
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8.508819
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3.754997
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7.527727
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H0: μ1 = μ2 = u3 = μr, all the means are the same
H1: two or more means are different from the others
Conclusion The sample means have different means and thus we reject he null hypothesis because the mean for group 1 is 81.8 while group 2 is 82.9 and group 3 is 85.
2) Mr. Rotondo is concerned about the level of knowledge possessed by various majors regarding United States history. Students from various majors in the three academic areas of the College were asked to complete a standardized U.S. history exam. The Academic area for students was also recorded. Data in terms of percent correct is recorded below for 30 students.
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Natural Science
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Social Science
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Humanities
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62
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82
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80
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81
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92
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87
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75
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81
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87
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|
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88
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80
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74
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|
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67
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72
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88
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68
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71
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79
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76
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68
|
82
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|
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86
|
76
|
85
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|
|
82
|
68
|
72
|
|
|
76
|
86
|
65
|
|
Mean
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76.1
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77.6
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79.9
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Standard deviation
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8.478076
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8.002777
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7.578478
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H0: μn = μs= uh = μr, all the means are the same
H1: two or more means are different from the others
Conclusion
The sample means have different means and thus we reject he null hypothesis because the mean for natural science is 76.1 while social science is 77.6 and humanities is 79.9 and the variance is different
3) A genetics engineer was attempting to cross a tiger and a cheetah. She crossed 150 tigers and cheetahs. Based on prior research she expected that 25% of the animals would have a phenotypic outcome of stripes only: 25% would have 3 spots only: and 50% would have both stripes and spots. When the cross was performed and she counted the individual cubs she found 45 with stripes only, 41 with spots only and 64 with both.
Chi-square = S (O-E)2/E
D.F. Value
1 3.841
2 5.991
3 7.815
Set up a table to keep track of the calculations:
Expected ratio Observed # Expected # O-E (O-E)2 (O-E)2/E
25% stripes 45 37.5 7.5 56.25 1.5
25% spots 41 37.5 3.5 12.25 0.33
50% stripes/spots 64 75 -11 121 1.61
100 total 150 total 150... total 0 total Sum = 3.44
25% * 150 = expected # of stripes = 37.5
25% * 150 = expected # of spots = 37.5
50% * 150 = expected # stripes/spots = 75
Degrees of Freedom = 3 - 1 = 2 (3 different characteristics - stripes, spots, or both) Since 3.44 is less than 5.991, we fail to reject the null hypothesis put forward by the engineer.