In: Statistics and Probability
A researcher at the college used a random sample of 30 students to investigate sleep patterns of two groups. The output below was constructed from this study for comparing average sleep hours of students who have early classes versus students who do not have early classes.
The hours of sleep for both group of students are stored in EarlyClass.xlsx. Download and open this file into Excel. Use Excel to undertake a suitable test to address the research question below.
NoEarlyClass | EarlyClass |
7.68 | 7.25 |
10.62 | 9.64 |
8.3 | 8.66 |
8.61 | 8.36 |
6.35 | 6.36 |
6.43 | 9.17 |
9.87 | 7.96 |
8.75 | 8.75 |
7.18 | 9.05 |
7.87 | 6.89 |
7.52 | 8.71 |
8.54 | 7.54 |
7.14 | 8.21 |
9.57 | 5.78 |
8.07 | 6.68 |
7.98 | 7.98 |
9.06 | 7.05 |
8.79 | 7.25 |
8.75 | 9.25 |
8.49 | 7.71 |
6.43 | 8.39 |
8.43 | 7.45 |
7.88 | 6.82 |
6.93 | 8.24 |
7.82 | 6.5 |
9.19 | 8 |
8.68 | 5.96 |
8.21 | 6.14 |
9 | 9.39 |
7.5 | 8.61 |
Is there a difference between the average hours of sleep for students with early classes versus students with no early classes?
Answer the following questions by choosing the most correct option or typing the answer:
(1 mark) The most appropriate test for these data is: AnswerA two-sample t-testA paired t-test
(1 mark) The normality assumption seems reasonable because: Answerboth sleep hours for students with early classes and no early classes may be drawn from normal distributions.the differences between sleep hours of students with no early classes and with early classes may be drawn from a normal distribution.
(1 mark) The assumption of equal variance seems reasonable because: Answercomparative boxplots suggest the variation in the two populations could be the same.comparative boxplots suggest the variation in the two populations is different.
For the remaining questions you may assume that any relevant assumptions have been met.
The absolute value of the test statistic is equal to (type your answer with 3 dp) Answer
(1 mark) The degrees of freedom is equal to (type your answer as an integer) Answer
(1 mark) The p-value is larger than 0.05 Answertruefalse
The test shows that the average hours of sleep was Answersignificantly greater for students with no early classes than for students with early classes.significantly greater for students with early classes than for students with no early classes.not significantly different for students with no early classes and for students with early classes.
please include working out if possible
Answer(1):
The two groups No Early Class and Early Class are independent groups because sleeping hours of No Early Class is not related to sleeping hours of Early Class.
Hence the correct option is A two-sample t-test
Answer(2):
The normality assumption seems reasonable because: both sleep hours for students with early classes and no early classes may be drawn from normal distributions.
Answer(3):
The boxplots of both groups are
The assumption of equal variance seems reasonable because: comparative boxplots suggest the variation in the two populations could be the same.
Answer(4):
NoEarlyClass(x1) |
x12 |
EarlyClass(x2) |
x22 |
|
7.68 |
58.982 |
7.25 |
52.5625 |
|
10.62 |
112.78 |
9.64 |
92.9296 |
|
8.3 |
68.89 |
8.66 |
74.9956 |
|
8.61 |
74.132 |
8.36 |
69.8896 |
|
6.35 |
40.323 |
6.36 |
40.4496 |
|
6.43 |
41.345 |
9.17 |
84.0889 |
|
9.87 |
97.417 |
7.96 |
63.3616 |
|
8.75 |
76.563 |
8.75 |
76.5625 |
|
7.18 |
51.552 |
9.05 |
81.9025 |
|
7.87 |
61.937 |
6.89 |
47.4721 |
|
7.52 |
56.55 |
8.71 |
75.8641 |
|
8.54 |
72.932 |
7.54 |
56.8516 |
|
7.14 |
50.98 |
8.21 |
67.4041 |
|
9.57 |
91.585 |
5.78 |
33.4084 |
|
8.07 |
65.125 |
6.68 |
44.6224 |
|
7.98 |
63.68 |
7.98 |
63.6804 |
|
9.06 |
82.084 |
7.05 |
49.7025 |
|
8.79 |
77.264 |
7.25 |
52.5625 |
|
8.75 |
76.563 |
9.25 |
85.5625 |
|
8.49 |
72.08 |
7.71 |
59.4441 |
|
6.43 |
41.345 |
8.39 |
70.3921 |
|
8.43 |
71.065 |
7.45 |
55.5025 |
|
7.88 |
62.094 |
6.82 |
46.5124 |
|
6.93 |
48.025 |
8.24 |
67.8976 |
|
7.82 |
61.152 |
6.5 |
42.25 |
|
9.19 |
84.456 |
8 |
64 |
|
8.68 |
75.342 |
5.96 |
35.5216 |
|
8.21 |
67.404 |
6.14 |
37.6996 |
|
9 |
81 |
9.39 |
88.1721 |
|
7.5 |
56.25 |
8.61 |
74.1321 |
|
Total |
245.64 |
2040.901 |
233.75 |
1855.397 |
We have to test the below hypotheses
H0: there is no difference in sleeping hours of two groups i.e. µ1=µ2
HA: there is significant difference in sleeping hours of two groups i.e. µ1 ≠ µ2
The sample mean for sample 1,
The sample mean for sample 2,
The test statistic to test the null hypothesis is
Where sp is pooled standard deviation
The test statistic is
The value of test statistic is 1.465
Answer(5):
Total degrees of freedom is
dftotal =n1 +n2 -2
dftotal =30+30-2
dftotal =58
Answer(6):
The p-value is larger than 0.05 – True
Answer(7):
The test shows that the average hours of sleep was -not significantly different for students with no early classes and for students with early classes.