Help with the following problem urget, bellow is the csv data.
Campus | Online |
153.83 | 148.66 |
212.21 | 214.58 |
191.65 | 191.24 |
142.49 | 141.73 |
154.71 | 152.62 |
223.36 | 224.79 |
93.93 | 92.87 |
112.21 | 106.41 |
182.58 | 185.15 |
179.55 | 177.33 |
187.51 | 185.17 |
168.75 | 167.08 |
SAMPLE 1 | SAMPLE 2 | difference , Di =sample1-sample2 | (Di - Dbar)² |
153.83 | 148.66 | 5.17 | 15.269 |
212.21 | 214.58 | -2.37 | 13.195 |
191.65 | 191.24 | 0.41 | 0.727 |
142.49 | 141.73 | 0.76 | 0.253 |
154.71 | 152.62 | 2.09 | 0.685 |
223.36 | 224.79 | -1.43 | 7.250 |
93.93 | 92.87 | 1.06 | 0.041 |
112.21 | 106.41 | 5.8 | 20.589 |
182.58 | 185.15 | -2.57 | 14.688 |
179.55 | 177.33 | 2.22 | 0.917 |
187.51 | 185.17 | 2.340 | 1.161 |
168.75 | 167.08 | 1.670 | 0.166 |
mean of difference , D̅ =ΣDi / n =
1.2625
std dev of difference , Sd = √ [ (Di-Dbar)²/(n-1) =
2.6101
============
a) option e)
Ho : µd= 0
Ha : µd > 0
b)
std error , SE = Sd / √n = 2.6101 /
√ 12 = 0.7535
t-statistic = (D̅ - µd)/SE = ( 1.2625
- 0 ) / 0.7535
= 1.6756
Degree of freedom, DF= n - 1 =
11
p-value =
0.0610 [excel function: =t.dist.rt(t-stat,df)
]
c)
Decision: p-value>α , Do not reject null
hypothesis
answer: not reject
d) option A)
Help with the following problem urget, bellow is the csv data. Campus Online 153.83 148.66 212.21...
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Answer the case study questions please and thank you
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