In: Statistics and Probability
a)
Expected frequency of a cell = sum of row*sum of column / total sum | |||||||
Expected Frequencies | |||||||
E | G | Total | |||||
N | 24*20/50=9.6 | 26*20/50=10.4 | 20 | ||||
M | 24*30/50=14.4 | 26*30/50=15.6 | 30 |
(fo-fe)^2/fe | ||||||
N | 0.267 | 0.246 | ||||
M | 0.178 | 0.164 |
Ho: given two variable are independent
H1: Given two variables are not independent
Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =
0.855
Level of Significance = 0.05
Number of Rows = 2
Number of Columns = 2
Degrees of Freedom=(#row - 1)(#column -1) = (2- 1 ) * ( 2- 1 )
= 1
p-Value = 0.3552 [Excel function:
=CHISQ.DIST.RT(χ²,df) ]
Decision: p value > α , do not reject
Ho
We can conclude the results of each coin were independent
c)
Expected frequency of a cell = sum of row*sum of column / total sum | |||||||
Expected Frequencies | |||||||
E | G | Total | |||||
N | 359*402/750=192.424 | 391*402/750=209.576 | 402 | ||||
M | 359*348/750=166.576 | 391*348/750=181.424 | 348 |
(fo-fe)^2/fe | ||||||
N | 0.102 | 0.093 | ||||
M | 0.117 | 0.108 |
Ho: given two variable are independent
H1: Given two variables are not independent
Chi-Square Test Statistic,χ² = Σ(fo-fe)^2/fe =
0.420
Level of Significance = 0.05
Number of Rows = 2
Number of Columns = 2
Degrees of Freedom=(#row - 1)(#column -1) = (2- 1 ) * ( 2- 1 )
= 1
p-Value = 0.5167 [Excel function:
=CHISQ.DIST.RT(χ²,df) ]
Decision: p value > α , do not reject
Ho
We can conclude the results of each coin were independent
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