In: Statistics and Probability
Consider Dataset C for answering the questions that follows below.
Team |
Gender |
Responses |
|||
1 |
A |
Male |
3.25 |
||
2 |
A |
Male |
3.54 |
||
3 |
A |
Male |
1.08 |
||
4 |
A |
Male |
2.14 |
||
5 |
A |
Male |
3.60 |
||
6 |
B |
Male |
4.36 |
||
7 |
B |
Male |
4.66 |
||
8 |
B |
Male |
1.52 |
||
9 |
B |
Male |
3.99 |
||
10 |
B |
Male |
3.60 |
||
11 |
C |
Female |
3.86 |
||
12 |
C |
Female |
4.89 |
||
13 |
C |
Female |
1.46 |
||
14 |
C |
Female |
4.74 |
||
15 |
C |
Female |
4.16 |
||
A | B | C | ||||
count, ni = | 5 | 5 | 5 | |||
mean , x̅ i = | 2.722 | 3.63 | 3.82 | |||
std. dev., si = | 1.090 | 1.243 | 1.385 | |||
sample variances, si^2 = | 1.188 | 1.544 | 1.920 | |||
total sum | 13.61 | 18.13 | 19.11 | 50.85 | (grand sum) | |
grand mean , x̅̅ = | Σni*x̅i/Σni = | 3.39 |
square of deviation of sample mean from grand mean,( x̅ - x̅̅)² | 0.446 | 0.056 | 0.187 | ||
TOTAL | |||||
SS(between)= SSB = Σn( x̅ - x̅̅)² = | 2.231 | 0.278 | 0.933 | 3.44272 | |
SS(within ) = SSW = Σ(n-1)s² = | 4.754 | 6.176 | 7.678 | 18.6081 |
no. of treatment , k = 3
df between = k-1 = 2
N = Σn = 15
df within = N-k = 12
mean square between groups , MSB = SSB/k-1 =
1.7214
mean square within groups , MSW = SSW/N-k =
1.5507
F-stat = MSB/MSW = 1.1101
SS | df | MS | F | p-value | F-critical | |
Between: | 3.44 | 2 | 1.72 | 1.11 | 0.3611 | 3.89 |
Within: | 18.61 | 12 | 1.55 | |||
Total: | 22.05 | 14 | ||||
α = | 0.05 | |||||
conclusion : | p-value>α , do not reject null hypothesis |
Ho: µ1=µ2=µ3
H1: not all means are equal
We can conclude that all means are equal
b)
Ho : µ1 - µ2 = 0
Ha : µ1-µ2 ╪ 0
Level of Significance , α =
0.01
Sample #1 ----> Male
mean of sample 1, x̅1= 3.17
standard deviation of sample 1, s1 =
1.20
size of sample 1, n1= 10
Sample #2 ----> Female
mean of sample 2, x̅2= 3.82
standard deviation of sample 2, s2 =
1.39
size of sample 2, n2= 5
difference in sample means = x̅1-x̅2 =
3.1740 - 3.8 =
-0.65
pooled std dev , Sp= √([(n1 - 1)s1² + (n2 -
1)s2²]/(n1+n2-2)) = N/A
std error , SE = Sp*√(1/n1+1/n2) =
0.7267
t-statistic = ((x̅1-x̅2)-µd)/SE = ( -0.6480
- 0 ) / 0.73
= -0.892
Degree of freedom, DF= n1+n2-2 =
7
t-critical value , t* =
3.4995 (excel formula =t.inv(α/2,df)
Decision: | t-stat | < | critical value |, so, Do
not Reject Ho
p-value = 0.402144
(excel function: =T.DIST.2T(t stat,df) )
Conclusion: p-value>α , Do not reject null
hypothesis
There is not enough evidence that male mean is different than femal
mean
Thanks in advance!
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