Excel output:
|
M car |
J car |
|
|
M car |
J car |
|
|
31 |
27 |
|
Mean |
30.33333 |
27.25 |
|
|
30 |
29 |
|
Variance |
7.623188 |
3.009259 |
|
|
29 |
27 |
|
Std.
dev |
2.761012 |
1.734722 |
|
|
30 |
28 |
|
|
|
|
|
|
33 |
28 |
|
|
|
|
|
|
36 |
29 |
|
t-Test:
Two-Sample Assuming Unequal Variances |
|
|
|
|
31 |
30 |
|
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|
|
|
|
29 |
28 |
|
|
Variable 1 |
Variable 2 |
|
|
28 |
30 |
|
Mean |
30.33333 |
27.25 |
|
|
34 |
25 |
|
Variance |
7.623188 |
3.009259 |
|
|
26 |
27 |
|
Observations |
24 |
28 |
|
|
32 |
25 |
|
Hypothesized Mean Difference |
0 |
|
|
|
28 |
28 |
|
df |
38 |
|
|
|
28 |
26 |
|
t
Stat |
4.729026 |
|
|
|
32 |
24 |
|
P(T<=t) one-tail |
1.54E-05 |
|
|
|
28 |
25 |
|
t
Critical one-tail |
1.685954 |
|
|
|
33 |
31 |
|
P(T<=t) two-tail |
3.08E-05 |
|
|
|
33 |
28 |
|
t
Critical two-tail |
2.024394 |
|
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|
28 |
26 |
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27 |
28 |
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35 |
25 |
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30 |
28 |
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26 |
27 |
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31 |
28 |
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27 |
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|
26 |
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|
28 |
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|
25 |
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You can use following functions and test to find answer:
Mean: =average()
variance: =var()
standard deviation: =stdev() Â Â
Now use \" t test: two sample Assuming unequal variance test \" to
find degrees of freedom, p value
1.
|
|
M car |
J car |
|
Mean |
30.33333 |
27.25 |
|
Variance |
7.623188 |
3.009259 |
| Std.
dev |
2.761012 |
1.734722 |
2.
Degrees of freedom = 38
3.
test statistics =t = 4.729026
p value = 0.0000308
4.
Since p value is smaller than 0.05 we reject null hypothesis of
equal mean at 5 % level of significance and conclude that two
population means are significantly different.
 Â
