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The t test
prepared by B.saikiran (12NA1E0036)
Introduction
ν The t-test assumes:
Applications
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occasions.
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1.Sample mean and population mean
ν Degrees of freedom = n - 1
−
t = X −μ
SE
Example
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The following data represents hemoglobin values in gm/dl for 10 patients:
10.5 9 6.5 8 11
7 7.5 8.5 9.5 12
Is the mean value for patients significantly differ from the mean value of general population
(12 gm/dl) . Evaluate the role of chance.
Solution
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ν
Mention all steps of testing hypothesis.
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Then compare with tabulated value, for 9 df, and 5% level of significance. It is = 2.262
The calculated value>tabulated value.
Reject Ho and conclude that there is a statistically significant difference
between the mean of sample and population mean, and this difference is unlikely due to chance.
1.80201
t = 8.95 − 12 = −5.352
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2.Two independent samples
The following data represents weight in Kg for 10 males and 12 females.
Males:
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| 80 | 75 | 95 | 55 | | 60 |
70 | 75 | 72 | 80 | | 65 | |
Females: | 60 | 70 | 50 | 85 | 45 | 60 |
| 80 | 65 | 70 | 62 | 77 | 82 |
2.Two independent samples, cont.
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1 1
2 2
1 1 2 2 (
n1 n2
+ )
n1 + n2 − 2
(n −1)S + (n −1)S
X 1 − X 2
t =
−
−
Results
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ν Mean1=72.7 Mean2=67.17
ν Variance1=128.46 Variance2=157.787
ν Df = n1+n2-2=20
ν t = 1.074
ν P>0.01
3.One sample in two occasions
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n
sd
t
−
=d
−
n n − 1
(∑ d )2
∑ d 2
sd =
Example: Blood pressure of 8 patients, before & after treatment
BP before | BP after | d d2 | |
180 | 140 | 40 | 1600 |
200 | 145 | 55 | 3025 |
230 | 150 | 80 | 6400 |
240 | 155 | 85 | 7225 |
170 | 120 | 50 | 2500 |
190 | 130 | 60 | 3600 |
200 | 140 | 60 | 3600 |
165 | 130 | 35 | 1225 |
Mean d=465/8=58.125
∑d=465
∑d2=2917513
Results and conclusion
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ν t=9.387
ν P<0.05.