Student's t test in Hindi | t test distribution in Hindi | classification of t test in Hindi
Working rule to solve t-test problem for one sample group:
for this test
tcal=|(x ̅-μ)/(S/√n)|
where S is a standard deviation and defined as
S=√((∑(x-x ̅)〗^2 )/(n-1))
solve the of the following problem :
Q1.)The annual rainfall in Lucknow city in normally distributed with mean 45 cm.The rainfall during the last five years are 48 cm, 42 cm, 40 cm, 44 cm and 43 cm respectively. Can we concluded that the average rainfall during the rainfall last five years, is less than the normal rainfall. Test at 5% Level of significance. The tabulated value
Q.2) 10 individuals are chosen at random from a population and their height are found to be inches 63,63,64,65,66,69,69,70,70 and 71. Discuss the suggestion that the mean height of universe is 65 for 9 degree of freedom t at.5% level of significance is 1.833.
Q.3)A sample of 20 items has mean 42 units and standard deviation 5 units. Test the hypothesis that it is a random sample from a normal population with mean 45 units. Given that t0.05=1.729
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t-test / student's t test for two sample groups
Q.1)Two
independent samples of size 7 and 6 has the following values
Examine
whether the samples have been drawn from normal papulation having same variance
Given
that tabular value at 5% level of significance is 1.796 for 11 degree of
freedom
Q.2)Two
independent samples of size 7 and 9 has the following values
Sample
A |
10 |
12 |
10 |
13 |
14 |
11 |
10 |
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Sample
B |
10 |
13 |
15 |
12 |
10 |
14 |
11 |
12 |
11 |
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Test
whether the difference between the mean is significant. |
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Given
that tabular value at 5% level of significance is 1.761 for 14 degree of
freedom
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