Tests of hypothesis Contents: Tests of significance for small samples Student’s t- test Properties of t-Distribution The t- table Application of the t-Distribution To test the significance of Mean of a random sample Means of two samples (independent) Means of two samples (dependent) Observed correlation coefficient Z-transformation Limitations of tests of significance

Tests of significance for small samples Student’s t-Distribution The t-distribution is used when sample size is 30 or less and the population standard deviation is unknown. The “t-statistic” is defined as :

Properties of t-Distribution The variable t- distribution ranges from minus infinity to plus infinity. The constant c is actually a function of , so that for particular value of , the distribution of (t) is completely specified. Thus (t) is a family of functions, one for each values of . The t-distribution is symmetrical and has a mean zero. The variance of it is greater than one.

Variance approaches one as the number of degree of freedom and the sample size becomes large. COMPARISON OF NORMAL & T-DISTRIBUTION

The t-Table The t-table is the probability integral of t- distribution. Over a range of values of , the probability of exceeding by chance value of t at different levels of significance. The t-distribution has a different value for each degree of freedom. When degrees of freedom are infinitely large, the t-distribution is equivalent to normal distribution & values given in the table are applicable.

Applications of t- distribution To test the significance of mean of random sample d= deviation from the assumed mean.

Example: a random sample of size 16 has 53 as mean Example: a random sample of size 16 has 53 as mean. The sum of squares of the deviations taken from mean is 135. can this sample be regarded as taken from the population having 56 as mean ? Solution. Let us take the hypothesis that there is no significant difference between the sample mean and hypothetical population mean. Applying t test : t = X - µ √n X=53 Sµ=56, n= 16 ∑(X-X) = 135 S= ∑(X-X) = 135 = 3 n-1 15 t= 53-56 √16 = 3*4 = 4 3 3 = 16-1 = 15. for = 16 , t 0.05 = 2.13. The calculated value of t  the table value. The hypothesis is rejected.

Testing difference between means of two samples (independent samples)

When actual means are in fraction the deviation should be taken from assumed mean The degree of freedom = (n1-n2-2).

If number of observations and standard deviation of two samples is given:

Example : the mean life of a sample of 10 electric light bulbs was found to be 1,456 hours with standard deviation of 423 hours. A second sample 17 bulbs chosen from a different batch showed a mean life of 1,280 hours with standard deviation of 398 hours. Is there a significant difference between the means of the two batches ? Solution : let us take the hypothesis that the means of two batches do not differ significantly. Applying t- test

Testing difference between means of two samples (dependent samples) The t- test based on paired observation is defined by the following formula Note : t is based on (n-1) degrees of freedom.

Example : a drug is given to 10 patients, and the increments in their blood pressure were recorded to be 3, 6, -2, 4, -3, 4, 6, 0, 0, 2. is it reasonable to believe that the drug has no side effect on change of blood pressure ? Solution : let us take the hypothesis that the drug has no effect on change of blood pressure. Applying the difference test:

The calculated value  the table value. The hypothesis is accepted . Hence it is reasonable to believe that drug has no effect on change of blood pressure.

Testing the significance of an observed correlation coefficient If we are to test the hypothesis that the correlation coefficient of population is zero, we have to apply the following test : Note : Here t is based on (n-2) degrees of freedom If calculated value of t exceeds t0.005 for (n-2) , we say that the value of r is significant at 5% level.

Example : A random sample of 27 pairs of observations from a normal population gives a correlation coefficient of 0.42. is it likely that the variables in the population are uncorrelated ? Solution : let us take the hypothesis that there is no significant difference in the sample correlation and correlation in population. Applying test: For =25, t0.005= 1.708. The calculated value > the table value. The hypothesis is rejected.

Z-test of the significance of correlation coefficient The Z- test given by Prof. Fisher is used to test: Whether an observed value of r differs significantly from some hypothetical value, or Whether two sample value of r differ significantly. The Z-test is defined as following:

Limitations of tests of significance They should not be used mechanically. Conclusions are to be given in terms of probabilities and not certainties. They do not tell us “why” the difference exists. Serious violation of of assumptions. Non-publication of non-significant results.

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