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b. Hypothesis testing for the ratio of Two Population variances
3. Tests for Proportions a. Hypothesis Testing for a Single Population Proportion For quantitative measurement of variables, the presence or absence of a particular characteristic is found out. If p indicates presence of a particular characteristic and q indicates the absence of a particular characteristic, the sampling distribution of p is approximated by normal distribution in accordance with the central limit theorem. The Test Hypothesis is H0 : p = p0 (null hypothesis) H1 : p ≠ p0 (2-sided alternative) H2 : p > p0 (1-sided alternative) H3 : p < p0 (1-sided alternative) where, p0 is the specified value of the sampling distribution p for a given sample . The test statistic used is
Decision Rule The test is reject H0, if the calculated value is greater than tabulated value i.e. (|Z|>Zα/2), then the test is significant. This means there is a significant difference between actual and estimated sample proportion. For example: A survey on people affected by cancer in a city found that 18 out of 423 were affected. You wish to know if you can conclude that fewer than 5 percent of people in the sampled population are affected by cancer at 5 percent level of significance. Solution: From the data it can be concluded that from the response of 423 individuals of which 18 possessed the characteristic of interest, p = 18/423 =0.426. The test hypothesis is H0 : p<0.05 HA : p≥0.05 The test statistic is
This shows that the population proportion of cancer affected people may be 0.05 or more. 4. Paired Comparisons Tests Definition: For a difference between two population means, it was assumed that the samples are dependent. This means that the values of observations in one sample depend on the other in any significant manner. In fact, the two samples may consist of pairs of observations made on the same object or individual. For example, to find out the effect of training on some employees or to find out the efficacy of a coaching class, a hypothesis test based on this type of data is known as a paired comparisons test. Given are two dependent random samples of size n 1 and n 2 from the same population having normal distribution. The test hypothesis used in this case is
Decision Rule The test is reject H0, if the calculated value is greater than the tabulated value i.e. ( |t| > tn-1,α/2), then the test is significant for a 2-sided alternative and for ( t > tn-1, α) the test is significant for a 1-sided alternative. This means that there is a difference between the actual mean and estimated mean. For example: 12 students were given intensive coaching and 4 tests were conducted in a month. The scores of tests 1 and 4 are given below. Do the scores from the 1 to 4 show an improvement?
Hence the course has been useful. 5. Goodness of Fit Tests Definition: The goodness of fit test enables one to ascertain how appropriately theoretical distributions of frequencies fit expected distributions from the sample. When an ideal frequency curve is fitted to the data, it is found out that how this curve fits with the observed facts. For a given random sample of observations drawn from a relevant statistical population, the test Hypothesis is H0 : Population is normally distributed HA : Population is not normally distributed The test Statistic is
When H0 is true and the assumptions are met, the test statistic is distributed as chi-square (X2) distribution with k-r degree of freedom. The level of significance α = 0.05 or 0.01 is known. The tabulated value is X2α/2,n-1 for an upper side and X21-α/2,n-1 for a lower side in a two-sided alternative. The tabulated value is X2α,n-1 for upper side and X21-α,n-1 for lower side for a one-sided alternative. Here, k = Number of groups for which observed and expected frequencies are available. r = Number of restrictions imposed on the comparison.
For Example: In an accounting department of a bank 50 accounts are selected at random and examined for errors. The following results have been obtained: No. of errors: 0 1 2 3 4 5 No. of accounts: 6 13 13 8 4 3 On the basis of this information, can it be concluded that the errors are distributed according to the Poisson process? Solution: To solve this question you have to obtain the Expected frequencies by supplying Poisson distribution and test the goodness of fit by X2 test. The probability of Poisson distribution is given by
Observed Frequency (O): 6 13 13 8 4 3 Expected Frequency (E): 6.24 13.52 13.52 9.01 4.50 1.80 Test Hypothesis H0 : Errors are distributed according to Poisson distribution. HA : Errors are not distributed according to Poisson distribution. The Test Statistic is
Since calculated value is smaller than tabulated value, test is accept H0 i.e. accepting the null hypothesis. Hence the given information verify that the errors are distributed according to Poisson distribution. |
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