{"id":218,"date":"2018-07-10T05:12:12","date_gmt":"2018-07-10T05:12:12","guid":{"rendered":"http:\/\/liscp10.epgpbooks.inflibnet.ac.in\/?post_type=chapter&p=218"},"modified":"2018-12-07T06:40:13","modified_gmt":"2018-12-07T06:40:13","slug":"basics-of-testing-of-hypothesis","status":"publish","type":"chapter","link":"https:\/\/ebooks.inflibnet.ac.in\/liscp10\/chapter\/basics-of-testing-of-hypothesis\/","title":{"rendered":"Basics of Testing of Hypothesis"},"content":{"raw":"
\"epgp<\/a>\r\n<\/span><\/div>\r\n
\r\n\r\n \r\n\r\n \r\n\r\n \r\n\r\nModule Structure<\/strong>\r\n\r\n \r\n\r\nObjectives\r\n\r\nSummary\r\n\r\n \r\n\r\n1.\u00a0\u00a0\u00a0\u00a0\u00a0 Introduction\r\n

1.1\u00a0\u00a0 Why Test Hypothesis?<\/p>\r\n

1.2\u00a0 \u00a0Type of Hypothesis and Notation<\/p>\r\n

1.3\u00a0\u00a0 Errors in Hypothesis and Testing<\/p>\r\n

1.4\u00a0\u00a0 Empirical Test of Hypothesis<\/p>\r\n2.\u00a0\u00a0\u00a0\u00a0\u00a0 Z-test: An Explanation\r\n\r\n3.\u00a0\u00a0\u00a0\u00a0\u00a0 Procedure Involved in Testing og Hypothesis\r\n\r\n4.\u00a0\u00a0\u00a0\u00a0\u00a0 Examples with Different Methods\r\n\r\n5.\u00a0\u00a0\u00a0\u00a0\u00a0 Use of p-Value\r\n\r\n6.\u00a0\u00a0\u00a0\u00a0\u00a0 Z-test: \u03c3 is Unknown\r\n

6.1\u00a0\u00a0 Procedure for t-Test<\/p>\r\n7.\u00a0\u00a0\u00a0\u00a0\u00a0 Difference between two Populations\r\n\r\n8.\u00a0\u00a0\u00a0\u00a0\u00a0 Tests about proportions\r\n

8.1\u00a0\u00a0 Difference between Two Proportions<\/p>\r\n9.\u00a0\u00a0\u00a0\u00a0\u00a0 Tests about Correlation Coefficient\r\n\r\n10. Non-Parametric Tests\r\n

10.1 Chi-Square Test Degrees of freedom<\/p>\r\n

10.2\u00a0 Measures of Association<\/p>\r\n

10.3\u00a0 Goodness-of-Fit Test<\/p>\r\n11.Conclusion References\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\nOBJECTIVES<\/strong>\r\n\r\n \r\n\r\n\u2022\u00a0\u00a0\u00a0\u00a0\u00a0 To study an overview of testing of hypotheses\r\n\r\n\u2022\u00a0\u00a0\u00a0\u00a0\u00a0 To study procedure\/steps in testing of hypotheses and related concepts\r\n\r\n \r\n\r\nSUMMARY<\/strong>\r\n\r\n \r\n

Most often, the research process is incomplete without testing of hypotheses. As such there are two types of hypotheses normally referred in statistical testing viz. Null Hypothesis and Alternative Hypothesis. A research process involves:<\/p>\r\n \r\n\r\n1.\u00a0\u00a0\u00a0\u00a0\u00a0 Identify the general problem(s);\r\n\r\n2.\u00a0\u00a0\u00a0\u00a0\u00a0 Conduct literature search;\r\n\r\n3.\u00a0\u00a0\u00a0\u00a0\u00a0 Decide the design methodology;\r\n\r\n4.\u00a0\u00a0\u00a0\u00a0\u00a0 Collect the data either for the population or for a sample;\r\n\r\n5.\u00a0\u00a0\u00a0\u00a0\u00a0 Analyze the data;\r\n\r\n6.\u00a0\u00a0\u00a0\u00a0\u00a0 Report the result; and\r\n\r\n7.\u00a0\u00a0\u00a0\u00a0\u00a0 Refine the hypotheses.\r\n\r\n \r\n

It is in step 1, we generally formulate the hypotheses and in step 5, we test the hypotheses. Based on the results of testing of hypotheses, we generalize the results. The Unit 19 discusses basically the z-test, t-test and the Chi Square test.<\/p>\r\n \r\n

1 INTRODUCTION<\/strong><\/p>\r\n \r\n

Statistical analysis aims at inferring about a population based on the information\/data contained in a sample. There are methods for making inferences which are usually based on statistical tests of hypotheses. Below we discuss only those aspects concerning testing of hypothesis.<\/span><\/p>\r\n \r\n

A hypothesis is a well-defined statement. However, the word hypothesis in science generally refers to a definite interpretation of a given set of facts, which is put forth as a tentative assumptions and remain partially or wholly unverified. A simple definition of hypothesis as given by Luniberg \u201cis a tentative generalisation, the validity of which remains to be tested. In this context testing of hypothesis, with relevant statistical data becomes important either to accept or reject the tentative assumption<\/span><\/p>\r\n \r\n

1.1 Why Test Hypothesis?<\/strong><\/p>\r\n \r\n

Science does not accept anything as valid knowledge, until satisfactory tests confirm its validity. Therefore they need to be tested through research process for their acceptance or rejection. The hypothesis is normally tested by making use of a pre-defined assertion, rule which is applied to sample data and direct the research process in deciding to accept or reject the hypothesis. The process of testing hypothesis embodies the major part of research process. A hypothesis is tested on the basis of facts.<\/span><\/p>\r\n \r\n

1.2\u00a0 Types of Hypotheses and Notations<\/strong><\/p>\r\n \r\n

The two hypotheses in a statistical test are normally referred to as:<\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\na)\u00a0\u00a0\u00a0\u00a0\u00a0 Null Hypothesis, and\r\n\r\nb)\u00a0\u00a0\u00a0\u00a0 Alternative Hypothesis.\r\n\r\n \r\n

a) The Null Hypothesis is a very useful tool in testing the significance of difference. In its simple form, the hypothesis asserts that there is no true difference in the sample and population in particular matter under consideration and that thedifference found is accidental, unimportant, arising out of fluctuations of sampling. A simple definition of hypothesis is that it is a hypothesis which is being tested.<\/p>\r\n \r\n

b) The Alternative hypothesis specifies those values that the researcher considers to be true, and hopes that the sample data leads to acceptance of this whole hypothesis as true. In other words, when a null hypothesis is rejected, then alternative hypothesis is likely to be accepted. For example,<\/span><\/p>\r\n\r\n<\/div>\r\n

<\/div>\r\n
\r\n\r\n \r\n\r\nH0: \u00b5 = \u00b50\r\n\r\nH1: \u00b5 \u2260 \u00b50\r\n

where \u00b5 is the population mean and \u00b50 is the hypothesised value of the population mean.<\/p>\r\n \r\n\r\n1.3\u00a0 Errors in Hypothesis Testing<\/strong>\r\n\r\n \r\n

When accepting or rejecting a null hypothesis, we may commit an error. For instance, we may reject H<\/strong>o<\/strong> when it is correct; and accept H<\/strong>o<\/strong> when it is not correct. These two errors are called Type I error and Type II error respectively. The probability of making a Type I error is denoted by \u03b1. The probability of making a Type II error is denoted as \u03b2. This is shown in the tabular form as below:<\/p>\r\n \r\n\r\n\r\n\r\n\r\n\r\n
Conclusion of the Test<\/strong><\/td>\r\nH<\/strong>0<\/strong>\u00a0 True<\/strong><\/td>\r\nH<\/strong>0<\/strong>\u00a0 False<\/strong><\/td>\r\n<\/tr>\r\n
Accept H0<\/td>\r\nCorrect<\/td>\r\nWrong (Type II Error)<\/td>\r\n<\/tr>\r\n
Reject H0<\/td>\r\nWrong (Type I Error)<\/td>\r\nCorrect<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n
<\/div>\r\n
1.4\u00a0 Empirical Test of Hypothesis<\/strong><\/div>\r\n
\r\n\r\n \r\n

For the purpose of understanding the testing of hypotheses, let us discuss an experimental situation. Consider a condition of verifying the manufacturer\u2019s statement about its product. For example, let us take a case of investigating the container weights specified on the labels of wheat products of the manufacturer. In order to demonstrate the hypothesis testing procedure, let us show how a test on label accuracy could be made for the company\u2019s 2 kg packet of wheat flour.<\/p>\r\n \r\n

The first assumption is that labels are correct. This assumption or hypothesis is subjected to a test by providing evidence regarding the truth of the claim or assumption. There are three possibilities in the case of 2 k.g. wheat flour packets. It is possible that the mean weight for the population of 2 kg packets could be;<\/p>\r\n \r\n\r\ni)\u00a0 \u00a0 \u00a0 \u2265 2 kg or\r\n\r\nii)\u00a0\u00a0\u00a0 \u2264\u00a0 2 kg or\r\n\r\niii)\u00a0 =\u00a0 2 kg.\r\n\r\n<\/div>\r\n

\r\n

In this situation, we have to determine whether or not the population mean (of wheat flour packets) \u00b5<\/strong> =<\/strong> \u00b5<\/strong>0<\/strong> (say, 2).<\/strong> How to determine? This is discussed below:<\/p>\r\n \r\n

A statement like \u00b5<\/strong> =<\/strong> \u00b5<\/strong>0<\/strong> or<\/em><\/strong> \u00b5<\/strong> \u2265<\/strong> \u00b5<\/strong>0<\/strong> or<\/strong> \u00b5<\/strong> \u2264<\/strong> \u00b5<\/strong>0<\/strong> is called a hypothesis. As said in section 1.2, a hypothesis that is being tested is called the Null hypothesis<\/strong>. It is denoted by H<\/strong>o<\/strong>.<\/strong> The hypothesis that we are willing to accept if we do not accept the null hypothesis is called the Alternative hypothesis.<\/strong> The two hypotheses, the Null Hypothesis (<\/em>H<\/strong> o<\/strong>)<\/strong> and the Alternative Hypothesis (H<\/strong>1<\/strong> )<\/strong> are so constructed that if one is correct the other is wrong. It is denoted by H<\/strong>1<\/strong> .<\/strong> Generally, the Null Hypothesis and Alternative Hypothesis for testing the mean will be shown in the following forms:<\/p>\r\n \r\n\r\n\r\n\r\n\r\n\r\n\r\n
<\/td>\r\n<\/td>\r\nNull Hypothesis<\/td>\r\n<\/td>\r\nAlternative Hypothesis<\/td>\r\n<\/tr>\r\n
\u2022<\/td>\r\nCase 1:<\/td>\r\nH0 : \u00b5<\/strong> \u2265<\/strong> \u00b5<\/strong>0<\/strong><\/td>\r\nor<\/strong><\/td>\r\nH1 : \u00b5<\/strong> <<\/strong> \u00b5<\/strong>0<\/strong><\/td>\r\n<\/tr>\r\n
\u2022<\/td>\r\nCase 2:<\/td>\r\nH0 : \u00b5<\/strong> \u2264<\/strong> \u00b5<\/strong>0<\/strong><\/td>\r\nor<\/strong><\/td>\r\nH1 : \u00b5<\/strong> ><\/strong> \u00b5<\/strong>0<\/strong><\/td>\r\n<\/tr>\r\n
\u2022<\/td>\r\nCase 3:<\/td>\r\nH0 : \u00b5<\/strong> =<\/strong> \u00b5<\/strong>0<\/strong><\/td>\r\nor<\/strong><\/td>\r\nH1 : \u00b5 \u2260 \u00b5<\/strong>0<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n

In establishing the critical value for a particular hypothesis testing situation, we always assume that the H<\/strong>o<\/strong> holds as equality<\/strong>. This allows us to control the maximum probability of Type I error. Thus, in cases 1-3 above, the null hypotheses may be treated as H0: \u00b5<\/strong> =<\/strong> \u00b5<\/strong>0.<\/strong> Now with an assumption that the null hypothesis is true, let us select a sample from the population. If the sample results do not differ significantly from the assumed null hypothesis, we accept H<\/strong>0<\/strong> as being true. If the<\/strong> sample results differ <\/strong>significantly from the hypothesis, we reject H<\/strong>0<\/strong> and conclude that the alternate hypothesis H<\/strong>1<\/strong> is true.<\/p>\r\n \r\n\r\n2\u00a0\u00a0\u00a0\u00a0 Z-TEST: AN EXPLANATION<\/strong>\r\n\r\n \r\n

In z-test, the distribution of the test statistic under the null hypothesis is approximated by a normal distribution. From the central limit theorem, we know that the sample mean \u00a0follows normal distribution with mean \u00b5 and standard<\/p>\r\n \r\n\r\n\"\"\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\n\"\"\r\n

So, while testing a hypothesis, if the computed value of z is greater than | z\u03b1\/2 | (if \u03b1 = 0.05, |z\u03b1\/2| = 1.96), the value of z falls in the critical region. Hence we reject Ho.<\/em> In other words, if the experiment is conducted a number of times and follow the test procedure mentioned above, it is likely that 5% of the times, we may commit Type I error \u2014 rejecting Ho when it should have been accepted. While computing the value of z from the sample data, we may use the sample standard deviation instead of z<\/em> which is usually unknown.<\/p>\r\n \r\n

Most often, as mentioned earlier, we test the null hypothesis H<\/strong>o<\/strong>: \u00b5 = \u00b5<\/strong>0<\/strong> against one of the following alternative hypothesis:<\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\n1)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong>H <\/strong>1<\/strong> : \u00b5 < \u00b5<\/strong>0<\/strong>\r\n\r\n2)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong>H <\/strong>1<\/strong> : \u00b5 > \u00b5<\/strong>0<\/strong>\r\n\r\n3)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong>H <\/strong>1<\/strong> : \u00b5 \u2260 \u00b5<\/strong>0<\/strong>\r\n\r\n \r\n\r\nIn such cases, the critical values (\u00b1z\u03b1 or \u00b1z\u03b1\/2) are given by:\r\n\r\n \r\n

Further, if H 1 : \u00b5 < \u00b50 or H 1 : \u00b5 > \u00b50 , the test is also called one-sided test; otherwise, it is called two-sided test. The critical regions for the above three alternative hypotheses are shown in the following figures.<\/p>\r\n \r\n\r\n\"\"\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\n\"\"\r\n\r\n\"\"\r\n\r\n \r\n\r\n \r\n\r\n2.1 A Confidence Interval Approach to Test a Hypothesis of the Form H<\/strong>0<\/strong>: \u03bc = \u03bc<\/strong>0<\/strong>\r\n\r\nH<\/strong>1<\/strong>: \u03bc \u2260 \u03bc<\/strong>0<\/strong>\r\n\r\n\"\"\r\n\r\n \r\n\r\n \r\n\r\n3\u00a0\u00a0\u00a0\u00a0 PROCEDURE INVOLVED IN TESTING OF HYPOTHESES<\/strong>\r\n\r\n \r\n\r\nThe following steps are involved in a test of significance:\r\n\r\n \r\n\r\nStep 1: <\/strong>Formulate the null and alternative hypotheses. For example:\r\n\r\n<\/div>\r\n
\r\n\r\n\r\n\r\n\r\n\r\n
a) H0 : \u03bc \u2265 \u03bc0<\/td>\r\nH1 : \u03bc < \u03bc0 or<\/td>\r\n<\/tr>\r\n
b) H0 : \u03bc \u2264 \u03bc0<\/td>\r\nH1 : \u03bc > \u03bc0 or<\/td>\r\n<\/tr>\r\n
c) H0 : \u03bc = \u03bc0<\/td>\r\nH1 : \u03bc \u2260 \u03bco.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n

Step 2: <\/strong>Fix the value of \u03b1; that is, deciding, the level of significance. Usually, we fix \u03b1 = 0.5 or \u03b1 = 0.01.<\/p>\r\n \r\n\r\n\"\"\r\n\r\n4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong>EXAMPLES WITH DIFFERENT METHODS:<\/strong>\r\n\r\n\u00a0<\/strong>\r\n\r\na)\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong>Example 1<\/strong>\r\n\r\n \r\n

Consider a sample of 36 Units; sample mean ( ) is 2.92 and is 0.18. Test whether H<\/strong>0<\/strong> :<\/strong> \u03bc \u2265 \u03bc<\/strong>0<\/strong> H<\/strong>1<\/strong> :<\/strong> \u03bc<\/strong> <<\/strong> \u03bc<\/strong>0<\/strong>; \u00b5<\/strong>0<\/strong> = 3.<\/strong><\/p>\r\n \r\n\r\nMethod 1:<\/em><\/strong>\r\n\r\n \r\n\r\n\"\"\r\n\r\n<\/div>\r\n

\r\n\r\n\u00a0 \u00a0 Basics of Testing of Hypotheses\r\n\r\n \r\n\r\nc = \u00b50 - z\u03b1\/2 \u00a0= 3.0 -2.33*0.03 = 2.93.\r\n\r\n \r\n\r\nSince \u00a0(= 2.92) < c (2.93), reject H0 that it is \u00b5 \u2265 3.\r\n\r\n \r\n\r\nMethod 2<\/em><\/strong>\r\n\r\n \r\n\r\nLet \u03b1 = 0.01 and |z\u03b1\/2| = 2.57; \u00a0= 0.18\/ = 0.03. Then\r\n\r\n \r\n\r\n\"\"\r\n\r\n5\u00a0\u00a0\u00a0\u00a0 USE OF P-VALUE<\/strong>\r\n\r\n \r\n

In statistical significance testing the p<\/em>-value<\/strong> is the probability of obtaining a test statistic at least as extreme as the one that was actually observed, assuming that the null hypothesis is true; the p-value is the smallest value of \u03b1 for which the given sample outcome would lead to accepting H0. The decision rule is to accept H0 if the p-value \u2265 \u03b1 and reject H0 if p- value < \u03b1. In the Example 1, above, the p-value is 0.0038 (i.e., P ( > 2.92) which is also equivalent to P(z \u2265 2.66)); the p-value is < \u03b1 and thus it leads to rejecting H0.<\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n\r\n \r\n\r\n6.1<\/strong>\r\n\r\nProcedure for t-test<\/strong>\r\n\r\n \r\n\r\nStep 1<\/em><\/strong>: <\/em>Formulate the null and alternative hypotheses. For example,\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\n
a) H0 : \u03bc \u2265 \u03bc0<\/td>\r\nH1 : \u03bc < \u03bc0<\/td>\r\n<\/tr>\r\n
b) H0 : \u03bc \u2264 \u03bc0<\/td>\r\nH1 : \u03bc > \u03bc0<\/td>\r\n<\/tr>\r\n
c) H0 : \u03bc = \u03bc0<\/td>\r\nH1 : \u03bc \u2260 \u03bc0.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n

Step 2<\/strong>:<\/em> Decide the level of significance (\u03b1) and determine the critical values for a given degree of freedom from the t<\/em>-table. For (a) and (b), the critical value is given by | t\u03b1| such that P(|t| \u2264 |t\u03b1|)= l - \u03b1 . For (a), the critical values are given by at | t \u03b1 \/ 2 | such that P<\/em>(t<\/em> \u2264| t<\/em>\u03b1<\/em> \/ 2 |)= 1 \u2212 \u03b1<\/em> .<\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n\r\n \r\n\r\nStep 4: Accept the null hypothesis if t<\/em> is less than the critical value; otherwise reject the null hypothesis.\r\n\r\n \r\n\r\n7.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Test of Difference between Two Population Means<\/strong>\r\n\r\n \r\n

Let us now deal with two populations for which standard deviations (\u03c31 and \u03c32) are known. However, the means (\u00b51 and \u00b5 2) are not known. Under the circumstances, can we test whether or not \u00b51 = \u00b5 2; i.e., \u00b51 - \u00b52 = 0? So in such cases, usually we would like to test<\/p>\r\n \r\n\r\nHo : \u00b51 - \u00b52 = D<\/em>o<\/em> against a specified alternative hypothesis. It may be any one of the following:\r\n\r\n \r\n\r\nH1 : \u00b51 - \u00b52 \u2260 Do\r\n\r\nH1 : \u00b51 - \u00b52 > Do\r\n\r\nH1 : \u00b51 - \u00b52 < Do\r\n\r\n \r\n

If D<\/em>o<\/em>=0, we are actually testing whether or not \u00b51 = \u00b52.<\/em> To test whether or not Ho is true against a specified H1 we consider the difference between x<\/em>1 and x<\/em>2 and their distribution in repeated samples. It has been shown in the probability theory that for repeated independent randomly drawn samples, x<\/em>1 \u2212 x<\/em>2 follows a normal<\/p>\r\n\r\n<\/div>\r\n\"\"\r\n

Thus, we can use the <\/span>z<\/em>-test to test the null hypothesis. The procedure is similar to the z-test explained earlier. If X1 and X2 follow normal distributions and if and \u03c31 and \u03c32 are unknown, for small samples, we can use the <\/span>t<\/em>-test. That is,\u00a0<\/span>t<\/em>-statistic in this case has n1 + n2 - 2 degrees of freedom. However, for large samples, even if \u03c31 and \u03c3 2 are unknown, we can use the <\/span>z<\/em>-test. In this case, we use s1 and s2 instead of \u03c31 and \u03c32. That is,<\/span><\/p>\r\n\r\n

\r\n\r\n\"\"\r\n\r\n \r\n\r\n8\u00a0\u00a0\u00a0\u00a0 TESTS ABOUT PROPORTIONS<\/strong>\r\n\r\n\"\"\r\n

is a standardized normal variate, wherein <\/span>p<\/em> is the proportion of successes in a sample. Hence, to test the null hypothesis, we can use z-test as discussed above. The procedure involved in testing <\/span>H<\/em>o<\/em> is given below.<\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\nStep 1<\/strong>: Formulate the null and alternative hypothesis; for instance\r\n\r\n \r\n\r\nH o : P = P o\u00a0 H 1 : P\u2260 P o\r\n\r\nH o : P \u2265 P o H 1 : P< P o\r\n\r\nH o : P \u2264 P o H 1 : P> P o\r\n\r\n \r\n

Step 2<\/strong>: Fix the \u03b1 value and then determine the critical value from the normal distribution table. | Z\u03b1\/2, | = 1.96 and 2.58 for \u03b1 = 0.05 and 0.01 respectively. | Z\u03b1 | = 1.64 and 2.33 for \u03b1 = 0.05 and 0.01 respectively.<\/p>\r\n \r\n\r\nStep 3<\/strong>: Compute p and q (= 1-<\/em>p) for the sample data.\r\n\r\n\"\"\r\n\r\n \r\n\r\nStep 5<\/strong>: For two-sided test, case (a). That is, reject H<\/em>o<\/em> if |z| > |z\u03b1\/2|.\r\n\r\nFor one-sided test, case (b): Reject H<\/em>o<\/em> if z > z\u03b1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 and\r\n\r\nFor one-sided test, case (c): Reject H<\/em>o<\/em> if z < - z\u03b1\r\n\r\n \r\n\r\n8.1 Difference between Two Proportions<\/strong>\r\n\r\n \r\n

A general hypothesis that is tested regarding the theoretical proportion of successes (in two sample cases) is that H o :P 1 - P 2 =P o against a specified alternative hypothesis. The alternative hypothesis may be any one of the following:<\/p>\r\n \r\n\r\n\r\n\r\n\r\n\r\n
H1 : P1<\/td>\r\n- P2 \u2260 P0<\/td>\r\n<\/tr>\r\n
H1 : P1<\/td>\r\n- P2 > P0<\/td>\r\n<\/tr>\r\n
H1 : P1<\/td>\r\n- P2 < P0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

To test whether or not H o is true against a specified H 1 , we consider the difference P 1 and P 2 and their distribution in repeated samples. It has been shown in the probability theory for repeated independent randomly drawn samples, that P 1 - P 2 follows a normal distribution with mean P 1 - P 2 and variance<\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n\r\n \r\n\r\n \r\n

n <\/em>is the size of the sample, x<\/em> and y<\/em> are the sample means of X <\/em>and Y <\/em>respectively. s<\/em>x<\/em> and s <\/em>y<\/em> are the sample standard deviations of X <\/em>and Y <\/em>respectively. We may like to test the null hypothesis that \u03c1 = 0 (where \u03c1 is the population correlation coefficient) against a specified alternative hypothesis. We are actually using r<\/em> to test the hypothesis about \u03c1 since r<\/em> is an estimate of \u03c1.<\/em> The testing is usually done by determining the calculated value of r<\/em> as significantly different from zero. This can be done by using a t<\/em>-test. For the purpose of testing the null hypothesis, the follow-ing t<\/em>-statistic is computed<\/p>\r\n \r\n\r\n\"\"\r\n

where the <\/span>t<\/em>-statistic is with (<\/span>n<\/em>\u20142) degrees of freedom. If the calculated value lies in the critical region, we have to reject the null hypothesis. If we accept the null hypothesis it means that there is no correlation between the two variables other than that due to chance.<\/span><\/p>\r\n \r\n

10\u00a0 NON-PARAMETRIC TESTS<\/strong><\/p>\r\n \r\n

The use of t-test or z-test requires an assumption that the sample data come from a normal or binomial population (or at least, the sample distribution must tend to a normal distribution for a large sample). Both the z-test and t-test are used to test the null hypothesis concerned with population means, variances and proportions. Hypotheses related to the independence of two criteria of classification, goodness-of-fit test, median of the population, etc. can be tested using the statistical tests called non-parametric tests. The non-parametric tests do not require many assumptions (like the <\/span>z<\/em>-test and <\/span>t<\/em>-test). A non-parametric tests is discussed below.<\/span><\/p>\r\n \r\n

10.1 <\/span>Chi-Square Test<\/strong><\/p>\r\n \r\n

The Chi Square test is normally applicable in situations in which determination of population parameters such as the mean and standard deviation are not an issue. The data in question falling into discrete categories and are presented in a contingency table. The entries in a contingency table are known cells. Let us consider the result of a survey of 100 adults (say, 50 females and 50 males). Let us say that it has been observed that among the 100 adults, 34 of them are library users and the rest are non-users. So, in this hypothetical example, we have two nominal variables sex and library use. On categorizing the 100 adults, using these two nominal variables, we get a contingency table like the one shown in Table below:<\/span><\/p>\r\n \r\n

A Typical 2 x 2 Table<\/strong><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n

Let us now try to find out whether or not the two categories\u2014gender (Male or Female) and library use\u2014are independent; let us assume that there is no rela-tionship between the gender and library use. Under this assumption, compute the frequencies in each of the cells. Such frequencies are called <\/span>theoretical frequencies.<\/em> They are usually referred to as the expected numbers or expected frequencies. The logic for computation of the theoretical frequencies for the data given in the above Table is as follows:<\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n

We have 50 each of males and females; the ratio is 1:1. So we would expect that half of the library users are males and also half of the non-users are males, that is, out of 100 adults (N<\/em>) we have 50 males (row total). Out of 34 users (column total), how many of them are males?<\/p>\r\n \r\n\r\nUsing the cross multiplication technique, we have:\r\n

The number of male users = 50 \u00d7 34<\/span> = 17 .<\/p>\r\n100\r\n\r\n \r\n

Similarly, we can compute the number of male non-users, female users and female non-users. The results are shown in bold face in the corresponding cells in Table 8.1. On generalizing the above logic, we can easily prove that the following formula can be used to obtain theoretical frequencies ( E<\/em> i j<\/em> )<\/em> in each of the cells:<\/p>\r\nE<\/em>\u00a0\u00a0\u00a0\u00a0\u00a0 =\u00a0 r<\/em>i<\/em> . \u00d7 c<\/em>. j<\/em><\/span>\r\n\r\nij<\/em>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 N<\/em>\r\n\r\n \r\n\r\nr<\/em>i<\/em>. <\/em>is the total number of observations in the ith row\r\n\r\n \r\n\r\ncij is the total number of observations in the jth column\r\n\r\n \r\n\r\nEij is the expected\/theoretical frequencies in the ijth cell (ith row and jth column).\r\n\r\n \r\n\r\nDegree of Freedom<\/strong>\r\n\r\n \r\n

Degrees of freedom are commonly discussed in relation to chi- square and other forms of hypothesis testing statistics. It is important to calculate the degree(s) of freedom when determining the significance of a chi square statistic and the validity of the null hypothesis. It is obvious that the theoretical frequencies need not necessarily be equal to the observed frequencies. If they are equal, one could perhaps conclude that there is no relationship between the two variables. If they\u00a0are not equal, the question is, \"Is the difference between the observed and expected frequencies statistically significant?\" To answer this question, we use a<\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n\r\n \r\n\r\n10.2\u00a0\u00a0\u00a0\u00a0 Measures of Association<\/strong>\r\n\r\n \r\n

The statistical significance of the null hypothesis depends both on the strength of the observed relationship and the size of the sample. Tests of statistical significance indicate only the likelihood that an observed relationship actually exists in the universe; but they do not reveal the fact as to how strong the relationship is. Further, a relationship may be statistically significant being substantially important.<\/p>\r\n\r\n<\/div>\r\n

\r\n\r\nThere are a few measures that will describe the strength of the association between two nominal variables. They are:\r\n\r\n\"\"\r\n\r\n \r\n\r\n10.3\u00a0\u00a0\u00a0\u00a0 Goodness-of-Fit Test<\/strong>\r\n\r\n \r\n

The chi-square statistic is also used to test the hypothesis that whether or not the probability distribution (of the population) is similar to that of the sample distribution. This type of test is often referred to as a goodness -of-fit test. The goodness-of-fit test is illustrated with following example. Examine whether or not the distribution of transactions follows a negative binomial distribution for the data shown below.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
x<\/td>\r\nf(x)<\/td>\r\nx<\/td>\r\nf(x)<\/td>\r\nx<\/td>\r\nf(x)<\/td>\r\n<\/tr>\r\n
0<\/td>\r\n324<\/td>\r\n3<\/td>\r\n16<\/td>\r\n6<\/td>\r\n2<\/td>\r\n<\/tr>\r\n
1<\/td>\r\n108<\/td>\r\n4<\/td>\r\n7<\/td>\r\n7<\/td>\r\n1<\/td>\r\n<\/tr>\r\n
2<\/td>\r\n43<\/td>\r\n5<\/td>\r\n4<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n
\r\n\r\nWe will use the goodness-of-fit test for this purpose. The procedure is:\r\n\r\n \r\n\r\nStep 1: <\/strong>Formulate the null and alternative hypothesis.\r\n\r\n \r\n\r\nH<\/em>o<\/em>: The sample data belongs to a population which follows a negative binomial distribution.\r\n\r\n \r\n\r\nH<\/em>1<\/em>: The sample data belongs to a population which does not follow a negative binomial distribution.\r\n\r\n \r\n

Step 2: <\/strong>Compute the parameters (such as mean, variance, etc.); estimate the parameters of the theoretical probability distribution (which is assumed in the null hypothesis). Use, as far as possible, the maximum likelihood estimators.<\/p>\r\n \r\n\r\nStep 3: <\/strong>Compute the probabilities under the assumption that the H<\/em>o<\/em> is true.\r\n\r\n \r\n

Step 4: <\/strong>Compute the theoretical or expected frequencies (use the formula, that expected frequency is equal to n<\/em>\u22c5<\/em>P(x),<\/em> where n<\/em> is the sample size, and P(x)<\/em> is the theoretical probability distribution function). In this case, P(x)<\/em> is the mass function of the negative binomial distribution.<\/p>\r\n \r\n

Step 5<\/strong>: Decide \u03b1 and determine the critical region (for \u03b1 = 0.05). Find out X<\/em>2<\/em> for (k \u2014 <\/em>1) degrees of freedom; k <\/em>is the number of frequency classes.<\/p>\r\n\r\n<\/div>\r\n\"\"\r\n

\r\n\r\n\"\"\r\n\r\n11.\u00a0<\/span>Conclusion<\/strong>\r\n\r\n<\/div>\r\n
\r\n\r\nIn this Unit, we have discussed the basics of z-test, t-test and chi square test.\r\n\r\n\r\n\r\n\r\n
you can view video on Basics of Testing of Hypothesis<\/strong><\/td>\r\n\"\"<\/a><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nREFERENCES<\/strong>\r\n\r\n \r\n