{"id":259,"date":"2018-10-30T12:17:00","date_gmt":"2018-10-30T12:17:00","guid":{"rendered":"http:\/\/mgmtp15.epgpbooks.inflibnet.ac.in\/?post_type=chapter&p=259"},"modified":"2018-10-30T12:34:46","modified_gmt":"2018-10-30T12:34:46","slug":"statistical-inference-with-two-populations-hypothesis-techniques-two-sample-tests-%cf%831-and-%cf%832-known-and-un-known","status":"publish","type":"chapter","link":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/chapter\/statistical-inference-with-two-populations-hypothesis-techniques-two-sample-tests-%cf%831-and-%cf%832-known-and-un-known\/","title":{"rendered":"Statistical Inference with two populations: Hypothesis Techniques-two sample tests: \u03c31 and \u03c32 (known and un-known)"},"content":{"raw":"
\r\n\r\n \r\n\r\n1.\u00a0\u00a0\u00a0\u00a0\u00a0 Introduction\r\n\r\n \r\n\r\n2.\u00a0\u00a0\u00a0\u00a0\u00a0 Test for two population means (Independent Samples)\r\n\r\n \r\n\r\n3.\u00a0\u00a0\u00a0\u00a0\u00a0 Test for two population means (dependent Samples)\r\n\r\n \r\n\r\n4.\u00a0\u00a0\u00a0\u00a0\u00a0 Two Sample t-test\r\n\r\n \r\n\r\n5.\u00a0\u00a0\u00a0\u00a0\u00a0 Hypothesis Testing of the Difference Between Two Means\r\n\r\n \r\n\r\n6.\u00a0\u00a0\u00a0\u00a0\u00a0 Hypothesis Testing For a Difference Between Means for Small Samples Using Pooled Standard Deviations (Optional)\r\n\r\n \r\n\r\n7.\u00a0\u00a0\u00a0\u00a0\u00a0 Hypothesis Testing for a Difference Between Proportions\r\n\r\n \r\n\r\n8.\u00a0\u00a0\u00a0\u00a0\u00a0 Paired Differences\r\n\r\n \r\n\r\n9.\u00a0\u00a0\u00a0\u00a0\u00a0 Summary\r\n\r\n \r\n\r\n10.\u00a0 Self-Check Exercise\r\n\r\n<\/div>\r\n
\r\n\r\n \r\n\r\nQuadrant-I<\/strong>\r\n\r\n \r\n\r\nStatistical Inference with two populations: Hypothesis Techniques-two sample tests: \u03c31 and \u03c32 (known and un-known)<\/strong>\r\n\r\n \r\n\r\nLearning Objectives:<\/strong>\r\n\r\n \r\n\r\nAfter the completion of this module the student will understand:\r\n\r\n \r\n\r\ni- Test for two population means (Independent Samples) ii- Test for two population means (Dependent Samples) iii- Two Sample t-test\r\n\r\n \r\n\r\niv- Hypothesis Testing of the Difference Between Two Means\r\n\r\n \r\n\r\nv-\u00a0\u00a0 Hypothesis Testing For a Difference Between Means for Small Samples Using Pooled Standard Deviations (Optional)\r\n\r\n \r\n\r\nvi- Hypothesis Testing for a Difference Between Proportions vii- Paired Differences\r\n\r\n \r\n\r\nIntroduction<\/strong>\r\n\r\n \r\n\r\nThis technique helps to test a claim to comparing parameters from two populations. It can be explain by this example\r\n\r\n \r\n\r\nExample<\/strong>\r\n\r\n \r\n

Suppose an entrepreneur wants to open a new startup business and is contemplating of doing so in either Business District- 1 or Business District- 2. The economic affluence of a business district is the key in finalizing the entrepreneur\u2019s decision. The entrepreneur wants to look at the household income as a key indicator of a business district\u2019s economic condition to settle the issue. Thus, from a statistical point of view, we are dealing with two populations, namely:<\/p>\r\n \r\n\r\nPopulation- 1: collection of households in the Business District- 1;\r\n\r\nPopulation- 1: collection of households in the Business District- 2;\r\n\r\n \r\n

The common variable under study (i.e., the variable applicable to both the population for comparison) is household income.<\/p>\r\n \r\n

If the entrepreneur wants to compare the populations in term of mean household income<\/em>, then the parameters of interest would be<\/p>\r\n \r\n

\u00b51 = mean household income of population \u2013 1 and \u00b52 = mean household income of population \u2013 2.<\/p>\r\n \r\n

First, one wants to know whether the populations are identical in terms of mean household income or not. To check this, one test the null hypothesis H<\/em>0:<\/em> \u00b51 = \u00b52 against the alternative H<\/em>A<\/em> :<\/p>\r\n \r\n\r\n\u00b51 \u2260 \u00b52.\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n

The idea is that if H<\/em>0<\/em> is accepted then it is concluded that the business districts are identical in term of mean household income, and the business person can then focus on the secondary factors (like transportation, tax structure etc. ) to make the final decision. If the above H<\/em>0<\/em> is rejected or there are some other reasons to believe that one population has higher mean than the other does, then one might be inclined to test<\/p>\r\n \r\n\r\nH<\/em>0<\/em>:\u00a0 <\/em>\u00b51 = \u00b52 (\u00b51 \u2264 \u00b52) against H<\/em>A:<\/em>\u00a0 <\/em>\u00b51 \u203a \u00b52)\r\n\r\n \r\n\r\nSimilarly, one can test\r\n\r\n \r\n\r\nH<\/em>0<\/em>: <\/em>\u00b51 = \u00b52 (\u00b51 \u2265 \u00b52) against H<\/em>A<\/em> :\u00a0 (\u00b51 \u203a \u00b52 )\r\n\r\n \r\n

On the other hand, if the entrepreneur wants to compare the populations in term of proportion of household having a minimum income level (say, K<\/em><\/strong> units per year), then the parameters of interest would be<\/p>\r\n \r\n

p1 = proportion of household in population \u2013 1 having a minimum specified income; and p1 = proportion of household in population \u2013 2 having a minimum specified income.<\/p>\r\n \r\n

One can now check whether the population are identical or not (in term of proportion of households having the minimum specified income) by testing<\/p>\r\n \r\n\r\nH<\/em>0<\/em>:\u00a0 <\/em>p1 = p2 against H<\/em>A<\/em>:\u00a0 <\/em>p1 \u2260 p2;\r\n\r\n \r\n\r\nDepending on the practicality, one can also test either\r\n\r\n \r\n\r\nH<\/em>0<\/em> :\u00a0 <\/em>p1 = p2 ( p 1 \u2264 p 2 ) against H<\/em>A<\/em> :\u00a0 <\/em>p 1 \u203a p 2 )\r\n\r\n \r\n\r\nSimilarly, one can test\r\n\r\n \r\n\r\nH<\/em>0<\/em> :\u00a0 <\/em>p1 = p2 ( p 1 \u2265 p 2 ) against H<\/em>A<\/em> : ( p 1 \u203a p 2 )\r\n\r\n \r\n\r\ns-11<\/strong>\r\n\r\nTest for two populations means (Independent Samples)<\/strong>\r\n\r\n \r\n\r\nFor two population say population-1 and population-2, let the mean of a common variable be \u00b51 and \u00b52\r\n

respectively, are also completely unknown. Our objective is to test<\/p>\r\n \r\n\r\nH<\/em>0<\/em>:\u00a0 <\/em>\u00b51 = \u00b52 against H<\/em>A<\/em> :<\/em>\r\n\r\n \r\n\r\nCreate three conditions\r\n\r\n \r\n\r\n(i)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00b51 \u203a \u00b52\r\n\r\nor\r\n\r\n(ii)\u00a0\u00a0\u00a0\u00a0 \u00b51 \u2039 \u00b52\r\n\r\nor\r\n\r\n(iii) \u00b51 \u2260 \u00b52\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 at a significance level\r\n\r\n<\/div>\r\n \r\n\r\ns-12<\/span>\r\n

\r\n\r\n \r\n

To test the above null hypothesis, random samples of size n1 and n2 are drawn from the above populations and the sample observations from the one population are independent of those from the other population. Let<\/p>\r\n \r\n

X1 = average of sample observation from population -1 X2 = average of sample observation from population -2<\/p>\r\n \r\n

s1 = standard deviation of sample observation from population -1 s2 = standard deviation of sample observation from population -2<\/p>\r\n \r\n\r\ns-13<\/strong>\r\n\r\nThis formula will apply in case if population standard deviation is unknown but equal<\/strong>\r\n\r\n\"\"\r\n\r\ns-14<\/strong>\r\n\r\nExample<\/strong>\r\n\r\n \r\n\r\nAn experiment is conducted to compare the mean lengths of time required for bodily absorption of two popular insulins, say from company-X and company-Y. Ten women are randomly selected and given a dose of company-X. Similarly, another group of ten randomly selected women are administrated company-Y. The length of time in minutes for the insulins to reach a specified level in the blood is recorded. The sample averages, standard deviation and sample size are given below in Table-1.\r\n\r\n \r\n\r\nTable-1\r\n\r\n\r\n\r\n\r\n\r\n\r\n
company-X<\/td>\r\ncompany-Y<\/td>\r\n<\/tr>\r\n
n1= 10<\/td>\r\nn2= 10<\/td>\r\n<\/tr>\r\n
mean 1=20.2<\/td>\r\nmean 2=17.9<\/td>\r\n<\/tr>\r\n
s1=8.1<\/td>\r\ns2=7.3<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\nUsing\u00a0 \u00a0= 0.05, test the claim that the insulins are identical in terms of mean time required for bodily\r\n\r\nabsorption.\r\n\r\n \r\n\r\ns-15<\/strong>\r\n\r\nSolution<\/strong>\r\n\r\n \r\n\r\nFirst, we need to identify the populations\r\n\r\n \r\n

Population -1 = collection of all users represented by the group of ten individuals who received from company-X; and<\/p>\r\n \r\n

Population -2 = collection of all users represented by the group of ten individuals who received from company-Y;<\/p>\r\n \r\n

The common variable under study = time required for bodily absorption of insulins.<\/span><\/p>\r\n \r\n

The parameter of interest:<\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\n\u00b51 = mean for the variable under study for Population-1 \u00b52 = mean for the variable under study for Population-2 The aim here is to test\r\n\r\nH<\/em>0<\/em>:\u00a0 <\/em>\u00b51 = \u00b5 2 against H<\/em>A<\/em>:\u00a0 <\/em>\u00b51 \u2260 \u00b5 2\r\n\r\ns-16\r\n\r\nAt level\u00a0 \u00a0= 0.05.\r\n\r\n\"\"\r\n\r\nWe conclude that the insulins are identical in terms of mean body absorption time.\r\n\r\n \r\n\r\ns-17<\/strong>\r\n\r\nTest for two population means (Dependent Samples)<\/strong>\r\n\r\n \r\n\r\n \r\n\r\nTwo samples are dependent if each member of one sample corresponds to a member of the other sample.\r\n\r\nDependent samples are also known as matched samples or paired samples. Use of such depend (paired)\r\n\r\nsamples will enable as to perform a more precise analysis, because they will allow us to control for\r\n

extraneous. With dependent samples we still follow the same basic procedure that we have followed in all<\/p>\r\nour hypothesis testing.\r\n\r\n \r\n\r\n \r\n\r\nS-18<\/strong>\r\n\r\n \r\n

i.e. sample of water before (inlet) and after (outlet) the Effluent Treatment Plant (ETP) is tested to measure the effectiveness of ETP considering that the two samples of water have different properties of pollution level once treated by ETP.<\/p>\r\n \r\n

s-19<\/strong><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\nHypothesis Testing of the Difference Between Two Means <\/strong>(Dependent sample)\r\n\r\n \r\n\r\nDo employees perform better at work with music playing. The music was turned on during the working hours of a business with 45 employees. There productivity level averaged 5.2 with a standard deviation of 2.4. On a different day the music was turned off and there were 40 workers. The workers' productivity level averaged 4.8 with a standard deviation of 1.2. What can we conclude at the 0.05 level?\r\n\r\n \r\n\r\nSolution<\/strong>\r\n\r\n \r\n\r\nWe first develop the hypotheses\r\n\r\n \r\n\r\n\r\n\r\n\r\n
H0:<\/td>\r\n1 -<\/td>\r\n2<\/td>\r\n= 0<\/td>\r\n<\/tr>\r\n
H1:<\/td>\r\n1 -<\/td>\r\n2<\/td>\r\n> 0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\ns-20\r\n\r\n \r\n\r\nNext we need to find the standard deviation. Recall from before, we had that the mean of the difference is\r\n\r\n \r\n\r\n\"\"\r\n

We can substitute the sample means and sample standard deviations for a point estimate of the population means and standard deviations. We have\u00a0<\/span>and\u00a0<\/span>Now we can calculate the z-score. We have<\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n\r\n \r\n\r\ns-21\r\n\r\n \r\n

Since this is a one tailed test, the critical value is 1.645 and 0.988 does not lie in the critical region. We fail to reject the null hypothesis and conclude that there is insufficient evidence to conclude that workers perform better at work when the music is on. Using the P-Value technique, we see that the P-value associated with 0.988 is<\/p>\r\n \r\n\r\nP = 1 - 0.8389 = 0.1611\r\n\r\n \r\n\r\nwhich is larger than 0.05.\u00a0 Yet another way of seeing that we fail to reject the null hypothesis.\r\n\r\n \r\n

Note: <\/strong>It would have been slightly more accurate had we used the t-table instead of the z-table. To calculate the degrees of freedom, we can take the smaller of the two numbers n1 - 1 and n2 - 1. So in this example, a better estimate would use 39 degrees of freedom. The t-table gives a value of 1.690 for the t.95 value. Notice that 0.988 is still smaller than 1.690 and the result is the same. This is an example that demonstrates that using the t-table and z-table for large samples results in practically the same results.<\/p>\r\n \r\n\r\ns-22<\/strong>\r\n\r\n \r\n

Hypothesis Testing For a Difference between Means for Small Samples Using Pooled Standard Deviations (Optional)<\/strong><\/p>\r\n \r\n\r\nRecall that for small samples we need to make the following assumptions:\r\n\r\n1.\u00a0\u00a0\u00a0\u00a0\u00a0 Random unbiased sample.<\/span>\r\n\r\n2.\u00a0\u00a0\u00a0\u00a0\u00a0 Both population distributions are normal.<\/span>\r\n\r\n3.\u00a0\u00a0\u00a0\u00a0\u00a0 The two standard deviations are equal.<\/span>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\ns-23\r\n\r\n\"\"\r\n\r\n \r\n\r\nPutting this together with hypothesis testing we can find the t-statistic. and use n1 + n2 - 2 degrees of freedom.\r\n\r\n \r\n\r\ns-24<\/strong>\r\n\r\n \r\n\r\nExample<\/strong>\r\n\r\n \r\n

Nine dogs and ten cats were tested to determine if there is a difference in the average number of days that the animal can survive without food. The dogs averaged 11 days with a standard\u00a0deviation of 2 days while the cats averaged 12 days with a standard deviation of 3 days. What can be concluded? (Use = .05)<\/span><\/p>\r\n\r\n<\/div>\r\n

\r\n\r\n \r\n\r\nSolution<\/strong>\r\n\r\n \r\n\r\nWe write:\r\n\r\n \r\n\r\nH0:\u00a0\u00a0\u00a0\u00a0\u00a0 dog -\u00a0\u00a0\u00a0 cat = 0\r\n\r\n \r\n\r\nH1:\u00a0\u00a0\u00a0\u00a0\u00a0 dog -\u00a0\u00a0\u00a0 cat 0\r\n\r\n \r\n\r\nWe have:\r\n\r\n \r\n\r\n\r\n\r\n\r\n\r\n
n1<\/td>\r\n= 9,<\/td>\r\nn2\u00a0 = 10<\/td>\r\n<\/tr>\r\n
x1<\/td>\r\n= 11,<\/td>\r\nx2<\/td>\r\n= 12<\/td>\r\n<\/tr>\r\n
s1<\/td>\r\n= 2,<\/td>\r\ns2<\/td>\r\n= 3<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\ns-25\r\n\r\n \r\n

The t-critical value corresponding to a = 0.05 with 10 + 9 - 2 = 17 degrees of freedom is 2.11 which is greater than 0.84. Hence we fail to reject the null hypothesis and conclude that there is not sufficient evidence to suggest that there is a difference between the mean starvation time for cats and dogs.<\/p>\r\n \r\n\r\nHypothesis Testing for a Difference between Proportions<\/strong>\r\n\r\n \r\n\r\nInferences on the Difference between Population Proportions<\/strong>\r\n\r\n \r\n\r\nIf two samples are counted independently of each other we use the test statistic:<\/span>\r\n\r\n<\/div>\r\n

\r\n\r\n\"\"\r\n\r\n \r\n\r\ns-26\r\n\r\n \r\n\r\nExample<\/strong>\r\n\r\n \r\n

Is the severity of the drug problem in high school the same for boys and girls? 85 boys and 70 girls were questioned and 34 of the boys and 14 of the girls admitted to having tried some sort of drug. What can be concluded at the 0.05 level?<\/p>\r\n \r\n\r\ns-27<\/strong>\r\n\r\n \r\n\r\nSolution<\/strong>\r\n\r\n \r\n\r\nThe hypothesis are\r\n\r\n \r\n\r\nH0: p1 - p2\u00a0 = 0\r\n\r\nH1:\u00a0\u00a0 p1 - p2<\/span>0<\/span>\r\n\r\n \r\n\r\nWe have<\/span>\r\n\r\n \r\n\r\np1<\/span>\r\n\r\n<\/div>\r\n

\r\n\r\n= 34\/85 = 0.4\r\n\r\np2\r\n\r\n= 14\/70 = 0.2\r\n\r\n<\/div>\r\n
\r\n\r\n \r\n\r\np = 48\/155 = 0.31\r\n\r\nq = 0.69\r\n\r\n<\/div>\r\n \r\n\r\nNow compute the z-score\r\n\r\n\"\"\r\n\r\n \r\n

Since we are using a significance level of .05 and it is a two tailed test, the critical value is 1.96. Clearly 2.68 is in the critical region, hence we can reject the null hypothesis and accept the alternative hypothesis and conclude that gender does make a difference for drug use. Notice that the P-Value is<\/p>\r\n \r\n\r\nP = 2(1 - .9963)\u00a0 = 0.0074 is less than 0.05. Yet another way to see that we reject the null hypothesis.\r\n\r\n \r\n\r\n \r\n\r\nSummary<\/strong>\r\n\r\n \r\n\r\n \r\n\r\nThis module help in understanding of sampling distributions when two sets of samples from the same or different populations. If they are random samples from the same population, then any differences\u00a0 across\u00a0 conditions\u00a0 or\u00a0 groups\u00a0 can\u00a0 be\u00a0 attributed\u00a0 to\u00a0 random\u00a0 sampling\u00a0 variability. However, if the two sets of scores are random samples from different populations, then we can attribute any difference between mean scores conditions to the independent variable or the treatment effect.\r\n\r\n \r\n\r\nSelf exercise question<\/strong>\r\n\r\n \r\n\r\nThe self exercise question is also mentioned on above describe pages of every topic.\r\n\r\n \r\n

Learn More:<\/strong><\/p>\r\n\r\n

    \r\n \t
  1. \r\n
      \r\n \t
    1. Sharma, J K (2014), Business Statistics, S Chand & Company, N Delhi.<\/li>\r\n \t
    2. Bajpai, N (2010) Business Statistics, Pearson, N Delhi.<\/li>\r\n \t
    3. Trevor Hastie, Robert Tibshirani, Jerome Friedman (2009), The Elements of Statistical Learning: Data Mining, Inference, and Prediction, 2nd Edition, Springer.<\/li>\r\n \t
    4. Darrell Huff (2010), How to Lie with Statistics,\u00a0 W. W. Norton, California.<\/li>\r\n \t
    5. K.R. Gupta (2012), Practical Statistics, Atlantic Publishers & Distributors (P) Ltd., N. Delhi<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n \r\n\r\n ","rendered":"
      \n

       <\/p>\n

      1.\u00a0\u00a0\u00a0\u00a0\u00a0 Introduction<\/p>\n

       <\/p>\n

      2.\u00a0\u00a0\u00a0\u00a0\u00a0 Test for two population means (Independent Samples)<\/p>\n

       <\/p>\n

      3.\u00a0\u00a0\u00a0\u00a0\u00a0 Test for two population means (dependent Samples)<\/p>\n

       <\/p>\n

      4.\u00a0\u00a0\u00a0\u00a0\u00a0 Two Sample t-test<\/p>\n

       <\/p>\n

      5.\u00a0\u00a0\u00a0\u00a0\u00a0 Hypothesis Testing of the Difference Between Two Means<\/p>\n

       <\/p>\n

      6.\u00a0\u00a0\u00a0\u00a0\u00a0 Hypothesis Testing For a Difference Between Means for Small Samples Using Pooled Standard Deviations (Optional)<\/p>\n

       <\/p>\n

      7.\u00a0\u00a0\u00a0\u00a0\u00a0 Hypothesis Testing for a Difference Between Proportions<\/p>\n

       <\/p>\n

      8.\u00a0\u00a0\u00a0\u00a0\u00a0 Paired Differences<\/p>\n

       <\/p>\n

      9.\u00a0\u00a0\u00a0\u00a0\u00a0 Summary<\/p>\n

       <\/p>\n

      10.\u00a0 Self-Check Exercise<\/p>\n<\/div>\n

      \n

       <\/p>\n

      Quadrant-I<\/strong><\/p>\n

       <\/p>\n

      Statistical Inference with two populations: Hypothesis Techniques-two sample tests: \u03c31 and \u03c32 (known and un-known)<\/strong><\/p>\n

       <\/p>\n

      Learning Objectives:<\/strong><\/p>\n

       <\/p>\n

      After the completion of this module the student will understand:<\/p>\n

       <\/p>\n

      i- Test for two population means (Independent Samples) ii- Test for two population means (Dependent Samples) iii- Two Sample t-test<\/p>\n

       <\/p>\n

      iv- Hypothesis Testing of the Difference Between Two Means<\/p>\n

       <\/p>\n

      v-\u00a0\u00a0 Hypothesis Testing For a Difference Between Means for Small Samples Using Pooled Standard Deviations (Optional)<\/p>\n

       <\/p>\n

      vi- Hypothesis Testing for a Difference Between Proportions vii- Paired Differences<\/p>\n

       <\/p>\n

      Introduction<\/strong><\/p>\n

       <\/p>\n

      This technique helps to test a claim to comparing parameters from two populations. It can be explain by this example<\/p>\n

       <\/p>\n

      Example<\/strong><\/p>\n

       <\/p>\n

      Suppose an entrepreneur wants to open a new startup business and is contemplating of doing so in either Business District- 1 or Business District- 2. The economic affluence of a business district is the key in finalizing the entrepreneur\u2019s decision. The entrepreneur wants to look at the household income as a key indicator of a business district\u2019s economic condition to settle the issue. Thus, from a statistical point of view, we are dealing with two populations, namely:<\/p>\n

       <\/p>\n

      Population- 1: collection of households in the Business District- 1;<\/p>\n

      Population- 1: collection of households in the Business District- 2;<\/p>\n

       <\/p>\n

      The common variable under study (i.e., the variable applicable to both the population for comparison) is household income.<\/p>\n

       <\/p>\n

      If the entrepreneur wants to compare the populations in term of mean household income<\/em>, then the parameters of interest would be<\/p>\n

       <\/p>\n

      \u00b51 = mean household income of population \u2013 1 and \u00b52 = mean household income of population \u2013 2.<\/p>\n

       <\/p>\n

      First, one wants to know whether the populations are identical in terms of mean household income or not. To check this, one test the null hypothesis H<\/em>0:<\/em> \u00b51 = \u00b52 against the alternative H<\/em>A<\/em> :<\/p>\n

       <\/p>\n

      \u00b51 \u2260 \u00b52.<\/p>\n<\/div>\n

      \n

       <\/p>\n

      The idea is that if H<\/em>0<\/em> is accepted then it is concluded that the business districts are identical in term of mean household income, and the business person can then focus on the secondary factors (like transportation, tax structure etc. ) to make the final decision. If the above H<\/em>0<\/em> is rejected or there are some other reasons to believe that one population has higher mean than the other does, then one might be inclined to test<\/p>\n

       <\/p>\n

      H<\/em>0<\/em>:\u00a0 <\/em>\u00b51 = \u00b52 (\u00b51 \u2264 \u00b52) against H<\/em>A:<\/em>\u00a0 <\/em>\u00b51 \u203a \u00b52)<\/p>\n

       <\/p>\n

      Similarly, one can test<\/p>\n

       <\/p>\n

      H<\/em>0<\/em>: <\/em>\u00b51 = \u00b52 (\u00b51 \u2265 \u00b52) against H<\/em>A<\/em> :\u00a0 (\u00b51 \u203a \u00b52 )<\/p>\n

       <\/p>\n

      On the other hand, if the entrepreneur wants to compare the populations in term of proportion of household having a minimum income level (say, K<\/em><\/strong> units per year), then the parameters of interest would be<\/p>\n

       <\/p>\n

      p1 = proportion of household in population \u2013 1 having a minimum specified income; and p1 = proportion of household in population \u2013 2 having a minimum specified income.<\/p>\n

       <\/p>\n

      One can now check whether the population are identical or not (in term of proportion of households having the minimum specified income) by testing<\/p>\n

       <\/p>\n

      H<\/em>0<\/em>:\u00a0 <\/em>p1 = p2 against H<\/em>A<\/em>:\u00a0 <\/em>p1 \u2260 p2;<\/p>\n

       <\/p>\n

      Depending on the practicality, one can also test either<\/p>\n

       <\/p>\n

      H<\/em>0<\/em> :\u00a0 <\/em>p1 = p2 ( p 1 \u2264 p 2 ) against H<\/em>A<\/em> :\u00a0 <\/em>p 1 \u203a p 2 )<\/p>\n

       <\/p>\n

      Similarly, one can test<\/p>\n

       <\/p>\n

      H<\/em>0<\/em> :\u00a0 <\/em>p1 = p2 ( p 1 \u2265 p 2 ) against H<\/em>A<\/em> : ( p 1 \u203a p 2 )<\/p>\n

       <\/p>\n

      s-11<\/strong><\/p>\n

      Test for two populations means (Independent Samples)<\/strong><\/p>\n

       <\/p>\n

      For two population say population-1 and population-2, let the mean of a common variable be \u00b51 and \u00b52<\/p>\n

      respectively, are also completely unknown. Our objective is to test<\/p>\n

       <\/p>\n

      H<\/em>0<\/em>:\u00a0 <\/em>\u00b51 = \u00b52 against H<\/em>A<\/em> :<\/em><\/p>\n

       <\/p>\n

      Create three conditions<\/p>\n

       <\/p>\n

      (i)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00b51 \u203a \u00b52<\/p>\n

      or<\/p>\n

      (ii)\u00a0\u00a0\u00a0\u00a0 \u00b51 \u2039 \u00b52<\/p>\n

      or<\/p>\n

      (iii) \u00b51 \u2260 \u00b52\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 at a significance level<\/p>\n<\/div>\n

       <\/p>\n

      s-12<\/span><\/p>\n

      \n

       <\/p>\n

      To test the above null hypothesis, random samples of size n1 and n2 are drawn from the above populations and the sample observations from the one population are independent of those from the other population. Let<\/p>\n

       <\/p>\n

      X1 = average of sample observation from population -1 X2 = average of sample observation from population -2<\/p>\n

       <\/p>\n

      s1 = standard deviation of sample observation from population -1 s2 = standard deviation of sample observation from population -2<\/p>\n

       <\/p>\n

      s-13<\/strong><\/p>\n

      This formula will apply in case if population standard deviation is unknown but equal<\/strong><\/p>\n

      \"\"<\/p>\n

      s-14<\/strong><\/p>\n

      Example<\/strong><\/p>\n

       <\/p>\n

      An experiment is conducted to compare the mean lengths of time required for bodily absorption of two popular insulins, say from company-X and company-Y. Ten women are randomly selected and given a dose of company-X. Similarly, another group of ten randomly selected women are administrated company-Y. The length of time in minutes for the insulins to reach a specified level in the blood is recorded. The sample averages, standard deviation and sample size are given below in Table-1.<\/p>\n

       <\/p>\n

      Table-1<\/p>\n\n\n\n\n\n\n
      company-X<\/td>\ncompany-Y<\/td>\n<\/tr>\n
      n1= 10<\/td>\nn2= 10<\/td>\n<\/tr>\n
      mean 1=20.2<\/td>\nmean 2=17.9<\/td>\n<\/tr>\n
      s1=8.1<\/td>\ns2=7.3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

       <\/p>\n

      Using\u00a0 \u00a0= 0.05, test the claim that the insulins are identical in terms of mean time required for bodily<\/p>\n

      absorption.<\/p>\n

       <\/p>\n

      s-15<\/strong><\/p>\n

      Solution<\/strong><\/p>\n

       <\/p>\n

      First, we need to identify the populations<\/p>\n

       <\/p>\n

      Population -1 = collection of all users represented by the group of ten individuals who received from company-X; and<\/p>\n

       <\/p>\n

      Population -2 = collection of all users represented by the group of ten individuals who received from company-Y;<\/p>\n

       <\/p>\n

      The common variable under study = time required for bodily absorption of insulins.<\/span><\/p>\n

       <\/p>\n

      The parameter of interest:<\/span><\/p>\n<\/div>\n

      \n

       <\/p>\n

      \u00b51 = mean for the variable under study for Population-1 \u00b52 = mean for the variable under study for Population-2 The aim here is to test<\/p>\n

      H<\/em>0<\/em>:\u00a0 <\/em>\u00b51 = \u00b5 2 against H<\/em>A<\/em>:\u00a0 <\/em>\u00b51 \u2260 \u00b5 2<\/p>\n

      s-16<\/p>\n

      At level\u00a0 \u00a0= 0.05.<\/p>\n

      \"\"<\/p>\n

      We conclude that the insulins are identical in terms of mean body absorption time.<\/p>\n

       <\/p>\n

      s-17<\/strong><\/p>\n

      Test for two population means (Dependent Samples)<\/strong><\/p>\n

       <\/p>\n

       <\/p>\n

      Two samples are dependent if each member of one sample corresponds to a member of the other sample.<\/p>\n

      Dependent samples are also known as matched samples or paired samples. Use of such depend (paired)<\/p>\n

      samples will enable as to perform a more precise analysis, because they will allow us to control for<\/p>\n

      extraneous. With dependent samples we still follow the same basic procedure that we have followed in all<\/p>\n

      our hypothesis testing.<\/p>\n

       <\/p>\n

       <\/p>\n

      S-18<\/strong><\/p>\n

       <\/p>\n

      i.e. sample of water before (inlet) and after (outlet) the Effluent Treatment Plant (ETP) is tested to measure the effectiveness of ETP considering that the two samples of water have different properties of pollution level once treated by ETP.<\/p>\n

       <\/p>\n

      s-19<\/strong><\/p>\n<\/div>\n

      \n

       <\/p>\n

      Hypothesis Testing of the Difference Between Two Means <\/strong>(Dependent sample)<\/p>\n

       <\/p>\n

      Do employees perform better at work with music playing. The music was turned on during the working hours of a business with 45 employees. There productivity level averaged 5.2 with a standard deviation of 2.4. On a different day the music was turned off and there were 40 workers. The workers’ productivity level averaged 4.8 with a standard deviation of 1.2. What can we conclude at the 0.05 level?<\/p>\n

       <\/p>\n

      Solution<\/strong><\/p>\n

       <\/p>\n

      We first develop the hypotheses<\/p>\n

       <\/p>\n\n\n\n\n
      H0:<\/td>\n1 –<\/td>\n2<\/td>\n= 0<\/td>\n<\/tr>\n
      H1:<\/td>\n1 –<\/td>\n2<\/td>\n> 0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

       <\/p>\n

      s-20<\/p>\n

       <\/p>\n

      Next we need to find the standard deviation. Recall from before, we had that the mean of the difference is<\/p>\n

       <\/p>\n

      \"\"<\/p>\n

      We can substitute the sample means and sample standard deviations for a point estimate of the population means and standard deviations. We have\u00a0<\/span>and\u00a0<\/span>Now we can calculate the z-score. We have<\/span><\/p>\n<\/div>\n

      \n

      \"\"<\/p>\n

       <\/p>\n

      s-21<\/p>\n

       <\/p>\n

      Since this is a one tailed test, the critical value is 1.645 and 0.988 does not lie in the critical region. We fail to reject the null hypothesis and conclude that there is insufficient evidence to conclude that workers perform better at work when the music is on. Using the P-Value technique, we see that the P-value associated with 0.988 is<\/p>\n

       <\/p>\n

      P = 1 – 0.8389 = 0.1611<\/p>\n

       <\/p>\n

      which is larger than 0.05.\u00a0 Yet another way of seeing that we fail to reject the null hypothesis.<\/p>\n

       <\/p>\n

      Note: <\/strong>It would have been slightly more accurate had we used the t-table instead of the z-table. To calculate the degrees of freedom, we can take the smaller of the two numbers n1 – 1 and n2 – 1. So in this example, a better estimate would use 39 degrees of freedom. The t-table gives a value of 1.690 for the t.95 value. Notice that 0.988 is still smaller than 1.690 and the result is the same. This is an example that demonstrates that using the t-table and z-table for large samples results in practically the same results.<\/p>\n

       <\/p>\n

      s-22<\/strong><\/p>\n

       <\/p>\n

      Hypothesis Testing For a Difference between Means for Small Samples Using Pooled Standard Deviations (Optional)<\/strong><\/p>\n

       <\/p>\n

      Recall that for small samples we need to make the following assumptions:<\/p>\n

      1.\u00a0\u00a0\u00a0\u00a0\u00a0 Random unbiased sample.<\/span><\/p>\n

      2.\u00a0\u00a0\u00a0\u00a0\u00a0 Both population distributions are normal.<\/span><\/p>\n

      3.\u00a0\u00a0\u00a0\u00a0\u00a0 The two standard deviations are equal.<\/span><\/p>\n<\/div>\n

      \n

       <\/p>\n

      s-23<\/p>\n

      \"\"<\/p>\n

       <\/p>\n

      Putting this together with hypothesis testing we can find the t-statistic. and use n1 + n2 – 2 degrees of freedom.<\/p>\n

       <\/p>\n

      s-24<\/strong><\/p>\n

       <\/p>\n

      Example<\/strong><\/p>\n

       <\/p>\n

      Nine dogs and ten cats were tested to determine if there is a difference in the average number of days that the animal can survive without food. The dogs averaged 11 days with a standard\u00a0deviation of 2 days while the cats averaged 12 days with a standard deviation of 3 days. What can be concluded? (Use = .05)<\/span><\/p>\n<\/div>\n

      \n

       <\/p>\n

      Solution<\/strong><\/p>\n

       <\/p>\n

      We write:<\/p>\n

       <\/p>\n

      H0:\u00a0\u00a0\u00a0\u00a0\u00a0 dog –\u00a0\u00a0\u00a0 cat = 0<\/p>\n

       <\/p>\n

      H1:\u00a0\u00a0\u00a0\u00a0\u00a0 dog –\u00a0\u00a0\u00a0 cat 0<\/p>\n

       <\/p>\n

      We have:<\/p>\n

       <\/p>\n\n\n\n\n\n
      n1<\/td>\n= 9,<\/td>\nn2\u00a0 = 10<\/td>\n<\/tr>\n
      x1<\/td>\n= 11,<\/td>\nx2<\/td>\n= 12<\/td>\n<\/tr>\n
      s1<\/td>\n= 2,<\/td>\ns2<\/td>\n= 3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

       <\/p>\n

      s-25<\/p>\n

       <\/p>\n

      The t-critical value corresponding to a = 0.05 with 10 + 9 – 2 = 17 degrees of freedom is 2.11 which is greater than 0.84. Hence we fail to reject the null hypothesis and conclude that there is not sufficient evidence to suggest that there is a difference between the mean starvation time for cats and dogs.<\/p>\n

       <\/p>\n

      Hypothesis Testing for a Difference between Proportions<\/strong><\/p>\n

       <\/p>\n

      Inferences on the Difference between Population Proportions<\/strong><\/p>\n

       <\/p>\n

      If two samples are counted independently of each other we use the test statistic:<\/span><\/p>\n<\/div>\n

      \n

      \"\"<\/p>\n

       <\/p>\n

      s-26<\/p>\n

       <\/p>\n

      Example<\/strong><\/p>\n

       <\/p>\n

      Is the severity of the drug problem in high school the same for boys and girls? 85 boys and 70 girls were questioned and 34 of the boys and 14 of the girls admitted to having tried some sort of drug. What can be concluded at the 0.05 level?<\/p>\n

       <\/p>\n

      s-27<\/strong><\/p>\n

       <\/p>\n

      Solution<\/strong><\/p>\n

       <\/p>\n

      The hypothesis are<\/p>\n

       <\/p>\n

      H0: p1 – p2\u00a0 = 0<\/p>\n

      H1:\u00a0\u00a0 p1 – p2<\/span>0<\/span><\/p>\n

       <\/p>\n

      We have<\/span><\/p>\n

       <\/p>\n

      p1<\/span><\/p>\n<\/div>\n

      \n

      = 34\/85 = 0.4<\/p>\n

      p2<\/p>\n

      = 14\/70 = 0.2<\/p>\n<\/div>\n

      \n

       <\/p>\n

      p = 48\/155 = 0.31<\/p>\n

      q = 0.69<\/p>\n<\/div>\n

       <\/p>\n

      Now compute the z-score<\/p>\n

      \"\"<\/p>\n

       <\/p>\n

      Since we are using a significance level of .05 and it is a two tailed test, the critical value is 1.96. Clearly 2.68 is in the critical region, hence we can reject the null hypothesis and accept the alternative hypothesis and conclude that gender does make a difference for drug use. Notice that the P-Value is<\/p>\n

       <\/p>\n

      P = 2(1 – .9963)\u00a0 = 0.0074 is less than 0.05. Yet another way to see that we reject the null hypothesis.<\/p>\n

       <\/p>\n

       <\/p>\n

      Summary<\/strong><\/p>\n

       <\/p>\n

       <\/p>\n

      This module help in understanding of sampling distributions when two sets of samples from the same or different populations. If they are random samples from the same population, then any differences\u00a0 across\u00a0 conditions\u00a0 or\u00a0 groups\u00a0 can\u00a0 be\u00a0 attributed\u00a0 to\u00a0 random\u00a0 sampling\u00a0 variability. However, if the two sets of scores are random samples from different populations, then we can attribute any difference between mean scores conditions to the independent variable or the treatment effect.<\/p>\n

       <\/p>\n

      Self exercise question<\/strong><\/p>\n

       <\/p>\n

      The self exercise question is also mentioned on above describe pages of every topic.<\/p>\n

       <\/p>\n

      Learn More:<\/strong><\/p>\n

        \n
      1. \n
          \n
        1. Sharma, J K (2014), Business Statistics, S Chand & Company, N Delhi.<\/li>\n
        2. Bajpai, N (2010) Business Statistics, Pearson, N Delhi.<\/li>\n
        3. Trevor Hastie, Robert Tibshirani, Jerome Friedman (2009), The Elements of Statistical Learning: Data Mining, Inference, and Prediction, 2nd Edition, Springer.<\/li>\n
        4. Darrell Huff (2010), How to Lie with Statistics,\u00a0 W. W. Norton, California.<\/li>\n
        5. K.R. Gupta (2012), Practical Statistics, Atlantic Publishers & Distributors (P) Ltd., N. Delhi<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

           <\/p>\n

           <\/p>\n","protected":false},"author":3,"menu_order":24,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":["prof-pankaj-madan"],"pb_section_license":""},"chapter-type":[],"contributor":[61],"license":[],"class_list":["post-259","chapter","type-chapter","status-publish","hentry","contributor-prof-pankaj-madan"],"part":3,"_links":{"self":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/wp-json\/pressbooks\/v2\/chapters\/259","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/wp-json\/wp\/v2\/users\/3"}],"version-history":[{"count":5,"href":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/wp-json\/pressbooks\/v2\/chapters\/259\/revisions"}],"predecessor-version":[{"id":271,"href":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/wp-json\/pressbooks\/v2\/chapters\/259\/revisions\/271"}],"part":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/wp-json\/pressbooks\/v2\/parts\/3"}],"metadata":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/wp-json\/pressbooks\/v2\/chapters\/259\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/wp-json\/wp\/v2\/media?parent=259"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/wp-json\/pressbooks\/v2\/chapter-type?post=259"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/wp-json\/wp\/v2\/contributor?post=259"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/ebooks.inflibnet.ac.in\/mgmtp15\/wp-json\/wp\/v2\/license?post=259"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}