{"id":287,"date":"2021-07-14T15:59:07","date_gmt":"2021-07-14T15:59:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/two-population-means-with-unknown-standard-deviations\/"},"modified":"2023-12-05T09:38:01","modified_gmt":"2023-12-05T09:38:01","slug":"two-population-means-with-unknown-standard-deviations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/two-population-means-with-unknown-standard-deviations\/","title":{"raw":"Testing for Two Population Means","rendered":"Testing for Two Population Means"},"content":{"raw":"
\r\n

Learning Outcomes<\/h3>\r\n
\r\n
    \r\n \t
  • Conduct a hypothesis test for a difference in two population means with unknown standard deviations and interpret the conclusion in context<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\n
      \r\n \t
    1. The two independent samples are simple random samples from two distinct populations.<\/li>\r\n \t
    2. For the two distinct populations:\r\n
        \r\n \t
      • if the sample sizes are small, the distributions are important (should be normal)<\/li>\r\n \t
      • if the sample sizes are large, the distributions are not important (need not be normal)<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\nNote:<\/strong> The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom formula was developed by Aspin-Welch.\r\n\r\nThe comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, [latex]\\displaystyle\\overline{{X}}_{{1}}-\\overline{{X}}_{{2}} [\/latex], and divide by the standard error in order to standardize the difference. The result is a t-score test statistic.\r\n\r\nBecause we do not know the population standard deviations, we estimate them using the two-sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error<\/strong>, of the difference in sample means, <\/strong>[latex]\\displaystyle\\overline{{X}}_{{1}}-\\overline{{X}}_{{2}} [\/latex].\r\n\r\nThe standard error is: [latex]\\displaystyle\\sqrt{\\frac{(s_1)^2}{n_1}+\\frac{(s_2)^2}{n_2}} [\/latex]\r\n
        \r\n

        Recall: Order of Operations<\/h3>\r\n
        \r\n\r\nWhen simplifying mathematical expressions perform the operations in the following order:\r\n1. P<\/strong>arentheses and other Grouping Symbols\r\n
          \r\n \t
        • Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first.<\/li>\r\n<\/ul>\r\n2. E<\/strong>xponents\r\n
            \r\n \t
          • Simplify all expressions with exponents.<\/li>\r\n<\/ul>\r\n3. M<\/strong>ultiplication and D<\/strong>ivision\r\n
              \r\n \t
            • Perform all multiplication and division in order from left to right. These operations have equal priority.<\/li>\r\n<\/ul>\r\n4. A<\/strong>ddition and S<\/strong>ubtraction\r\n
                \r\n \t
              • Perform all addition and subtraction in order from left to right. These operations have equal priority.<\/li>\r\n<\/ul>\r\n<\/div>\r\nFor the standard error formula, you would follow the following steps:\r\n\r\nFirst, calculate [latex]\\frac{(s_1)^2}{n_1}[\/latex] by removing the parentheses by squaring the standard deviation of the first data set and then divide by n of the first data set.\r\n\r\nSecond, calculate [latex]\\frac{(s_2)^2}{n_2}[\/latex] by removing the parentheses by squaring the standard deviation of the second data set and then divide by n of the second data set.\r\n\r\nThird, add what you got in both previous steps and then take the square root of the sum.\r\n\r\n<\/div>\r\nThe test statistic (t<\/em>-score) is calculated as follows: [latex]t= \\dfrac{(\\overline{x}_1-\\overline{x}_2)-(\\mu_1-\\mu_2)}{\\displaystyle\\sqrt{\\frac{(s_1)^2}{n_1}+\\frac{(s_2)^2}{n_2}}} [\/latex]\r\n\r\nWhere:<\/strong>\r\n
                  \r\n \t
                • s<\/em>1<\/sub> and s<\/em>2<\/sub>, the sample standard deviations, are estimates of \u03c3<\/em>1<\/sub> and \u03c3<\/em>2<\/sub>, respectively.<\/li>\r\n \t
                • \u03c3<\/em>1<\/sub> and \u03c3<\/em>1<\/sub> are the unknown population standard deviations.<\/li>\r\n \t
                • [latex]\\displaystyle\\overline{{x}}_{{1}} [\/latex] and [latex]\\overline{{x}}_{{2}} [\/latex] are the sample means.<\/li>\r\n \t
                • [latex]\\mu_1 [\/latex] and [latex]\\mu_2[\/latex] are the population means.<\/li>\r\n<\/ul>\r\nThe number of degrees of freedom (df<\/em>)<\/strong> requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The df<\/em> are not always a whole number. The test statistic calculated previously is approximated by the Student's t<\/em>-distribution with df<\/em> as follows:\r\n\r\n[latex]\\displaystyle{df}=\\dfrac{((\\dfrac{(s_1)^2}{n_1})+(\\dfrac{(s_2)^2}{n_2}))^2}{(\\dfrac{1}{n_1-1})(\\dfrac{(s_1)^2}{n_1})^2+(\\dfrac{1}{n_2-1})(\\dfrac{(s_2)^2}{n_2})^2} [\/latex]\r\n\r\nWhen both sample sizes n<\/em>1<\/sub> and n<\/em>2<\/sub> are five or larger, the Student's t<\/em> approximation is very good. Notice that the sample variances (s<\/em>1<\/sub>)2<\/sup> and (s<\/em>2<\/sub>)2<\/sup> are not pooled. (If the question comes up, do not pool the variances.)\r\n\r\nNote:<\/strong> It is not necessary to compute this by hand. A calculator or computer easily computes it.\r\n
                  \r\n

                  Example 1<\/h3>\r\nIndependent groups<\/strong>\r\n\r\nThe average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in the table below. Each populations has a normal distribution.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                  <\/th>\r\nSample Size<\/th>\r\nAverage Number of Hours Playing Sports Per Day<\/th>\r\nSample Standard Deviation<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
                  Girls<\/td>\r\n9<\/td>\r\n2<\/td>\r\n0.866<\/td>\r\n<\/tr>\r\n
                  Boys<\/td>\r\n16<\/td>\r\n3.2<\/td>\r\n1.00<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIs there a difference in the mean amount of time boys and girls aged seven to 11 play sports each day? Test at the 5% level of significance.\r\n\r\n[reveal-answer q=\"67452\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"67452\"]\r\n\r\nThe population standard deviations are not known. <\/strong>Let g<\/em> be the subscript for girls and b<\/em> be the subscript for boys. Then, \u03bcg<\/sub><\/em> is the population mean for girls and \u03bcb<\/sub><\/em> is the population mean for boys. This is a test of two independent groups<\/strong>, two population means<\/strong>.\r\n\r\nRandom variable: <\/strong>[latex]\\displaystyle\\overline{{X}}_{{{g}}}-\\overline{{X}}_{{b}} [\/latex] = difference in the sample mean amount of time girls and boys play sports each day.\r\n\r\n[latex]H_0:\\mu_g=\\mu_b [\/latex]; [latex]H_0:\\mu_g-\\mu_b=0 [\/latex]\r\n\r\n[latex]H_a:\\mu_g\\neq\\mu_b [\/latex]; [latex]H_a:\\mu_g-\\mu_b\\neq{0} [\/latex]\r\n\r\nThe words \"the same<\/strong>\" tell you H0<\/sub><\/em> has an equal sign. Since there are no other words to indicate Ha<\/sub><\/em>, assume it says \"is different<\/strong>.\" This is a two-tailed test.\r\n\r\nDistribution for the test:<\/strong> Use tdf<\/sub><\/em> where df<\/em> is calculated using the df <\/em>formula for independent groups, two population means. Using a calculator, df<\/em> is approximately 18.8462. Do not pool the variances.<\/strong>\r\n\r\nCalculate the p<\/em>-value using a Student's t<\/em>-distribution:<\/strong> p<\/em>-value = 0.0054\r\n\r\nGraph:<\/strong>\r\n\r\n\"This\r\n\r\nsg<\/sub><\/em> = 0.866\r\n\r\nsb<\/sub><\/em> = 1\r\n\r\nSo, [latex]\\displaystyle\\overline{x}_g-\\overline{x}_b=2-3.2=-1.2 [\/latex]\r\n\r\nHalf the p<\/em>-value is below \u20131.2 and half is above 1.2.\r\n\r\nMake a decision:<\/strong> Since \u03b1<\/em> > p<\/em>-value, reject H0<\/sub><\/em>. This means you reject \u03bcg<\/sub><\/em> = \u03bcb<\/sub><\/em>. The means are different.\r\n\r\n
                  \r\n

                  USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\n<\/header>
                  \r\n
                  <\/div>\r\n<\/section>\r\n
                    \r\n \t
                  • Press STAT<\/code>.<\/li>\r\n \t
                  • Arrow over to TESTS<\/code> and press 4:2-SampTTest<\/code>.<\/li>\r\n \t
                  • Arrow over to Stats and press ENTER<\/code>.<\/li>\r\n \t
                  • Arrow down and enter 2<\/code> for the first sample mean, [latex]0.866[\/latex] for Sx1, 9<\/code> for n1, 3.2<\/code> for the second sample mean, 1<\/code> for Sx2, and 16<\/code> for n2.<\/li>\r\n \t
                  • Arrow down to \u03bc1: and arrow to does not equal<\/code> \u03bc2.<\/li>\r\n \t
                  • Press ENTER<\/code>.<\/li>\r\n \t
                  • Arrow down to Pooled: andNo<\/code>.<\/li>\r\n \t
                  • Press ENTER<\/code>.<\/li>\r\n \t
                  • Arrow down to Calculate<\/code> and press ENTER<\/code>.<\/li>\r\n \t
                  • The p<\/em>-value is <\/span>p<\/em> = 0.0054, the <\/span>dfs<\/em> are approximately 18.8462, and the test statistic is \u20133.14.<\/span><\/li>\r\n \t
                  • Do the procedure again but instead of Calculate do Draw.<\/li>\r\n<\/ul>\r\nConclusion: <\/strong>At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (the mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per day is greater than the mean number of hours played by boys).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
                    \r\n

                    try it 1<\/h3>\r\nTwo samples are shown in the table. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                    <\/th>\r\nSample Size<\/th>\r\nSample Mean<\/th>\r\nSample Standard Deviation<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
                    Population A<\/td>\r\n25<\/td>\r\n5<\/td>\r\n1<\/td>\r\n<\/tr>\r\n
                    Population B<\/td>\r\n16<\/td>\r\n4.7<\/td>\r\n1.2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"14467\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"14467\"]\r\n\r\nThe p<\/em>-value is 0.4125, which is much higher than 0.05, so we decline to reject the null hypothesis. There is not sufficient evidence to conclude that the means of the two populations are not the same.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNote:<\/strong> When the sum of the sample sizes is larger than 30 (n<\/em>1<\/sub> + n<\/em>2<\/sub> > 30) you can use the normal distribution to approximate the Student's t<\/em>.\r\n
                    \r\n

                    Example 2<\/h3>\r\nA study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is four math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of one math class. The community group believes that a student who graduates from College A has taken more math classes<\/strong>, on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions.\r\n
                      \r\n \t
                    1. Is this a test of two means or two proportions?<\/li>\r\n \t
                    2. Are the populations standard deviations known or unknown?<\/li>\r\n \t
                    3. Which distribution do you use to perform the test?<\/li>\r\n \t
                    4. What is the random variable?<\/li>\r\n \t
                    5. What are the null and alternate hypotheses?<\/li>\r\n \t
                    6. Is this test right-, left-, or two-tailed?<\/li>\r\n \t
                    7. What is the p<\/em>-value?<\/li>\r\n \t
                    8. Do you reject or not reject the null hypothesis?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"493064\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"493064\"]\r\n
                        \r\n \t
                      1. two means<\/li>\r\n \t
                      2. unknown<\/li>\r\n \t
                      3. Student's t<\/em><\/li>\r\n \t
                      4. [latex]\\displaystyle\\overline{{X}}_{{{A}}}-\\overline{{X}}_{{B}} [\/latex]<\/li>\r\n \t
                      5. H0<\/sub><\/em>: \u03bcA<\/sub><\/em>\u2264 \u03bcB<\/sub><\/em>\r\n Ha<\/sub><\/em>: \u03bcA<\/sub><\/em>> \u03bcB<\/sub><\/em><\/li>\r\n \t
                      6. right\r\n\"This<\/li>\r\n \t
                      7. 0.1928<\/li>\r\n \t
                      8. do not reject<\/li>\r\n<\/ol>\r\nConclusion: <\/strong>At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from College A has taken more math classes, on the average, than a student who graduates from College B.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
                        \r\n

                        try it 2<\/h3>\r\nA study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed.\r\n
                          \r\n \t
                        1. Are the population standard deviations known?<\/li>\r\n \t
                        2. Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion?<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"196475\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"196475\"]\r\n
                            \r\n \t
                          1. They are unknown.<\/li>\r\n \t
                          2. The p<\/em>-value = 0.0878. At the 5% level of significance, there is insufficient evidence to conclude that the workers of Company A stay longer with the company.<\/span><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nhttps:\/\/www.youtube.com\/embed\/mvye6X_0upA\r\n
                            \r\n

                            Example 3<\/h3>\r\nA professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in the two tables below:\r\n\r\nOnline Class:\r\n\r\n\r\n\r\n\r\n\r\n
                            67.6<\/td>\r\n41.2<\/td>\r\n85.3<\/td>\r\n55.9<\/td>\r\n82.4<\/td>\r\n91.2<\/td>\r\n73.5<\/td>\r\n94.1<\/td>\r\n64.7<\/td>\r\n64.7<\/td>\r\n<\/tr>\r\n
                            70.6<\/td>\r\n38.2<\/td>\r\n61.8<\/td>\r\n88.2<\/td>\r\n70.6<\/td>\r\n58.8<\/td>\r\n91.2<\/td>\r\n73.5<\/td>\r\n82.4<\/td>\r\n35.5<\/td>\r\n<\/tr>\r\n
                            94.1<\/td>\r\n88.2<\/td>\r\n64.7<\/td>\r\n55.9<\/td>\r\n88.2<\/td>\r\n97.1<\/td>\r\n85.3<\/td>\r\n61.8<\/td>\r\n79.4<\/td>\r\n79.4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFace-to-face Class:\r\n\r\n\r\n\r\n\r\n\r\n
                            77.9<\/td>\r\n95.3<\/td>\r\n81.2<\/td>\r\n74.1<\/td>\r\n98.8<\/td>\r\n88.2<\/td>\r\n85.9<\/td>\r\n92.9<\/td>\r\n87.1<\/td>\r\n88.2<\/td>\r\n<\/tr>\r\n
                            69.4<\/td>\r\n57.6<\/td>\r\n69.4<\/td>\r\n67.1<\/td>\r\n97.6<\/td>\r\n85.9<\/td>\r\n88.2<\/td>\r\n91.8<\/td>\r\n78.8<\/td>\r\n71.8<\/td>\r\n<\/tr>\r\n
                            98.8<\/td>\r\n61.2<\/td>\r\n92.9<\/td>\r\n90.6<\/td>\r\n97.6<\/td>\r\n100<\/td>\r\n95.3<\/td>\r\n83.5<\/td>\r\n92.9<\/td>\r\n89.4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIs the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? Test at a 5% significance level. Answer the following questions:\r\n
                              \r\n \t
                            1. Is this a test of two means or two proportions?<\/li>\r\n \t
                            2. Are the population standard deviations known or unknown?<\/li>\r\n \t
                            3. Which distribution do you use to perform the test?<\/li>\r\n \t
                            4. What is the random variable?<\/li>\r\n \t
                            5. What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols.<\/li>\r\n \t
                            6. Is this test right, left, or two tailed?<\/li>\r\n \t
                            7. What is the p<\/em>-value?<\/li>\r\n \t
                            8. Do you reject or not reject the null hypothesis?<\/li>\r\n \t
                            9. At the ___ level of significance, from the sample data, there ______ (is\/is not) sufficient evidence to conclude that ______.<\/li>\r\n<\/ol>\r\n(Review the conclusion in Example 2, and write yours in a similar fashion)\r\n\r\nBe careful not to mix up the information for Group 1 and Group 2!\r\n\r\n[reveal-answer q=\"984147\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"984147\"]\r\n\r\n
                              \r\n

                              USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\r\n<\/header>First put the data for each group into two lists (such as L1 and L2). Press STAT. Arrow over to TESTS and press 4:2SampTTest. Make sure Data is highlighted and press ENTER. Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to\r\n\u03bc<\/em>1<\/sub>: and arrow to \u2260 \u03bc<\/em>2<\/sub> (does not equal). Press ENTER. Arrow down to Pooled: No. Press ENTER. Arrow down to Calculate and press ENTER.\r\n
                                \r\n \t
                              1. two means<\/li>\r\n \t
                              2. unknown<\/li>\r\n \t
                              3. Student's t<\/em><\/li>\r\n \t
                              4. [latex]\\displaystyle\\overline{{X}}_{{1}}-\\overline{{X}}_{{2}} [\/latex]\r\n
                                  \r\n \t
                                1. H0<\/sub><\/em>: \u03bc1<\/sub><\/em> = \u03bc2<\/sub><\/em> Null hypothesis: the means of the final exam scores are equal for the online and face-to-face statistics classes.<\/li>\r\n \t
                                2. Ha<\/sub><\/em>: \u03bc1<\/sub><\/em> < \u03bc2<\/sub><\/em> Alternative hypothesis: the mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                3. left-tailed<\/li>\r\n \t
                                4. p<\/em>-value = 0.0011\r\n\"This<\/li>\r\n \t
                                5. reject the null hypothesis<\/li>\r\n \t
                                6. The professor was correct. The evidence shows that the mean of the final exam scores for the online class is lower than that of the face-to-face class. At the 5%<\/span> level of significance, from the sample data, there is<\/span> (is\/is not) sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the face-to-face class.<\/span><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"
                                  \n

                                  Learning Outcomes<\/h3>\n
                                  \n
                                    \n
                                  • Conduct a hypothesis test for a difference in two population means with unknown standard deviations and interpret the conclusion in context<\/li>\n<\/ul>\n<\/section>\n<\/div>\n
                                      \n
                                    1. The two independent samples are simple random samples from two distinct populations.<\/li>\n
                                    2. For the two distinct populations:\n
                                        \n
                                      • if the sample sizes are small, the distributions are important (should be normal)<\/li>\n
                                      • if the sample sizes are large, the distributions are not important (need not be normal)<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n

                                        Note:<\/strong> The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom formula was developed by Aspin-Welch.<\/p>\n

                                        The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, [latex]\\displaystyle\\overline{{X}}_{{1}}-\\overline{{X}}_{{2}}[\/latex], and divide by the standard error in order to standardize the difference. The result is a t-score test statistic.<\/p>\n

                                        Because we do not know the population standard deviations, we estimate them using the two-sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error<\/strong>, of the difference in sample means, <\/strong>[latex]\\displaystyle\\overline{{X}}_{{1}}-\\overline{{X}}_{{2}}[\/latex].<\/p>\n

                                        The standard error is: [latex]\\displaystyle\\sqrt{\\frac{(s_1)^2}{n_1}+\\frac{(s_2)^2}{n_2}}[\/latex]<\/p>\n

                                        \n

                                        Recall: Order of Operations<\/h3>\n
                                        \n

                                        When simplifying mathematical expressions perform the operations in the following order:
                                        \n1. P<\/strong>arentheses and other Grouping Symbols<\/p>\n

                                          \n
                                        • Simplify all expressions inside the parentheses or other grouping symbols, working on the innermost parentheses first.<\/li>\n<\/ul>\n

                                          2. E<\/strong>xponents<\/p>\n

                                            \n
                                          • Simplify all expressions with exponents.<\/li>\n<\/ul>\n

                                            3. M<\/strong>ultiplication and D<\/strong>ivision<\/p>\n

                                              \n
                                            • Perform all multiplication and division in order from left to right. These operations have equal priority.<\/li>\n<\/ul>\n

                                              4. A<\/strong>ddition and S<\/strong>ubtraction<\/p>\n

                                                \n
                                              • Perform all addition and subtraction in order from left to right. These operations have equal priority.<\/li>\n<\/ul>\n<\/div>\n

                                                For the standard error formula, you would follow the following steps:<\/p>\n

                                                First, calculate [latex]\\frac{(s_1)^2}{n_1}[\/latex] by removing the parentheses by squaring the standard deviation of the first data set and then divide by n of the first data set.<\/p>\n

                                                Second, calculate [latex]\\frac{(s_2)^2}{n_2}[\/latex] by removing the parentheses by squaring the standard deviation of the second data set and then divide by n of the second data set.<\/p>\n

                                                Third, add what you got in both previous steps and then take the square root of the sum.<\/p>\n<\/div>\n

                                                The test statistic (t<\/em>-score) is calculated as follows: [latex]t= \\dfrac{(\\overline{x}_1-\\overline{x}_2)-(\\mu_1-\\mu_2)}{\\displaystyle\\sqrt{\\frac{(s_1)^2}{n_1}+\\frac{(s_2)^2}{n_2}}}[\/latex]<\/p>\n

                                                Where:<\/strong><\/p>\n

                                                  \n
                                                • s<\/em>1<\/sub> and s<\/em>2<\/sub>, the sample standard deviations, are estimates of \u03c3<\/em>1<\/sub> and \u03c3<\/em>2<\/sub>, respectively.<\/li>\n
                                                • \u03c3<\/em>1<\/sub> and \u03c3<\/em>1<\/sub> are the unknown population standard deviations.<\/li>\n
                                                • [latex]\\displaystyle\\overline{{x}}_{{1}}[\/latex] and [latex]\\overline{{x}}_{{2}}[\/latex] are the sample means.<\/li>\n
                                                • [latex]\\mu_1[\/latex] and [latex]\\mu_2[\/latex] are the population means.<\/li>\n<\/ul>\n

                                                  The number of degrees of freedom (df<\/em>)<\/strong> requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The df<\/em> are not always a whole number. The test statistic calculated previously is approximated by the Student’s t<\/em>-distribution with df<\/em> as follows:<\/p>\n

                                                  [latex]\\displaystyle{df}=\\dfrac{((\\dfrac{(s_1)^2}{n_1})+(\\dfrac{(s_2)^2}{n_2}))^2}{(\\dfrac{1}{n_1-1})(\\dfrac{(s_1)^2}{n_1})^2+(\\dfrac{1}{n_2-1})(\\dfrac{(s_2)^2}{n_2})^2}[\/latex]<\/p>\n

                                                  When both sample sizes n<\/em>1<\/sub> and n<\/em>2<\/sub> are five or larger, the Student’s t<\/em> approximation is very good. Notice that the sample variances (s<\/em>1<\/sub>)2<\/sup> and (s<\/em>2<\/sub>)2<\/sup> are not pooled. (If the question comes up, do not pool the variances.)<\/p>\n

                                                  Note:<\/strong> It is not necessary to compute this by hand. A calculator or computer easily computes it.<\/p>\n

                                                  \n

                                                  Example 1<\/h3>\n

                                                  Independent groups<\/strong><\/p>\n

                                                  The average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in the table below. Each populations has a normal distribution.<\/p>\n\n\n\n\n\n\n
                                                  <\/th>\nSample Size<\/th>\nAverage Number of Hours Playing Sports Per Day<\/th>\nSample Standard Deviation<\/th>\n<\/tr>\n<\/thead>\n
                                                  Girls<\/td>\n9<\/td>\n2<\/td>\n0.866<\/td>\n<\/tr>\n
                                                  Boys<\/td>\n16<\/td>\n3.2<\/td>\n1.00<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                                  Is there a difference in the mean amount of time boys and girls aged seven to 11 play sports each day? Test at the 5% level of significance.<\/p>\n

                                                  Show Answer<\/span><\/p>\n
                                                  \n

                                                  The population standard deviations are not known. <\/strong>Let g<\/em> be the subscript for girls and b<\/em> be the subscript for boys. Then, \u03bcg<\/sub><\/em> is the population mean for girls and \u03bcb<\/sub><\/em> is the population mean for boys. This is a test of two independent groups<\/strong>, two population means<\/strong>.<\/p>\n

                                                  Random variable: <\/strong>[latex]\\displaystyle\\overline{{X}}_{{{g}}}-\\overline{{X}}_{{b}}[\/latex] = difference in the sample mean amount of time girls and boys play sports each day.<\/p>\n

                                                  [latex]H_0:\\mu_g=\\mu_b[\/latex]; [latex]H_0:\\mu_g-\\mu_b=0[\/latex]<\/p>\n

                                                  [latex]H_a:\\mu_g\\neq\\mu_b[\/latex]; [latex]H_a:\\mu_g-\\mu_b\\neq{0}[\/latex]<\/p>\n

                                                  The words “the same<\/strong>” tell you H0<\/sub><\/em> has an equal sign. Since there are no other words to indicate Ha<\/sub><\/em>, assume it says “is different<\/strong>.” This is a two-tailed test.<\/p>\n

                                                  Distribution for the test:<\/strong> Use tdf<\/sub><\/em> where df<\/em> is calculated using the df <\/em>formula for independent groups, two population means. Using a calculator, df<\/em> is approximately 18.8462. Do not pool the variances.<\/strong><\/p>\n

                                                  Calculate the p<\/em>-value using a Student’s t<\/em>-distribution:<\/strong> p<\/em>-value = 0.0054<\/p>\n

                                                  Graph:<\/strong><\/p>\n

                                                  \"This<\/p>\n

                                                  sg<\/sub><\/em> = 0.866<\/p>\n

                                                  sb<\/sub><\/em> = 1<\/p>\n

                                                  So, [latex]\\displaystyle\\overline{x}_g-\\overline{x}_b=2-3.2=-1.2[\/latex]<\/p>\n

                                                  Half the p<\/em>-value is below \u20131.2 and half is above 1.2.<\/p>\n

                                                  Make a decision:<\/strong> Since \u03b1<\/em> > p<\/em>-value, reject H0<\/sub><\/em>. This means you reject \u03bcg<\/sub><\/em> = \u03bcb<\/sub><\/em>. The means are different.<\/p>\n

                                                  \n

                                                  USING THE TI-83, 83+, 84, 84+ CALCULATOR<\/span><\/h4>\n<\/header>\n
                                                  \n
                                                  <\/div>\n<\/section>\n
                                                    \n
                                                  • Press STAT<\/code>.<\/li>\n
                                                  • Arrow over to TESTS<\/code> and press 4:2-SampTTest<\/code>.<\/li>\n
                                                  • Arrow over to Stats and press ENTER<\/code>.<\/li>\n
                                                  • Arrow down and enter 2<\/code> for the first sample mean, [latex]0.866[\/latex] for Sx1, 9<\/code> for n1, 3.2<\/code> for the second sample mean, 1<\/code> for Sx2, and 16<\/code> for n2.<\/li>\n
                                                  • Arrow down to \u03bc1: and arrow to does not equal<\/code> \u03bc2.<\/li>\n
                                                  • Press ENTER<\/code>.<\/li>\n
                                                  • Arrow down to Pooled: andNo<\/code>.<\/li>\n
                                                  • Press ENTER<\/code>.<\/li>\n
                                                  • Arrow down to Calculate<\/code> and press ENTER<\/code>.<\/li>\n
                                                  • The p<\/em>-value is <\/span>p<\/em> = 0.0054, the <\/span>dfs<\/em> are approximately 18.8462, and the test statistic is \u20133.14.<\/span><\/li>\n
                                                  • Do the procedure again but instead of Calculate do Draw.<\/li>\n<\/ul>\n

                                                    Conclusion: <\/strong>At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (the mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per day is greater than the mean number of hours played by boys).<\/p>\n<\/div>\n<\/div>\n<\/div>\n

                                                    \n

                                                    try it 1<\/h3>\n

                                                    Two samples are shown in the table. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5% level of significance.<\/p>\n\n\n\n\n\n\n
                                                    <\/th>\nSample Size<\/th>\nSample Mean<\/th>\nSample Standard Deviation<\/th>\n<\/tr>\n<\/thead>\n
                                                    Population A<\/td>\n25<\/td>\n5<\/td>\n1<\/td>\n<\/tr>\n
                                                    Population B<\/td>\n16<\/td>\n4.7<\/td>\n1.2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                                                    Show Answer<\/span><\/p>\n
                                                    \n

                                                    The p<\/em>-value is 0.4125, which is much higher than 0.05, so we decline to reject the null hypothesis. There is not sufficient evidence to conclude that the means of the two populations are not the same.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

                                                    Note:<\/strong> When the sum of the sample sizes is larger than 30 (n<\/em>1<\/sub> + n<\/em>2<\/sub> > 30) you can use the normal distribution to approximate the Student’s t<\/em>.<\/p>\n

                                                    \n

                                                    Example 2<\/h3>\n

                                                    A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is four math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of one math class. The community group believes that a student who graduates from College A has taken more math classes<\/strong>, on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions.<\/p>\n

                                                      \n
                                                    1. Is this a test of two means or two proportions?<\/li>\n
                                                    2. Are the populations standard deviations known or unknown?<\/li>\n
                                                    3. Which distribution do you use to perform the test?<\/li>\n
                                                    4. What is the random variable?<\/li>\n
                                                    5. What are the null and alternate hypotheses?<\/li>\n
                                                    6. Is this test right-, left-, or two-tailed?<\/li>\n
                                                    7. What is the p<\/em>-value?<\/li>\n
                                                    8. Do you reject or not reject the null hypothesis?<\/li>\n<\/ol>\n
                                                      Show Answer<\/span><\/p>\n
                                                      \n
                                                        \n
                                                      1. two means<\/li>\n
                                                      2. unknown<\/li>\n
                                                      3. Student’s t<\/em><\/li>\n
                                                      4. [latex]\\displaystyle\\overline{{X}}_{{{A}}}-\\overline{{X}}_{{B}}[\/latex]<\/li>\n
                                                      5. H0<\/sub><\/em>: \u03bcA<\/sub><\/em>\u2264 \u03bcB<\/sub><\/em>
                                                        \n Ha<\/sub><\/em>: \u03bcA<\/sub><\/em>> \u03bcB<\/sub><\/em><\/li>\n
                                                      6. right
                                                        \n\"This<\/li>\n
                                                      7. 0.1928<\/li>\n
                                                      8. do not reject<\/li>\n<\/ol>\n

                                                        Conclusion: <\/strong>At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from College A has taken more math classes, on the average, than a student who graduates from College B.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

                                                        \n

                                                        try it 2<\/h3>\n

                                                        A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is five years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed.<\/p>\n

                                                          \n
                                                        1. Are the population standard deviations known?<\/li>\n
                                                        2. Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion?<\/li>\n<\/ol>\n
                                                          Show Answer<\/span><\/p>\n
                                                          \n
                                                            \n
                                                          1. They are unknown.<\/li>\n
                                                          2. The p<\/em>-value = 0.0878. At the 5% level of significance, there is insufficient evidence to conclude that the workers of Company A stay longer with the company.<\/span><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n