{"id":290,"date":"2021-07-14T15:59:07","date_gmt":"2021-07-14T15:59:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/matched-or-paired-samples\/"},"modified":"2023-12-05T09:39:54","modified_gmt":"2023-12-05T09:39:54","slug":"matched-or-paired-samples","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/introstatscorequisite\/chapter\/matched-or-paired-samples\/","title":{"raw":"Hypothesis Testing with Paired Samples","rendered":"Hypothesis Testing with Paired Samples"},"content":{"raw":"
\r\n

Learning Outcomes<\/h3>\r\n
\r\n
    \r\n \t
  • Conduct a hypothesis test for matched or paired data and interpret the conclusion in context<\/li>\r\n<\/ul>\r\n<\/section><\/div>\r\nWhen using a hypothesis test for matched or paired samples, the following characteristics should be present:\r\n
      \r\n \t
    1. Simple random sampling is used.<\/li>\r\n \t
    2. Sample sizes are often small.<\/li>\r\n \t
    3. Two measurements (samples) are drawn from the same pair of individuals or objects.<\/li>\r\n \t
    4. Differences are calculated from the matched or paired samples.<\/li>\r\n \t
    5. The differences form the sample that is used for the hypothesis test.<\/li>\r\n \t
    6. Either the matched pairs have differences that come from a population that is normal or the number of differences is sufficiently large so that distribution of the sample mean of differences is approximately normal.<\/li>\r\n<\/ol>\r\nIn a hypothesis test for matched or paired samples, subjects are matched in pairs, and differences are calculated. The differences are the data. The population mean for the differences, \u03bcd<\/sub><\/em>, is then tested using a Student's t<\/em>-test for a single population mean with n<\/em> \u2013 1 degrees of freedom, where n<\/em> is the number of differences.\r\n\r\nThe test statistic (t<\/em>-score) is:\r\n\r\n[latex]\\displaystyle{t}=\\dfrac{{\\overline{{x}}_{{d}}-{\\mu}_{{d}}}}{{{(\\dfrac{{s}_{{d}}}{\\sqrt{{n}}})}}}[\/latex]\r\n
      \r\n

      Recall: ORDER OF OPERATIONS<\/h3>\r\n
      \r\n\r\n\r\n\r\n\r\n\r\n
      Please<\/strong><\/td>\r\nExcuse<\/strong><\/td>\r\nMy<\/strong><\/td>\r\nDear<\/strong><\/td>\r\nAunt<\/strong><\/td>\r\nSally<\/strong><\/td>\r\n<\/tr>\r\n
      parentheses<\/td>\r\nexponents<\/td>\r\nmultiplication<\/td>\r\ndivision<\/td>\r\naddition<\/td>\r\nsubtraction<\/td>\r\n<\/tr>\r\n
      [latex]( \\ )[\/latex]<\/td>\r\n[latex]x^2[\/latex]<\/td>\r\n[latex]\\times \\ \\mathrm{or} \\ \\div[\/latex]<\/td>\r\n[latex]+ \\ \\mathrm{or} \\ -[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n

      To calculate the test statistic (t<\/em>-score) follow these steps:<\/strong><\/h2>\r\nFirst, find the numerator. Calculate the difference between the matched or paired sample mean and the population mean for the differences, [latex](\\overline{x} _d - \\mu _d)[\/latex].<\/strong>\r\n\r\nSecond, find the denominator.<\/strong>\r\n\r\nStep 1: Find the square root of the sample size (number of differences), [latex]\\sqrt{n}[\/latex].\r\n\r\nStep 2: Take the population standard deviation for the differences and divide by the value you found in Step 1.\r\n\r\nThird, take the numerator and divide by the denominator.<\/strong>\r\n\r\n<\/div>\r\n<\/div>\r\nhttps:\/\/www.youtube.com\/embed\/5ABpqVSx33I\r\n
      \r\n

      Example 1<\/h3>\r\nA study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in the table below. A lower score indicates less pain. The \"before\" value is matched to an \"after\" value and the differences are calculated. The differences have a normal distribution. Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Subject:<\/th>\r\nA<\/th>\r\nB<\/th>\r\nC<\/th>\r\nD<\/th>\r\nE<\/th>\r\nF<\/th>\r\nG<\/th>\r\nH<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      Before<\/td>\r\n6.6<\/td>\r\n6.5<\/td>\r\n9.0<\/td>\r\n10.3<\/td>\r\n11.3<\/td>\r\n8.1<\/td>\r\n6.3<\/td>\r\n11.6<\/td>\r\n<\/tr>\r\n
      After<\/td>\r\n6.8<\/td>\r\n2.4<\/td>\r\n7.4<\/td>\r\n8.5<\/td>\r\n8.1<\/td>\r\n6.1<\/td>\r\n3.4<\/td>\r\n2.0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"651770\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"651770\"]\r\n\r\nCorresponding \"before\" and \"after\" values form matched pairs. (Calculate \"after\" \u2013 \"before.\")\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\n
      After Data<\/th>\r\nBefore Data<\/th>\r\nDifference<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      6.8<\/td>\r\n6.6<\/td>\r\n0.2<\/td>\r\n<\/tr>\r\n
      2.4<\/td>\r\n6.5<\/td>\r\n-4.1<\/td>\r\n<\/tr>\r\n
      7.4<\/td>\r\n9<\/td>\r\n-1.6<\/td>\r\n<\/tr>\r\n
      8.5<\/td>\r\n10.3<\/td>\r\n-1.8<\/td>\r\n<\/tr>\r\n
      8.1<\/td>\r\n11.3<\/td>\r\n-3.2<\/td>\r\n<\/tr>\r\n
      6.1<\/td>\r\n8.1<\/td>\r\n-2<\/td>\r\n<\/tr>\r\n
      3.4<\/td>\r\n6.3<\/td>\r\n-2.9<\/td>\r\n<\/tr>\r\n
      2<\/td>\r\n11.6<\/td>\r\n-9.6<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe data for the test<\/strong> are the differences: {0.2, \u20134.1, \u20131.6, \u20131.8, \u20133.2, \u20132, \u20132.9, \u20139.6}\r\n\r\nThe sample mean and sample standard deviation of the differences are: [latex]\\displaystyle\\overline{{x}}_{{d}}=-{3.13}{\\quad\\text{and}\\quad}{s}_{{d}}={2.91}[\/latex]. Verify these values.\r\n\r\nLet [latex]{\\mu}_{d}[\/latex] be the population mean for the differences. We use the subscript to denote \"differences.\"\r\n\r\nRandom variable:<\/strong> [latex]\\displaystyle\\overline{{X}}_{{d}}[\/latex] = the mean difference of the sensory measurements\r\n\r\nH0<\/sub><\/em>: \u03bcd<\/sub><\/em> \u2265 0\r\n\r\nThe null hypothesis is zero or positive, meaning that there is the same or more pain felt after hypnotism. That means the subject shows no improvement. \u03bcd<\/sub><\/em> is the population mean of the differences.)\r\n\r\nHa<\/sub><\/em>: \u03bcd<\/sub><\/em> < 0\r\n\r\nThe alternative hypothesis is negative, meaning there is less pain felt after hypnotism. That means the subject shows improvement. The score should be lower after hypnotism, so the difference ought to be negative to indicate improvement.\r\n\r\nDistribution for the test: <\/strong>The distribution is a Student's t<\/em> with df<\/em> = n <\/em>\u2013 1 = 8 \u2013 1 = 7. Use t<\/em>7<\/sub>. (Notice that the test is for a single population mean.)<\/strong>\r\n\r\nCalculate the p<\/em>-value using the Student's t<\/em>-distribution: <\/strong>p<\/em>-value = 0.0095\r\n\r\nGraph:<\/strong>\r\n\r\n\"Normal\r\n\r\n[latex]\\displaystyle\\overline{{X}}_{{d}}[\/latex] is the random variable for the differences.\r\n\r\nThe sample mean and sample standard deviation of the differences are:\r\n\r\n[latex]\\displaystyle\\overline{{x}}_{{d}}=-{3.13}[\/latex]\r\n\r\n[latex]\\displaystyle{s}_{{d}}={2.91}[\/latex]\r\n\r\nCompare \u03b1<\/em> and the p<\/em>-value:<\/strong> \u03b1<\/em> = 0.05 and p<\/em>-value = 0.0095. \u03b1<\/em> > p<\/em>-value.\r\n\r\nMake a decision:<\/strong>\u00a0since \u03b1<\/em> > p<\/em>-value, reject H0<\/sub><\/em>. This means that \u03bcd<\/sub><\/em> < 0 and there is improvement.\r\n\r\nConclusion:<\/strong>\u00a0at a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears to be effective in reducing pain.\r\n\r\n[\/hidden-answer]\r\n

      Note<\/h4>\r\nFor the TI-83+ and TI-84 calculators, you can either calculate the differences ahead of time (after \u2013 before) and put the differences into a list or you can put the after data into a first list and the before data into a second list. Then, go to a third list and arrow up to the name. Enter 1st list name \u2013 2nd list name. The calculator will do the subtraction, and you will have the differences in the third list.\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
      • Use your list of differences as the data.<\/li>\r\n \t
      • Press\u00a0STAT<\/code> and arrow over to TESTS<\/code>.<\/li>\r\n \t
      • Press 2:T-Test<\/code>.<\/li>\r\n \t
      • Arrow over to Data<\/code> and press ENTER<\/code>.<\/li>\r\n \t
      • Arrow down and enter 0<\/code> for the name of the list where you put the data, and 1<\/code> for Freq:.<\/li>\r\n \t
      • Arrow down to \u03bc<\/code>: and arrow over to <<\/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 0.0094, and the test statistic is \u20133.04.<\/li>\r\n \t
      • Do these instructions again except, arrow to Draw<\/code> (instead of Calculate<\/code>). Press ENTER<\/code>.<\/li>\r\n<\/ul>\r\n<\/div>\r\n
        \r\n

        try it 1<\/h3>\r\nA study was conducted to investigate how effective a new diet was in lowering cholesterol. Results for the randomly selected subjects are shown in the table. The differences have a normal distribution. Are the subjects' cholesterol levels lower on average after the diet? Test at the 5% level.\r\n\r\n\r\n\r\n\r\n\r\n
        Subject<\/td>\r\nA<\/td>\r\nB<\/td>\r\nC<\/td>\r\nD<\/td>\r\nE<\/td>\r\nF<\/td>\r\nG<\/td>\r\nH<\/td>\r\nI<\/td>\r\n<\/tr>\r\n
        Before<\/td>\r\n209<\/td>\r\n210<\/td>\r\n205<\/td>\r\n198<\/td>\r\n216<\/td>\r\n217<\/td>\r\n238<\/td>\r\n240<\/td>\r\n222<\/td>\r\n<\/tr>\r\n
        After<\/td>\r\n199<\/td>\r\n207<\/td>\r\n189<\/td>\r\n209<\/td>\r\n217<\/td>\r\n202<\/td>\r\n211<\/td>\r\n223<\/td>\r\n201<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"391953\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"391953\"]\r\n\r\nThe p<\/em>-value is 0.0130, so we can reject the null hypothesis. There is enough evidence to suggest that the diet lowers cholesterol.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

        <\/h3>\r\n
        \r\n

        Example 2<\/h3>\r\nA college football coach was interested in whether the college's strength development class increased his players' maximum lift (in pounds) on the bench press exercise. He asked four of his players to participate in a study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows:\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        Weight (in pounds)<\/th>\r\nPlayer 1<\/th>\r\nPlayer 2<\/th>\r\nPlayer 3<\/th>\r\nPlayer 4<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        Amount of weight lifted prior to the class<\/td>\r\n205<\/td>\r\n241<\/td>\r\n338<\/td>\r\n368<\/td>\r\n<\/tr>\r\n
        Amount of weight lifted after the class<\/td>\r\n295<\/td>\r\n252<\/td>\r\n330<\/td>\r\n360<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThe coach wants to know if the strength development class makes his players stronger, on average.\r\n[reveal-answer q=\"741609\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"741609\"]\r\n\r\nRecord the differences<\/strong> data. Calculate the differences by subtracting the amount of weight lifted prior to the class from the weight lifted after completing the class. The data for the differences are: {90, 11, -8, -8}. Assume the differences have a normal distribution.\r\n\r\nUsing the differences data, calculate the sample mean and the sample standard deviation.\r\n\r\n[latex]\\displaystyle\\overline{{x}}_{{d}}={21.3},{s}_{{d}}={46.7}[\/latex]\r\n

        Note<\/h4>\r\nThe data given here would indicate that the distribution is actually right-skewed. The difference 90 may be an extreme outlier? It is pulling the sample mean to be 21.3 (positive). The means of the other three data values are actually negative.\r\n\r\nUsing the difference data, this becomes a test of a single __________ (fill in the blank).\r\n\r\nDefine the random variable:<\/strong> [latex]\\displaystyle\\overline{{X}}_{{d}}[\/latex] mean difference in the maximum lift per player.\r\n\r\nThe distribution for the hypothesis test is t3<\/sub><\/em>.\r\n\r\nH0<\/sub><\/em>: \u03bcd<\/sub><\/em> \u2264 0, Ha<\/sub><\/em>: \u03bcd<\/sub><\/em> > 0\r\n\r\nGraph:<\/strong>\r\n\r\n\"Normal\r\n\r\nCalculate the p<\/em>-value:<\/strong>\u00a0the p<\/em>-value is 0.2150\r\n\r\nDecision:<\/strong>\u00a0if the level of significance is 5%, the decision is not to reject the null hypothesis, because \u03b1 < p<\/em>-value.\r\n\r\nConclusion: a<\/strong>t a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped to make the players stronger, on average.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
        \r\n

        try it 2<\/h3>\r\nA new prep class was designed to improve SAT test scores. Five students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data recorded in this table. Are the scores, on average, higher after the class? Test at a 5% level.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        SAT Scores<\/th>\r\nStudent 1<\/th>\r\nStudent 2<\/th>\r\nStudent 3<\/th>\r\nStudent 4<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        Score before class<\/td>\r\n1840<\/td>\r\n1960<\/td>\r\n1920<\/td>\r\n2150<\/td>\r\n<\/tr>\r\n
        Score after class<\/td>\r\n1920<\/td>\r\n2160<\/td>\r\n2200<\/td>\r\n2100<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"380723\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"380723\"]\r\n\r\nThe p<\/em>-value is 0.0874, so we decline to reject the null hypothesis. The data do not support that the class improves SAT scores significantly.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
        \r\n

        Example 3<\/h3>\r\nSeven eighth-graders at Kennedy Middle School measured how far they could push the shot-put with their dominant (writing) hand and their weaker (non-writing) hand. They thought that they could push equal distances with either hand. The data were collected and recorded in this table.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        Distance (in feet) using<\/th>\r\nStudent 1<\/th>\r\nStudent 2<\/th>\r\nStudent 3<\/th>\r\nStudent 4<\/th>\r\nStudent 5<\/th>\r\nStudent 6<\/th>\r\nStudent 7<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        Dominant Hand<\/td>\r\n30<\/td>\r\n26<\/td>\r\n34<\/td>\r\n17<\/td>\r\n19<\/td>\r\n26<\/td>\r\n20<\/td>\r\n<\/tr>\r\n
        Weaker Hand<\/td>\r\n28<\/td>\r\n14<\/td>\r\n27<\/td>\r\n18<\/td>\r\n17<\/td>\r\n26<\/td>\r\n16<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nConduct a hypothesis test to determine whether the mean difference in distances between the children's dominant versus weaker hands is significant.\r\n\r\n[reveal-answer q=\"58390\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"58390\"]\r\n\r\nRecord the differences<\/strong> data. Calculate the differences by subtracting the distances with the weaker hand from the distances with the dominant hand. The data for the differences are: {2, 12, 7, \u20131, 2, 0, 4}. The differences have a normal distribution.\r\n\r\nUsing the differences data, calculate the sample mean and the sample standard deviation. [latex]\\displaystyle\\overline{{x}}_{{d}}={3.71},{s}_{{d}}={4.5}[\/latex].\r\n\r\nRandom variable:<\/strong> [latex]\\displaystyle\\overline{{X}}_{{d}}[\/latex] = mean difference in the distances between the hands.\r\n\r\nDistribution for the hypothesis test:\u00a0<\/strong>t6<\/sub><\/em>\r\n\r\nH<\/em>0<\/sub>: \u03bcd<\/sub><\/em> = 0\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0H<\/em>a<\/sub><\/em>: \u03bc<\/em>d<\/sub><\/em> \u2260 0\r\n\r\nGraph:<\/strong>\r\n\r\n\"This\r\n\r\nCalculate the p<\/em>-value:<\/strong>\u00a0the p<\/em>-value is 0.0716 (using the data directly).\r\n\r\n(test statistic = 2.18. p<\/em>-value = 0.0719 using ([latex]{\\overline x} = 3.71, {s_d} = 4.5[\/latex].)\r\n\r\nDecision:<\/strong>\u00a0assume \u03b1<\/em> = 0.05. Since \u03b1<\/em> < p<\/em>-value, Do not reject H0<\/sub><\/em>.\r\n\r\nConclusion:<\/strong>\u00a0at the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the children's weaker and dominant hands to push the shot-put.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
        \r\n

        try it 3<\/h3>\r\nFive ball players think they can throw the same distance with their dominant hand (throwing) and off-hand (catching hand). The data were collected and recorded in the table below. Conduct a hypothesis test to determine whether the mean difference in distances between the dominant and off-hand is significant. Test at the 5% level.\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        <\/td>\r\nPlayer 1<\/strong><\/td>\r\nPlayer 2<\/strong><\/td>\r\nPlayer 3<\/strong><\/th>\r\nPlayer 4<\/strong><\/td>\r\nPlayer 5<\/strong><\/td>\r\n<\/tr>\r\n<\/thead>\r\n
        Dominant Hand<\/strong><\/td>\r\n120<\/td>\r\n111<\/td>\r\n135<\/td>\r\n140<\/td>\r\n125<\/td>\r\n<\/tr>\r\n
        Off-hand<\/strong><\/td>\r\n105<\/td>\r\n109<\/td>\r\n98<\/td>\r\n111<\/td>\r\n99<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"578994\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"578994\"]\r\n\r\nThe p<\/em>-level is 0.0230, so we can reject the null hypothesis. The data show that the players do not throw the same distance with their off-hands as they do with their dominant hands.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"
        \n

        Learning Outcomes<\/h3>\n
        \n
          \n
        • Conduct a hypothesis test for matched or paired data and interpret the conclusion in context<\/li>\n<\/ul>\n<\/section>\n<\/div>\n

          When using a hypothesis test for matched or paired samples, the following characteristics should be present:<\/p>\n

            \n
          1. Simple random sampling is used.<\/li>\n
          2. Sample sizes are often small.<\/li>\n
          3. Two measurements (samples) are drawn from the same pair of individuals or objects.<\/li>\n
          4. Differences are calculated from the matched or paired samples.<\/li>\n
          5. The differences form the sample that is used for the hypothesis test.<\/li>\n
          6. Either the matched pairs have differences that come from a population that is normal or the number of differences is sufficiently large so that distribution of the sample mean of differences is approximately normal.<\/li>\n<\/ol>\n

            In a hypothesis test for matched or paired samples, subjects are matched in pairs, and differences are calculated. The differences are the data. The population mean for the differences, \u03bcd<\/sub><\/em>, is then tested using a Student’s t<\/em>-test for a single population mean with n<\/em> \u2013 1 degrees of freedom, where n<\/em> is the number of differences.<\/p>\n

            The test statistic (t<\/em>-score) is:<\/p>\n

            [latex]\\displaystyle{t}=\\dfrac{{\\overline{{x}}_{{d}}-{\\mu}_{{d}}}}{{{(\\dfrac{{s}_{{d}}}{\\sqrt{{n}}})}}}[\/latex]<\/p>\n

            \n

            Recall: ORDER OF OPERATIONS<\/h3>\n
            \n\n\n\n\n\n
            Please<\/strong><\/td>\nExcuse<\/strong><\/td>\nMy<\/strong><\/td>\nDear<\/strong><\/td>\nAunt<\/strong><\/td>\nSally<\/strong><\/td>\n<\/tr>\n
            parentheses<\/td>\nexponents<\/td>\nmultiplication<\/td>\ndivision<\/td>\naddition<\/td>\nsubtraction<\/td>\n<\/tr>\n
            [latex]( \\ )[\/latex]<\/td>\n[latex]x^2[\/latex]<\/td>\n[latex]\\times \\ \\mathrm{or} \\ \\div[\/latex]<\/td>\n[latex]+ \\ \\mathrm{or} \\ -[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

            To calculate the test statistic (t<\/em>-score) follow these steps:<\/strong><\/h2>\n

            First, find the numerator. Calculate the difference between the matched or paired sample mean and the population mean for the differences, [latex](\\overline{x} _d - \\mu _d)[\/latex].<\/strong><\/p>\n

            Second, find the denominator.<\/strong><\/p>\n

            Step 1: Find the square root of the sample size (number of differences), [latex]\\sqrt{n}[\/latex].<\/p>\n

            Step 2: Take the population standard deviation for the differences and divide by the value you found in Step 1.<\/p>\n

            Third, take the numerator and divide by the denominator.<\/strong><\/p>\n<\/div>\n<\/div>\n