{"id":2875,"date":"2022-05-09T15:29:02","date_gmt":"2022-05-09T19:29:02","guid":{"rendered":"https:\/\/openbooks.macewan.ca\/rcommander\/?post_type=chapter&p=2875"},"modified":"2024-01-29T12:29:36","modified_gmt":"2024-01-29T17:29:36","slug":"13-11-review-questions","status":"publish","type":"chapter","link":"https:\/\/openbooks.macewan.ca\/introstats\/chapter\/13-11-review-questions\/","title":{"raw":"13.11 Review Questions","rendered":"13.11 Review Questions"},"content":{"raw":"Researchers examined the controversial issue of the human vomeronasal organ regarding its structure, function, and identity. The following table shows the age of fetuses ([latex]x[\/latex]) in weeks and the length of crown-rump ([latex]y[\/latex]) in millimeters.\r\n
\r\n\r\n\r\n\r\n\r\n
Age ([latex]x[\/latex])<\/th>\r\n10<\/td>\r\n10<\/td>\r\n13<\/td>\r\n13<\/td>\r\n18<\/td>\r\n19<\/td>\r\n19<\/td>\r\n23<\/td>\r\n25<\/td>\r\n28<\/td>\r\n<\/tr>\r\n
Length ([latex]y[\/latex])<\/th>\r\n66<\/td>\r\n66<\/td>\r\n108<\/td>\r\n106<\/td>\r\n161<\/td>\r\n166<\/td>\r\n177<\/td>\r\n228<\/td>\r\n235<\/td>\r\n280<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nThe summaries of the data are given by [latex] n = 10, \\sum x_i = 178, \\sum x_i^2 = 3522, \\sum y_i = 1593, \\sum y_i^2 = 302027, \\sum x_iy_i = 32476. [\/latex]\r\n
    \r\n \t
  1. Given the summaries of the data, find the least-squares regression equation.<\/li>\r\n \t
  2. Graph the regression equation and the data points.<\/li>\r\n \t
  3. Interpret the slope of the regression equation obtained in part (a) in the context of the study.<\/li>\r\n \t
  4. Calculate [latex]r[\/latex], the correlation coefficient between [latex]y[\/latex] and [latex]x[\/latex]. Interpret the number.<\/li>\r\n \t
  5. Calculate the coefficient of determination [latex]r^2[\/latex]. Interpret the number.<\/li>\r\n \t
  6. Test at the 1% significant level whether the age of fetuses is a useful predictor for the length of the crown-rump. You could use [latex]s_e = 5.518.[\/latex]<\/li>\r\n \t
  7. Predict the crown-rump length of a 19-week-old fetus.<\/li>\r\n \t
  8. What is the residual for the last observation with response [latex]y = 280[\/latex] and [latex]x = 28[\/latex]?<\/li>\r\n<\/ol>\r\n
    Show\/Hide Answer<\/summary>\r\n
      \r\n \t
    1. [latex]\\begin{align*} S_{xx}&=\\sum x_i^2-\\frac{(\\sum x_i)^2}{n} = 3522-\\frac{(178)^2}{10} =353.6; \\\\\r\nS_{xy}&=\\sum x_iy_i-\\frac{(\\sum x_i)(\\sum y_i)}{n} =32476- \\frac{(178)(1593)}{10}=4120.6; \\\\\r\nS_{yy}&=\\sum y_i^2-\\frac{(\\sum y_i)^2}{n} = 302027- \\frac{(1593)^2}{10} = 48262.1; \\\\\r\nb_1&=\\frac{S_{xy}}{S_{xx}}=\\frac{4120.6}{353.6}=11.65328;\\\\ b_0&=\\frac{\\sum y_i}{n}-b_1\\times \\frac{\\sum x_i}{n}=\\frac{1593}{10}-11.65328\\times \\frac{178}{10}=-48.12838. \\end{align*}[\/latex]\r\nTherefore, the least-squares regression equation [latex]\\hat y=b_0+b_1 x=-48.12838+11.65328x[\/latex] or [latex]\\widehat {length}=-48.12838+11.65328age[\/latex].<\/li>\r\n \t
    2. Left as an exercise for the reader.<\/li>\r\n \t
    3. The slope is [latex]b_1=11.65328[\/latex].\r\nInterpretation: The average length of the crown rump increases by 11.65328 millimeters when the age of the fetus increases by 1 week. In other words, for each week the fetus ages, the expected increase in crown-rump length is 11.65328 mm.<\/li>\r\n \t
    4. The correlation coefficient [latex]r[\/latex] is given by [latex]r=\\frac{S_{xy}}{\\sqrt{S_{xx}S_{yy}}}=\\frac{4120.6}{\\sqrt{(353.6)(48262.1)}}=0.9974732.[\/latex]\r\nInterpretation: there is a very strong, positive, linear association between the length of crown-rump [latex](y)[\/latex] and the age [latex](x)[\/latex] of the fetus.<\/li>\r\n \t
    5. The coefficient of determination is [latex]r^2=0.9974732^2=0.9949528.[\/latex]\r\nInterpretation: 99.50% of the variation in the length of the crown rump is due to the age of the fetus. Or 99.50% of the variation in the length of crown-rump can be explained by the age of the fetus through the fitted regression line [latex]\\hat y=b_0+b_1 x=-48.12838+11.65328x.[\/latex]<\/li>\r\n \t
    6. We assume all assumptions for inference on simple linear regression are satisfied.\r\nStep 1: Hypotheses. [latex]H_0: \\beta_1=0[\/latex] versus [latex]H_a: \\beta_1\\ne 0.[\/latex]\r\nStep 2: Significance level [latex]\\alpha=0.01.[\/latex]\r\nStep 3: Test statistic [latex]t_o=\\frac{b_1}{(\\frac{s_e}{\\sqrt{S_{xx}}})}=\\frac{11.65328}{(\\frac{5.518}{\\sqrt{353.6}})}=39.71208[\/latex] with [latex]df=n-2=10-2=8.[\/latex]\r\nStep 4: P-value. It is a two-tailed test, [latex]\\text{p-value}= 2P(t\\ge |t_o|)=2P(t\\ge 39.71208)<2\\times 0.0005=0.001.[\/latex]\r\nStep 5: Decision. We reject [latex]H_0[\/latex] since [latex]\\text{p-value}<0.001<0.01(\\alpha).[\/latex]\r\nStep 6: Conclusion. At the 1% significant level, we have sufficient evidence that the age of fetuses is a useful predictor for the crown-rump length.<\/li>\r\n \t
    7. [latex]\\hat y=b_0+b_1 x=-48.12838+11.65328x=-48.12838+11.65328\\times 19=173.2839[\/latex]\r\nThe predicted crown-rump length of a 19-week-old fetus is 173.2839 mm.<\/li>\r\n \t
    8. Residual [latex]e=y-\\hat y=y-(b_0+b_1x)=280-(-48.12838+11.65328\\times 28)=280-278.1635=1.8365.[\/latex]<\/li>\r\n<\/ol>\r\n \r\n\r\n<\/details>","rendered":"

      Researchers examined the controversial issue of the human vomeronasal organ regarding its structure, function, and identity. The following table shows the age of fetuses ([latex]x[\/latex]) in weeks and the length of crown-rump ([latex]y[\/latex]) in millimeters.<\/p>\n

      \n\n\n\n\n
      Age ([latex]x[\/latex])<\/th>\n10<\/td>\n10<\/td>\n13<\/td>\n13<\/td>\n18<\/td>\n19<\/td>\n19<\/td>\n23<\/td>\n25<\/td>\n28<\/td>\n<\/tr>\n
      Length ([latex]y[\/latex])<\/th>\n66<\/td>\n66<\/td>\n108<\/td>\n106<\/td>\n161<\/td>\n166<\/td>\n177<\/td>\n228<\/td>\n235<\/td>\n280<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

      The summaries of the data are given by [latex]n = 10, \\sum x_i = 178, \\sum x_i^2 = 3522, \\sum y_i = 1593, \\sum y_i^2 = 302027, \\sum x_iy_i = 32476.[\/latex]<\/p>\n

        \n
      1. Given the summaries of the data, find the least-squares regression equation.<\/li>\n
      2. Graph the regression equation and the data points.<\/li>\n
      3. Interpret the slope of the regression equation obtained in part (a) in the context of the study.<\/li>\n
      4. Calculate [latex]r[\/latex], the correlation coefficient between [latex]y[\/latex] and [latex]x[\/latex]. Interpret the number.<\/li>\n
      5. Calculate the coefficient of determination [latex]r^2[\/latex]. Interpret the number.<\/li>\n
      6. Test at the 1% significant level whether the age of fetuses is a useful predictor for the length of the crown-rump. You could use [latex]s_e = 5.518.[\/latex]<\/li>\n
      7. Predict the crown-rump length of a 19-week-old fetus.<\/li>\n
      8. What is the residual for the last observation with response [latex]y = 280[\/latex] and [latex]x = 28[\/latex]?<\/li>\n<\/ol>\n
        \nShow\/Hide Answer<\/summary>\n
          \n
        1. [latex]\\begin{align*} S_{xx}&=\\sum x_i^2-\\frac{(\\sum x_i)^2}{n} = 3522-\\frac{(178)^2}{10} =353.6; \\\\ S_{xy}&=\\sum x_iy_i-\\frac{(\\sum x_i)(\\sum y_i)}{n} =32476- \\frac{(178)(1593)}{10}=4120.6; \\\\ S_{yy}&=\\sum y_i^2-\\frac{(\\sum y_i)^2}{n} = 302027- \\frac{(1593)^2}{10} = 48262.1; \\\\ b_1&=\\frac{S_{xy}}{S_{xx}}=\\frac{4120.6}{353.6}=11.65328;\\\\ b_0&=\\frac{\\sum y_i}{n}-b_1\\times \\frac{\\sum x_i}{n}=\\frac{1593}{10}-11.65328\\times \\frac{178}{10}=-48.12838. \\end{align*}[\/latex]
          \nTherefore, the least-squares regression equation [latex]\\hat y=b_0+b_1 x=-48.12838+11.65328x[\/latex] or [latex]\\widehat {length}=-48.12838+11.65328age[\/latex].<\/li>\n
        2. Left as an exercise for the reader.<\/li>\n
        3. The slope is [latex]b_1=11.65328[\/latex].
          \nInterpretation: The average length of the crown rump increases by 11.65328 millimeters when the age of the fetus increases by 1 week. In other words, for each week the fetus ages, the expected increase in crown-rump length is 11.65328 mm.<\/li>\n
        4. The correlation coefficient [latex]r[\/latex] is given by [latex]r=\\frac{S_{xy}}{\\sqrt{S_{xx}S_{yy}}}=\\frac{4120.6}{\\sqrt{(353.6)(48262.1)}}=0.9974732.[\/latex]
          \nInterpretation: there is a very strong, positive, linear association between the length of crown-rump [latex](y)[\/latex] and the age [latex](x)[\/latex] of the fetus.<\/li>\n
        5. The coefficient of determination is [latex]r^2=0.9974732^2=0.9949528.[\/latex]
          \nInterpretation: 99.50% of the variation in the length of the crown rump is due to the age of the fetus. Or 99.50% of the variation in the length of crown-rump can be explained by the age of the fetus through the fitted regression line [latex]\\hat y=b_0+b_1 x=-48.12838+11.65328x.[\/latex]<\/li>\n
        6. We assume all assumptions for inference on simple linear regression are satisfied.
          \nStep 1: Hypotheses. [latex]H_0: \\beta_1=0[\/latex] versus [latex]H_a: \\beta_1\\ne 0.[\/latex]
          \nStep 2: Significance level [latex]\\alpha=0.01.[\/latex]
          \nStep 3: Test statistic [latex]t_o=\\frac{b_1}{(\\frac{s_e}{\\sqrt{S_{xx}}})}=\\frac{11.65328}{(\\frac{5.518}{\\sqrt{353.6}})}=39.71208[\/latex] with [latex]df=n-2=10-2=8.[\/latex]
          \nStep 4: P-value. It is a two-tailed test, [latex]\\text{p-value}= 2P(t\\ge |t_o|)=2P(t\\ge 39.71208)<2\\times 0.0005=0.001.[\/latex]\nStep 5: Decision. We reject [latex]H_0[\/latex] since [latex]\\text{p-value}<0.001<0.01(\\alpha).[\/latex]\nStep 6: Conclusion. At the 1% significant level, we have sufficient evidence that the age of fetuses is a useful predictor for the crown-rump length.<\/li>\n
        7. [latex]\\hat y=b_0+b_1 x=-48.12838+11.65328x=-48.12838+11.65328\\times 19=173.2839[\/latex]
          \nThe predicted crown-rump length of a 19-week-old fetus is 173.2839 mm.<\/li>\n
        8. Residual [latex]e=y-\\hat y=y-(b_0+b_1x)=280-(-48.12838+11.65328\\times 28)=280-278.1635=1.8365.[\/latex]<\/li>\n<\/ol>\n

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