{"id":46,"date":"2021-01-12T22:19:31","date_gmt":"2021-01-12T22:19:31","guid":{"rendered":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/contingency-tables\/"},"modified":"2021-09-07T03:56:31","modified_gmt":"2021-09-07T03:56:31","slug":"contingency-tables","status":"publish","type":"chapter","link":"https:\/\/textbooks.jaykesler.net\/introstats\/chapter\/contingency-tables\/","title":{"rendered":"Contingency Tables"},"content":{"raw":"\r\n[latexpage]\r\n<\/span>\r\n
\r\n

A contingency table<\/span> provides a way of portraying data that can facilitate calculating probabilities. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner.<\/p>\r\n\r\n

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Example <\/span>3.20<\/span><\/h3>\r\n<\/header>
\r\n

Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:<\/p>\r\n\r\n

\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
<\/th>\r\nSpeeding violation in the last year<\/th>\r\nNo speeding violation in the last year<\/th>\r\nTotal<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
Uses cell phone while driving<\/td>\r\n25<\/td>\r\n280<\/td>\r\n305<\/td>\r\n<\/tr>\r\n
Does not use cell phone while driving<\/td>\r\n45<\/td>\r\n405<\/td>\r\n450<\/td>\r\n<\/tr>\r\n
Total<\/td>\r\n70<\/td>\r\n685<\/td>\r\n755<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
Table <\/span>3.2<\/span><\/div>\r\n<\/div>\r\n

The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755.<\/p>\r\n

Calculate the following probabilities using the table.<\/p>\r\n \r\n

    \r\n \t
  1. Find P<\/em>(Driver is a cell phone user).\r\n<\/span><\/li>\r\n \t
  2. Find P<\/em>(driver had no violation in the last year).<\/li>\r\n \t
  3. Find P<\/em>(Driver had no violation in the last year AND was a cell phone user).<\/li>\r\n \t
  4. Find P<\/em>(Driver is a cell phone user OR driver had no violation in the last year).<\/li>\r\n \t
  5. Find P<\/em>(Driver is a cell phone user GIVEN driver had a violation in the last year).<\/li>\r\n \t
  6. Find P<\/em>(Driver had no violation last year GIVEN driver was not a cell phone user)<\/li>\r\n<\/ol>\r\n

    Solutions:<\/span><\/h4>\r\n
      \r\n \t
    1. $\\frac{\\text{number of cell phone users}}{\\text{total number in study}}= \\frac{305}{755} = 0.404$\r\n<\/span><\/li>\r\n \t
    2. $\\frac{\\text{number that had no violation}}{\\text{total number in study}}= \\frac{685}{755} = 0.9073$<\/li>\r\n \t
    3. $\\frac{280}{755} = 0.3709$<\/li>\r\n \t
    4. $\\left(\\frac{305}{755}+\\frac{685}{755}\\right)-\\frac{280}{755} = \\frac{710}{755}=0.9404$<\/li>\r\n \t
    5. $\\frac{25}{70}=0.3571$<\/li>\r\n \t
    6. $\\frac{405}{450}=0.9$<\/li>\r\n<\/ol>\r\n<\/section><\/div>\r\n
      \r\n

      Try It <\/span>3.20<\/span><\/h3>\r\n<\/header>
      \r\n
      <\/header>
      \r\n
      \r\n
      \r\n

      Table 3.3<\/a> below shows the number of athletes who stretch before exercising and how many had injuries within the past year.<\/p>\r\n\r\n

      \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      <\/th>\r\nInjury in last year<\/th>\r\nNo injury in last year<\/th>\r\nTotal<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      Stretches<\/td>\r\n55<\/td>\r\n295<\/td>\r\n350<\/td>\r\n<\/tr>\r\n
      Does not stretch<\/td>\r\n231<\/td>\r\n219<\/td>\r\n450<\/td>\r\n<\/tr>\r\n
      Total<\/td>\r\n286<\/td>\r\n514<\/td>\r\n800<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
      Table <\/span>3.3<\/span><\/div>\r\n<\/div>\r\n
        \r\n \t
      1. What is P<\/em>(athlete stretches before exercising)?<\/li>\r\n \t
      2. What is P<\/em>(athlete stretches before exercising|no injury in the last year)?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n
        \r\n

        Example <\/span>3.21<\/span><\/h3>\r\n<\/header>
        \r\n

        Table 3.4<\/a> below shows a random sample of 100 hikers and the areas of hiking they prefer.<\/p>\r\n\r\n

        \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        Sex<\/th>\r\nThe Coastline<\/th>\r\nNear Lakes and Streams<\/th>\r\nOn Mountain Peaks<\/th>\r\nTotal<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
        Female<\/td>\r\n18<\/td>\r\n16<\/td>\r\n___<\/td>\r\n45<\/td>\r\n<\/tr>\r\n
        Male<\/td>\r\n___<\/td>\r\n___<\/td>\r\n14<\/td>\r\n55<\/td>\r\n<\/tr>\r\n
        Total<\/td>\r\n___<\/td>\r\n41<\/td>\r\n___<\/td>\r\n___<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
        Table <\/span>3.4<\/span> Hiking Area Preference<\/span><\/div>\r\n<\/div>\r\n
        <\/header>
        \r\n
        \r\n
        \r\n
          \r\n \t
        1. Complete the table.<\/li>\r\n \t
        2. Are the events \"being female\" and \"preferring the coastline\" independent events?\r\nLet F<\/em> = being female and let C<\/em> = preferring the coastline.\r\n
            \r\n \t
          1. Find P<\/em>(F<\/em> AND C<\/em>).<\/li>\r\n \t
          2. Find P<\/em>(F<\/em>)P<\/em>(C<\/em>)<\/li>\r\n<\/ol>\r\n

            Are these two numbers the same? If they are, then F<\/em> and C<\/em> are independent. If they are not, then F<\/em> and C<\/em> are not independent.<\/p>\r\n<\/li>\r\n \t

          3. \r\n

            Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let M<\/em> = being male, and let L<\/em> = prefers hiking near lakes and streams.<\/p>\r\n\r\n

              \r\n \t
            1. What word tells you this is a conditional?<\/li>\r\n \t
            2. Fill in the blanks and calculate the probability: P<\/em>(___|___) = ___.<\/li>\r\n \t
            3. Is the sample space for this problem all 100 hikers? If not, what is it?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
            4. Find the probability that a person is female or prefers hiking on mountain peaks. Let F<\/em> = being female, and let P<\/em> = prefers mountain peaks.\r\n
                \r\n \t
              1. Find P<\/em>(F<\/em>).<\/li>\r\n \t
              2. Find P<\/em>(P<\/em>).<\/li>\r\n \t
              3. Find P<\/em>(F<\/em> AND P<\/em>).<\/li>\r\n \t
              4. Find P<\/em>(F<\/em> OR P<\/em>).<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n
                \r\n
                \r\n
                <\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n
                \r\n

                Try It <\/span>3.21<\/span><\/h3>\r\n<\/header>
                \r\n
                <\/header>
                \r\n
                \r\n
                \r\n

                Table 3.6<\/a> below shows a random sample of 200 cyclists and the routes they prefer. Let M<\/em> = males and H<\/em> = hilly path.<\/p>\r\n\r\n

                \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                Gender<\/th>\r\nLake Path<\/th>\r\nHilly Path<\/th>\r\nWooded Path<\/th>\r\nTotal<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
                Female<\/td>\r\n45<\/td>\r\n38<\/td>\r\n27<\/td>\r\n110<\/td>\r\n<\/tr>\r\n
                Male<\/td>\r\n26<\/td>\r\n52<\/td>\r\n12<\/td>\r\n90<\/td>\r\n<\/tr>\r\n
                Total<\/td>\r\n71<\/td>\r\n90<\/td>\r\n39<\/td>\r\n200<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
                Table <\/span>3.6<\/span><\/div>\r\n<\/div>\r\n
                  \r\n \t
                1. Out of the males, what is the probability that the cyclist prefers a hilly path?<\/li>\r\n \t
                2. Are the events \u201cbeing male\u201d and \u201cpreferring the hilly path\u201d independent events?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/div>\r\n<\/section><\/div>\r\n
                  \r\n

                  Example <\/span>3.22<\/span><\/h3>\r\n<\/header>
                  Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is $\\frac{1}{5}$ and the probability he is not caught is $\\frac{4}{5}$ . If he goes out the second door, the probability he gets caught by Alissa is $\\frac{1}{4}$ and the probability he is not caught is $\\frac{3}{4}$. The probability that Alissa catches Muddy coming out of the third door is $\\frac{1}{2}$\u00a0 and the probability she does not catch Muddy is $\\frac{1}{2}$ . It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is $\\frac{1}{3}$.\r\n
                  \r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
                  Caught or Not<\/th>\r\nDoor One<\/th>\r\nDoor Two<\/th>\r\nDoor Three<\/th>\r\nTotal<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
                  Caught<\/td>\r\n$\\frac{1}{15}$<\/td>\r\n$\\frac{1}{12}$<\/td>\r\n$\\frac{1}{6}$<\/td>\r\n____<\/td>\r\n<\/tr>\r\n
                  Not Caught<\/td>\r\n$\\frac{4}{15}$<\/td>\r\n$\\frac{3}{12}=\\frac{1}{4}$<\/td>\r\n$\\frac{1}{6}$<\/td>\r\n____<\/td>\r\n<\/tr>\r\n
                  Total<\/td>\r\n____<\/td>\r\n____<\/td>\r\n____<\/td>\r\n1<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
                  Table <\/span>3.7<\/span> Door Choice<\/span><\/div>\r\n<\/div>\r\n