{"id":312,"date":"2020-06-29T15:42:38","date_gmt":"2020-06-29T19:42:38","guid":{"rendered":"https:\/\/openbooks.macewan.ca\/rcommander\/?post_type=chapter&#038;p=312"},"modified":"2024-02-08T13:09:08","modified_gmt":"2024-02-08T18:09:08","slug":"2-6-z-score-as-a-measure-of-relative-standing","status":"publish","type":"chapter","link":"https:\/\/openbooks.macewan.ca\/introstats\/chapter\/2-6-z-score-as-a-measure-of-relative-standing\/","title":{"raw":"2.6 Z-Score as a Measure of Relative Standing","rendered":"2.6 Z-Score as a Measure of Relative Standing"},"content":{"raw":"A college admissions officer is looking at the files of two international candidates, one with a computer-based TOEFL score of 210 and the other with an IELTS score of 7.5. Which score is better? How can we compare measurements with different scales? The solution is the z-score, which gives a relative standing among the population.\r\n\r\nSuppose variable [latex]X[\/latex] follows a distribution with a mean [latex]\\mu[\/latex] and a standard deviation [latex]\\sigma[\/latex], the corresponding standardized variable is defined as\r\n<p style=\"text-align: center;\">[latex] Z = \\frac{X - \\mu}{\\sigma}.[\/latex]<\/p>\r\nIt can be shown that the standardized variable [latex]Z[\/latex] has a mean 0 and a standard deviation 1.\r\n\r\nFor a given value of [latex]X[\/latex], denoted as the corresponding lower-case [latex]x[\/latex], the value of the standardized variable is called the z-score of [latex]x[\/latex], which is given by\r\n<p style=\"text-align: center;\" align=\"center\">[latex] z = \\frac{x-\\mu}{\\sigma}.[\/latex]<\/p>\r\nIf the population parameters [latex]\\mu[\/latex] and [latex]\\sigma[\/latex] are unknown, use the sample mean and standard deviation [latex] \\bar{x}[\/latex]\u00a0and [latex]s[\/latex] to estimate them, then the z-score becomes\r\n<p style=\"text-align: center;\" align=\"center\">[latex] z = \\frac{x - \\bar{x}}{s}.[\/latex]<\/p>\r\nProperties of the z-score:\r\n<ul>\r\n \t<li>It measures how far an individual is away from the mean using standard deviation as the unit (ruler).<\/li>\r\n \t<li>It represents a relative standing of an observation.<\/li>\r\n \t<li>A z-score &gt; 0 means the observation x is above the mean; z-score &lt; 0 means the observation x is below the mean; z-score = 0 means the observation x is equal to the mean.<\/li>\r\n \t<li>A z-score has no unit.<\/li>\r\n<\/ul>\r\n&nbsp;\r\n<div class=\"textbox textbox--examples\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Example: Z-score<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n\r\nSuppose that the TOEFL score for admission at MacEwan has a mean 200 and a standard deviation 10, and the IELTS score has a mean 6 and a standard deviation 1. Two international candidates, one with a TOEFL score of 210 and the other with an IELTS score of 7.5. Which score is relatively better?\r\n\r\nCalculate the z-score for the student who took the TOEFL exam:\r\n<p style=\"text-align: center;\" align=\"center\">[latex] z_T = \\frac{x - \\mu}{\\sigma} = \\frac{210 - 200}{10} = 1.[\/latex]<\/p>\r\nThis z-score means that the student who took the TOEFL exam is one standard deviation <strong>above the average<\/strong>.\r\n\r\nThe z-score for the student who took the IELTS exam:\r\n<p style=\"text-align: center;\" align=\"center\">[latex] z_I = \\frac{x - \\mu}{\\sigma} = \\frac{7.5 - 6}{1} = 1.5.[\/latex]<\/p>\r\nThis z-score means that the student who took the IELTS exam is 1.5 standard deviation <strong>above the average<\/strong>. Therefore, the student who took the IELTS exam is better because the z-score is larger. This tells us that the IELTS student score is relatively further above the mean than the TOEFL student score is above its mean.\r\n\r\n<\/div>\r\n<\/div>\r\n<div style=\"height: 55px; margin-top: 2.1428571429em;\">\r\n\r\n<img class=\"size-full wp-image-99 alignleft\" src=\"https:\/\/openbooks.macewan.ca\/rcommander\/wp-content\/uploads\/sites\/8\/2020\/06\/activity.png\" alt=\"\" width=\"250\" height=\"50\" \/>\r\n\r\n<\/div>\r\n<div class=\"textbox textbox--exercises\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Exercise: Women Heptathlon Champion<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<div>Women\u2019s heptathlon in Olympics includes seven track and field events\u2014200-m and 800-m runs, 100-m high hurdles, shot put, javelin, high jump, and long jump. How do you determine the champion?<\/div>\r\n<\/div>\r\n<details><summary style=\"padding-left: 40px;\">Show\/Hide Answer<\/summary>\r\n<div>\r\n<p style=\"padding-left: 40px;\">The events are measured in different units, some are in metres and some are in seconds. Moreover, for those runs and high hurdles, results are the smaller the better; for jumps, shot put, and javelin, the larger the better. In order to combine the results of the seven events to a single score to determine the winner, one possible solution is the z-score. The idea is as follows:<\/p>\r\n<p style=\"padding-left: 40px;\">For each athlete, calculate her z-score in each of those seven events:<\/p>\r\n<p style=\"padding-left: 40px;\">Put a negative sign in front of those z-scores whose results are the smaller the better, i.e., for 200-m and 800-m runs, 100-m high hurdles. Note that a positive z-score in 100-m run indicates the athlete is below average, since she spends more time than the mean to finish the run. The athlete has the smallest z-score, which is a negative z-score, is the winner of this event. Putting a negative sign in front of the smallest z-score will make it positive and the largest z-score for this event. This makes sense in the way that the larger the z-score the better performance.<\/p>\r\n<p style=\"padding-left: 40px;\">Add the seven z-scores together and get a single value. The one has the largest value is the winner.<\/p>\r\n\r\n<\/div>\r\n<\/details><\/div>","rendered":"<p>A college admissions officer is looking at the files of two international candidates, one with a computer-based TOEFL score of 210 and the other with an IELTS score of 7.5. Which score is better? How can we compare measurements with different scales? The solution is the z-score, which gives a relative standing among the population.<\/p>\n<p>Suppose variable [latex]X[\/latex] follows a distribution with a mean [latex]\\mu[\/latex] and a standard deviation [latex]\\sigma[\/latex], the corresponding standardized variable is defined as<\/p>\n<p style=\"text-align: center;\">[latex]Z = \\frac{X - \\mu}{\\sigma}.[\/latex]<\/p>\n<p>It can be shown that the standardized variable [latex]Z[\/latex] has a mean 0 and a standard deviation 1.<\/p>\n<p>For a given value of [latex]X[\/latex], denoted as the corresponding lower-case [latex]x[\/latex], the value of the standardized variable is called the z-score of [latex]x[\/latex], which is given by<\/p>\n<p style=\"text-align: center; text-align: center;\">[latex]z = \\frac{x-\\mu}{\\sigma}.[\/latex]<\/p>\n<p>If the population parameters [latex]\\mu[\/latex] and [latex]\\sigma[\/latex] are unknown, use the sample mean and standard deviation [latex]\\bar{x}[\/latex]\u00a0and [latex]s[\/latex] to estimate them, then the z-score becomes<\/p>\n<p style=\"text-align: center; text-align: center;\">[latex]z = \\frac{x - \\bar{x}}{s}.[\/latex]<\/p>\n<p>Properties of the z-score:<\/p>\n<ul>\n<li>It measures how far an individual is away from the mean using standard deviation as the unit (ruler).<\/li>\n<li>It represents a relative standing of an observation.<\/li>\n<li>A z-score &gt; 0 means the observation x is above the mean; z-score &lt; 0 means the observation x is below the mean; z-score = 0 means the observation x is equal to the mean.<\/li>\n<li>A z-score has no unit.<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<div class=\"textbox textbox--examples\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Example: Z-score<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<p>Suppose that the TOEFL score for admission at MacEwan has a mean 200 and a standard deviation 10, and the IELTS score has a mean 6 and a standard deviation 1. Two international candidates, one with a TOEFL score of 210 and the other with an IELTS score of 7.5. Which score is relatively better?<\/p>\n<p>Calculate the z-score for the student who took the TOEFL exam:<\/p>\n<p style=\"text-align: center; text-align: center;\">[latex]z_T = \\frac{x - \\mu}{\\sigma} = \\frac{210 - 200}{10} = 1.[\/latex]<\/p>\n<p>This z-score means that the student who took the TOEFL exam is one standard deviation <strong>above the average<\/strong>.<\/p>\n<p>The z-score for the student who took the IELTS exam:<\/p>\n<p style=\"text-align: center; text-align: center;\">[latex]z_I = \\frac{x - \\mu}{\\sigma} = \\frac{7.5 - 6}{1} = 1.5.[\/latex]<\/p>\n<p>This z-score means that the student who took the IELTS exam is 1.5 standard deviation <strong>above the average<\/strong>. Therefore, the student who took the IELTS exam is better because the z-score is larger. This tells us that the IELTS student score is relatively further above the mean than the TOEFL student score is above its mean.<\/p>\n<\/div>\n<\/div>\n<div style=\"height: 55px; margin-top: 2.1428571429em;\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-99 alignleft\" src=\"https:\/\/openbooks.macewan.ca\/rcommander\/wp-content\/uploads\/sites\/8\/2020\/06\/activity.png\" alt=\"\" width=\"250\" height=\"50\" srcset=\"https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2020\/06\/activity.png 250w, https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2020\/06\/activity-65x13.png 65w, https:\/\/openbooks.macewan.ca\/introstats\/wp-content\/uploads\/sites\/8\/2020\/06\/activity-225x45.png 225w\" sizes=\"auto, (max-width: 250px) 100vw, 250px\" \/><\/p>\n<\/div>\n<div class=\"textbox textbox--exercises\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Exercise: Women Heptathlon Champion<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<div>Women\u2019s heptathlon in Olympics includes seven track and field events\u2014200-m and 800-m runs, 100-m high hurdles, shot put, javelin, high jump, and long jump. How do you determine the champion?<\/div>\n<\/div>\n<details>\n<summary style=\"padding-left: 40px;\">Show\/Hide Answer<\/summary>\n<div>\n<p style=\"padding-left: 40px;\">The events are measured in different units, some are in metres and some are in seconds. Moreover, for those runs and high hurdles, results are the smaller the better; for jumps, shot put, and javelin, the larger the better. In order to combine the results of the seven events to a single score to determine the winner, one possible solution is the z-score. The idea is as follows:<\/p>\n<p style=\"padding-left: 40px;\">For each athlete, calculate her z-score in each of those seven events:<\/p>\n<p style=\"padding-left: 40px;\">Put a negative sign in front of those z-scores whose results are the smaller the better, i.e., for 200-m and 800-m runs, 100-m high hurdles. Note that a positive z-score in 100-m run indicates the athlete is below average, since she spends more time than the mean to finish the run. The athlete has the smallest z-score, which is a negative z-score, is the winner of this event. Putting a negative sign in front of the smallest z-score will make it positive and the largest z-score for this event. This makes sense in the way that the larger the z-score the better performance.<\/p>\n<p style=\"padding-left: 40px;\">Add the seven z-scores together and get a single value. The one has the largest value is the winner.<\/p>\n<\/div>\n<\/details>\n<\/div>\n","protected":false},"author":19,"menu_order":6,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-312","chapter","type-chapter","status-publish","hentry"],"part":209,"_links":{"self":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/312","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/users\/19"}],"version-history":[{"count":31,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/312\/revisions"}],"predecessor-version":[{"id":5275,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/312\/revisions\/5275"}],"part":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/parts\/209"}],"metadata":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapters\/312\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/media?parent=312"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/pressbooks\/v2\/chapter-type?post=312"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/contributor?post=312"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/openbooks.macewan.ca\/introstats\/wp-json\/wp\/v2\/license?post=312"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}