{"id":452,"date":"2015-07-27T22:39:49","date_gmt":"2015-07-27T21:39:49","guid":{"rendered":"http:\/\/pcool.dyndns.org:8080\/statsbook\/?page_id=452"},"modified":"2025-06-30T16:52:08","modified_gmt":"2025-06-30T15:52:08","slug":"z-and-t-scores","status":"publish","type":"page","link":"https:\/\/pcool.dyndns.org\/index.php\/z-and-t-scores\/","title":{"rendered":"Z and T scores"},"content":{"rendered":"\n<h1 class=\"wp-block-heading\"><strong>Z-Scores<\/strong><\/h1>\n\n\n\n<p>The mean height is different for children of different age. Children are growing and their height increases with age. It is important to know whether a child grows sufficiently as compared to its peers. Therefore, standardisation is required. Z-scores were introduced for this purpose. The child\u2019s height is expressed as how many standard deviations it is above or below the mean.<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Zscore = \\frac{Height &#8211; MeanHeight}{StandardDeviation} \\)<script src=\"https:\/\/pcool.dyndns.org\/wp-includes\/js\/dist\/hooks.min.js?ver=dd5603f07f9220ed27f1\" id=\"wp-hooks-js\"><\/script>\n<script src=\"https:\/\/pcool.dyndns.org\/wp-includes\/js\/dist\/i18n.min.js?ver=c26c3dc7bed366793375\" id=\"wp-i18n-js\"><\/script>\n<script id=\"wp-i18n-js-after\">\nwp.i18n.setLocaleData( { 'text direction\\u0004ltr': [ 'ltr' ] } );\n\/\/# sourceURL=wp-i18n-js-after\n<\/script>\n<script  async src=\"https:\/\/cdnjs.cloudflare.com\/ajax\/libs\/mathjax\/2.7.7\/MathJax.js?config=TeX-MML-AM_CHTML\" id=\"mathjax-js\"><\/script>\n<\/div>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\"><\/div>\n\n\n\n<p>The growth charts show that the average height for a twelve year old girl is 150 cm.<\/p>\n\n\n\n<p>A child of average height has a z-score of 0:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Zscore = \\frac{150-150}{Standard Devation} = \\frac{0}{Standard Devation} = 0 \\)<\/div>\n\n\n\n<p>What about twelve year old girl who is 160 cm?<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Zscore = \\frac{160-150}{Standard Deviation} \\)<\/div>\n\n\n\n<p>The standard deviation can be estimated from the growth chart. It was <a href=\"https:\/\/pcool.dyndns.org\/index.php\/percentiles-and-quantiles\/\" data-type=\"page\" data-id=\"455\">shown<\/a> that the mean \u00b1 one standard deviation is between the 16<sup>th<\/sup> and 84<sup>th<\/sup> percentile. The 16<sup>th<\/sup> percentile for a twelve year old girl is approximately 142 cm (read from chart). Therefore, the standard deviation is 150 \u2013 142 = 8 cm. Consequently, the Z-score of a 12 year old girls who is 160 cm is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Zscore = \\frac{160-150}{8} = \\frac{10}{8}=1.25 \\)<\/div>\n\n\n\n<p>She is one and a quarter standard deviations <strong>above<\/strong> the mean for her age.<\/p>\n\n\n\n<p>Similarly, a twelve year old girl who is 125 cm has a Z-score of:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Zscore = \\frac{125-150}{8} = \\frac{-25}{8}=3.125 \\)<\/div>\n\n\n\n<p>She is more than 3 standard deviations <strong>below<\/strong> the average for her age.<\/p>\n\n\n\n<p>Z-scores are very useful in follow up of children and comparing their growth rate with their peers. Assume a five year old child has a Z-score of -1 (one standard deviation below the average height for a five year old). If the child has a Z-score of -2 when 10 years old, the rate of growth has been less than its peers and this could indicate malnutrition or disease. But, if the child has a Z-score of 0 when 10 years old, the rate of growth has exceeded that of its peers. If the child has a Z-score of -1 when 10 years old, the rate of growth has been the same as its peers. The child however remains smaller than average and one standard deviation below the mean.<\/p>\n\n\n\n<p>In summary, a change in Z-score indicates a difference in growth rate as compared to the normal population. An increase in Z-score indicates a growth rate greater than average and a decrease in Z-score a growth rate less than average. If the Z-score remains the same, the rate of growth is the same as the normal population.<\/p>\n\n\n\n<h1 class=\"wp-block-heading\"><strong>T-Scores<\/strong><\/h1>\n\n\n\n<p>Z-scores can also be used to express bone mineral density. The bone mineral density is referenced to the mean for that age:<\/p>\n\n\n\n<p><strong>A Z-score is the number of standard deviations the bone mineral density measurement is above or below the <em>age-matched mean<\/em> bone mineral density.<\/strong><\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Zscore = \\frac{BMD &#8211; AgeMatched Mean BMD}{Standard Deviation} \\)<\/div>\n\n\n\n<p class=\"is-style-text-annotation is-style-text-annotation--1\"><em>BMD = bone mineral density<\/em><\/p>\n\n\n\n<p>However, to define osteoporosis the bone mineral density in a <strong>young normal<\/strong> is used as reference. This is the T-score:<\/p>\n\n\n\n<p><strong>A T-score is the number of standard deviations the bone mineral density measurement is above or below the <em>young-normal mean<\/em> bone mineral density.<\/strong><\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Tscore = \\frac{BMD -YoungNormalMeanBMD}{Standard Deviation} \\)<\/div>\n\n\n\n<p class=\"is-style-text-annotation is-style-text-annotation--2\"><em>BMD = bone mineral density<\/em><\/p>\n\n\n\n<p>The World Health Organisation defined osteoporosis \/ osteopenia in 1994. Four diagnostic categories were defined based on bone mineral density. The T-score, using young adult women as the referent group, defines the diagnostic categories:<\/p>\n\n\n\n<table id=\"tablepress-43\" class=\"tablepress tablepress-id-43\">\n<thead>\n<tr class=\"row-1\">\n\t<td class=\"column-1\"><\/td><th class=\"column-2\">T Score<\/th>\n<\/tr>\n<\/thead>\n<tbody class=\"row-striping row-hover\">\n<tr class=\"row-2\">\n\t<td class=\"column-1\">Normal<\/td><td class=\"column-2\">-1 or above<\/td>\n<\/tr>\n<tr class=\"row-3\">\n\t<td class=\"column-1\">Osteopenia<\/td><td class=\"column-2\">between -1 and -2.5<\/td>\n<\/tr>\n<tr class=\"row-4\">\n\t<td class=\"column-1\">Osteoporosis<\/td><td class=\"column-2\">less than -2.5<\/td>\n<\/tr>\n<tr class=\"row-5\">\n\t<td class=\"column-1\">Severe osteoporosis<\/td><td class=\"column-2\">less than -2.5 and fragility fracture<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<!-- #tablepress-43 from cache -->\n\n\n<p>The graph below shows the forearm bone mineral density as function of the age:<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"467\" height=\"350\" src=\"https:\/\/pcool.dyndns.org\/wp-content\/uploads\/2025\/06\/TZ1.png\" alt=\"\" class=\"wp-image-3864\" srcset=\"https:\/\/pcool.dyndns.org\/wp-content\/uploads\/2025\/06\/TZ1.png 467w, https:\/\/pcool.dyndns.org\/wp-content\/uploads\/2025\/06\/TZ1-300x225.png 300w\" sizes=\"auto, (max-width: 467px) 100vw, 467px\" \/><\/figure>\n\n\n\n<p>The left hand y-axis indicates the bone mineral density in gram per centimetre square. The corresponding T-score is based on the bone mineral density in a 20 year old and shown on the right hand y-axis. The solid line indicates the mean bone mineral density and the dashed lines one standard deviation away from the mean. Therefore, 68% of the people have a bone mineral density between the two dashed lines.<\/p>\n\n\n\n<p>The two horizontal dotted lines show the osteopenia limit (T-score = -1) and the osteoporosis limit (t-score = -2.5). Z-scores are less commonly used than T-scores, but may be helpful in identifying persons who should undergo a work-up for secondary causes of osteoporosis. Z-scores and T-scores can be converted into each other using a reference table provided the age, gender, race and skeletal site are known.<\/p>\n\n\n\n<p>Neither the Z-score not the T-score can predict fracture risk on its own. It is also necessary to know the age of the patient. Because T-scores and Z-scores can be converted into each other, fracture risk can be predicted with either.<\/p>\n\n\n\n<p>Consider the patient in the graph below:<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"467\" height=\"350\" src=\"https:\/\/pcool.dyndns.org\/wp-content\/uploads\/2025\/06\/TZ2.png\" alt=\"\" class=\"wp-image-3865\" srcset=\"https:\/\/pcool.dyndns.org\/wp-content\/uploads\/2025\/06\/TZ2.png 467w, https:\/\/pcool.dyndns.org\/wp-content\/uploads\/2025\/06\/TZ2-300x225.png 300w\" sizes=\"auto, (max-width: 467px) 100vw, 467px\" \/><\/figure>\n\n\n\n<p><a href=\"http:\/\/pcool.dyndns.org:8080\/statsbook\/wp-content\/uploads\/TZ2.png\"><\/a><strong>Patient<\/strong><\/p>\n\n\n\n<p>Age: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 70 years<\/p>\n\n\n\n<p>Bone mineral density: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp;0.465 g\/cm^2<\/p>\n\n\n\n<p>Z-score:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp;+1<\/p>\n\n\n\n<p><strong>Young normal (20 years)<\/strong><\/p>\n\n\n\n<p>Mean bone density:&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;0.49 g\/cm^2<\/p>\n\n\n\n<p>Standard deviation:                   &nbsp; 0.49 \u2013 0.43 = 0.06 g\/cm^2<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Tscore=\\frac{0.465 &#8211; 0.49}{0.06} = \\frac{-0.025}{0.06} = -0.42 \\)<\/div>\n\n\n\n<p><strong>WHO category:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Normal&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <\/strong><\/p>\n\n\n\n<p>Also, the bone density measurement of the 90 years old lady with a fracture neck of femur:<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"467\" height=\"350\" src=\"https:\/\/pcool.dyndns.org\/wp-content\/uploads\/2025\/06\/TZ3.png\" alt=\"\" class=\"wp-image-3866\" srcset=\"https:\/\/pcool.dyndns.org\/wp-content\/uploads\/2025\/06\/TZ3.png 467w, https:\/\/pcool.dyndns.org\/wp-content\/uploads\/2025\/06\/TZ3-300x225.png 300w\" sizes=\"auto, (max-width: 467px) 100vw, 467px\" \/><\/figure>\n\n\n\n<p><a href=\"http:\/\/pcool.dyndns.org:8080\/statsbook\/wp-content\/uploads\/TZ3.png\"><\/a><strong>Patient<\/strong><\/p>\n\n\n\n<p>Age: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 90 years<\/p>\n\n\n\n<p>Bone mineral density: \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00.31 g\/cm^2<\/p>\n\n\n\n<p>Z-score:\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2248 + 0.5<\/p>\n\n\n\n<p><strong>Young normal (20 years)<\/strong><\/p>\n\n\n\n<p>Mean bone density:\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00.49 g\/cm^2<\/p>\n\n\n\n<p>Standard deviation:\u00a0                     0.49 \u2013 0.43 = 0.06 g\/cm^2<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Tscore=\\frac{0.31 &#8211; 0.49}{0.06} = \\frac{-0.18}{0.06} = -3 \\)<\/div>\n\n\n\n<p><strong>WHO category:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Severe osteoporosis<\/strong><\/p>\n\n\n\n<p><strong>In R:<\/strong><\/p>\n\n\n\n<p>The Normal distribution is central to parametric statistics. Data that can be modelled by a normal distribution are commonly described in terms of their mean and standard deviation. It has been demonstrated that people&#8217;s heights can be modelled with a Normal distribution. It is important to distinguish between the sample descriptives (mean and standard deviation) and population descriptives (mean\u00a0\u03bc and standard deviation\u00a0\u03c3).<\/p>\n\n\n\n<p>The data set <a href=\"https:\/\/pcool.dyndns.org:\/wp-content\/data_files\/heights.rda\" target=\"_blank\" rel=\"noreferrer noopener\">heights.rda<\/a> contains the heights of two groups of people. To obtain the descriptives:<\/p>\n\n\n\n<pre class=\"wp-block-code has-small-font-size\"><code><em><mark style=\"background-color:rgba(0, 0, 0, 0);color:#f80505\" class=\"has-inline-color\">heights %>% <\/mark><\/em>\n<em><mark style=\"background-color:rgba(0, 0, 0, 0);color:#f80505\" class=\"has-inline-color\">dplyr::select('Group1') %>% <\/mark><\/em>\n<em><mark style=\"background-color:rgba(0, 0, 0, 0);color:#f80505\" class=\"has-inline-color\">summarise(n=n(), mean=mean(Group1), sd=sd(Group1), min=min(Group1), max=max(Group1))\n<\/mark><mark style=\"background-color:rgba(0, 0, 0, 0);color:#0518f7\" class=\"has-inline-color\">     n     mean       sd      min      max\n1 1000 164.9281 4.888238 147.5271 179.0782<\/mark><mark style=\"background-color:rgba(0, 0, 0, 0);color:#f80505\" class=\"has-inline-color\">\nheights %>% <\/mark><\/em>\n<em><mark style=\"background-color:rgba(0, 0, 0, 0);color:#f80505\" class=\"has-inline-color\">dplyr::select('Group2') %>% <\/mark><\/em>\n<em><mark style=\"background-color:rgba(0, 0, 0, 0);color:#f80505\" class=\"has-inline-color\">summarise(n=n(), mean=mean(Group2), sd=sd(Group2), min=min(Group2), max=max(Group2))\n<\/mark><mark style=\"background-color:rgba(0, 0, 0, 0);color:#0510f7\" class=\"has-inline-color\">     n     mean       sd      min      max\n1 1000 178.5239 15.28449 125.4151 232.9516<\/mark><\/em><\/code><\/pre>\n\n\n\n<p>To obtain a histogram showing both group of data, Group1 in red and Group2 in blue is straight forward in plot builder, or alternatively directly in the console:<\/p>\n\n\n\n<pre class=\"wp-block-code has-small-font-size\"><code><em><span style=\"color: #ff0000;\">ggplot() + <\/span><\/em>\n<em><span style=\"color: #ff0000;\">geom_histogram(aes(y = ..count..,x = Group1), data=heights, colour = \"#990000\", fill = \"#ff0066\") + <\/span><\/em>\n<em><span style=\"color: #ff0000;\">geom_histogram(aes(x = Group2), data=heights, <\/span><\/em>\n<em><span style=\"color: #ff0000;\">colour = \"#000099\",fill = \"#3399ff\") +<\/span><\/em>\n<em><span style=\"color: #ff0000;\">xlab(label = \"Height &#91;cm]\") + <\/span><\/em>\n<em><span style=\"color: #ff0000;\">ggtitle(label = \"Heights\") + <\/span><\/em>\n<em><span style=\"color: #ff0000;\">theme_bw()<\/span><\/em><\/code><\/pre>\n\n\n\n<figure class=\"wp-block-image size-large\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"768\" src=\"https:\/\/pcool.dyndns.org\/wp-content\/uploads\/2025\/06\/heightsgroups-1024x768.png\" alt=\"\" class=\"wp-image-3201\" srcset=\"https:\/\/pcool.dyndns.org\/wp-content\/uploads\/2025\/06\/heightsgroups-1024x768.png 1024w, https:\/\/pcool.dyndns.org\/wp-content\/uploads\/2025\/06\/heightsgroups-300x225.png 300w, https:\/\/pcool.dyndns.org\/wp-content\/uploads\/2025\/06\/heightsgroups-768x576.png 768w, https:\/\/pcool.dyndns.org\/wp-content\/uploads\/2025\/06\/heightsgroups.png 1355w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n\n\n\n<p>The Z-score can be calculated by the following equation:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Zscore = \\frac{x-mean}{sd} \\)<\/div>\n\n\n\n<p>So, the Z-score of a person in Group 2 whose height is 200 cm is:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Zscore = \\frac{200-178.5}{15.3} \\approx 1.41 \\)<\/div>\n\n\n\n<p>The positive Z score indicates the height is above the mean; by 1.41 standard deviations. R can be used to calculate the probability of having a height of 200 cm or less using the pnorm function:<\/p>\n\n\n\n<pre class=\"wp-block-code has-small-font-size\"><code><span style=\"color: #ff0000;\"><em>pnorm(1.41)<\/em><\/span>\n<span style=\"color: #0000ff;\"><em>&#91;1] 0.9207302<\/em><\/span><\/code><\/pre>\n\n\n\n<p>So, 92%.<\/p>\n\n\n\n<p>If the probability of having a height of <strong>more than<\/strong> or equal to 210 cm is required:<\/p>\n\n\n\n<div class=\"wp-block-mathml-mathmlblock\">\\(Zscore = \\frac{210 &#8211; 178.5}{15.3} \\approx 2.06 \\)<\/div>\n\n\n\n<pre class=\"wp-block-code has-small-font-size\"><code><span style=\"color: #ff0000;\"><em>1-pnorm(2.06)<\/em><\/span>\n<span style=\"color: #0000ff;\"><em>&#91;1] 0.01969927<\/em><\/span><\/code><\/pre>\n\n\n\n<p>Or 2%.<\/p>\n\n\n\n<p>The probability of the height being between 200 and 210 cm is:<\/p>\n\n\n\n<pre class=\"wp-block-code has-small-font-size\"><code><span style=\"color: #ff0000;\"><em>pnorm(2.06)-pnorm(1.41)<\/em><\/span>\n<span style=\"color: #0000ff;\"><em>&#91;1] 0.05957057<\/em><\/span><\/code><\/pre>\n\n\n\n<p>Or 6%.<\/p>\n\n\n\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Z-Scores The mean height is different for children of different age. Children are growing and their height increases with age. It is important to know whether a child grows sufficiently as compared to its peers. Therefore, standardisation is required. Z-scores were introduced for this purpose. The child\u2019s height is expressed as how many standard deviations [&hellip;]<\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"inline_featured_image":false,"footnotes":""},"class_list":["post-452","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/pages\/452","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/comments?post=452"}],"version-history":[{"count":4,"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/pages\/452\/revisions"}],"predecessor-version":[{"id":4615,"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/pages\/452\/revisions\/4615"}],"wp:attachment":[{"href":"https:\/\/pcool.dyndns.org\/index.php\/wp-json\/wp\/v2\/media?parent=452"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}