{"id":2723,"date":"2024-10-16T15:40:39","date_gmt":"2024-10-16T13:40:39","guid":{"rendered":"https:\/\/clooma.ai\/?post_type=ressource-pedagogiqu&#038;p=2723"},"modified":"2024-10-24T16:48:48","modified_gmt":"2024-10-24T14:48:48","slug":"konfidenciaintervallum","status":"publish","type":"ressource-pedagogiqu","link":"https:\/\/clooma.ai\/hu\/oktatasi-forras\/konfidenciaintervallum\/","title":{"rendered":"Bizonoss\u00e1gi intervallum"},"content":{"rendered":"<p>A konfidenciaintervallum egy statisztikai param\u00e9ter egy statisztikai mint\u00e1b\u00f3l becs\u00fclt, val\u00f3sz\u00edn\u0171s\u00edthet\u0151 \u00e9rt\u00e9ktartom\u00e1nya. A param\u00e9ter becsl\u00e9s\u00e9nek pontoss\u00e1g\u00e1r\u00f3l ad k\u00e9pet. A konfidenciaintervallumot \u00e1ltal\u00e1ban a hozz\u00e1 tartoz\u00f3 konfidenciaszinttel fejezik ki, amely annak a val\u00f3sz\u00edn\u0171s\u00e9g\u00e9t jelenti, hogy az intervallum val\u00f3ban tartalmazza a val\u00f3di popul\u00e1ci\u00f3s param\u00e9tert.&nbsp;<\/p>\n\n\n\n<p><img decoding=\"async\" src=\"blob:https:\/\/clooma.ai\/5c5f689c-38e0-4233-8b65-b2363db5418e\"><\/p>\n\n\n\n<h2 class=\"wp-block-heading\">\u00c1tlagos :&nbsp;<\/h2>\n\n\n\n<p>Az \u00e1tlag konfidenciaintervalluma egy olyan statisztikai intervallum, amely egy val\u00f3sz\u00edn\u0171s\u00edthet\u0151 becsl\u00e9st ad arra az intervallumra, amelyen bel\u00fcl a popul\u00e1ci\u00f3 val\u00f3di \u00e1tlaga fekszik. Ezt az intervallumot a sokas\u00e1g mint\u00e1j\u00e1b\u00f3l sz\u00e1rmaz\u00f3 adatok felhaszn\u00e1l\u00e1s\u00e1val \u00e1ll\u00edtj\u00e1k \u00f6ssze.&nbsp;&nbsp;<\/p>\n\n\n\n<p>Term\u00e9szetesen az \u00e1tlagra vonatkoz\u00f3 konfidenciaintervallum l\u00e9trehoz\u00e1sa lehets\u00e9ges a k\u00f6zponti hat\u00e1r\u00e9rt\u00e9kt\u00e9telnek k\u00f6sz\u00f6nhet\u0151en. Megfelel\u0151en nagy mint\u00e1k eset\u00e9n (n\u226530), f\u00fcggetlen\u00fcl a popul\u00e1ci\u00f3 eloszl\u00e1s\u00e1nak alakj\u00e1t\u00f3l, ha v\u00e9letlenszer\u0171en t\u00f6bb \"n\" m\u00e9ret\u0171 mint\u00e1t vesz\u00fcnk, e mint\u00e1k \u00e1tlagai <span class=\"wp-katex-eq\" data-display=\"false\">\\left{overline{X} \\right}<\/span> megk\u00f6zel\u00edt\u0151leg norm\u00e1lis eloszl\u00e1s\u00faak. Ez lehet\u0151v\u00e9 teszi, hogy megb\u00edzhat\u00f3 konfidenciaintervallumokat alkossunk a sokas\u00e1g val\u00f3di \u00e1tlag\u00e1nak becsl\u00e9s\u00e9re.&nbsp;<\/p>\n\n\n\n<p>Az \u00e1tlagra vonatkoz\u00f3 konfidenciaintervallum megalkot\u00e1sa a Student-f\u00e9le t-eloszl\u00e1s vagy a norm\u00e1lis eloszl\u00e1s alkalmaz\u00e1s\u00e1n alapul, a minta m\u00e9ret\u00e9t\u0151l \u00e9s a popul\u00e1ci\u00f3 sz\u00f3r\u00e1s\u00e1r\u00f3l val\u00f3 ismereteinkt\u0151l f\u00fcgg\u0151en.&nbsp;<\/p>\n\n\n\n<p>Mivel ez a sz\u00e1m\u00edt\u00e1s egy k\u00f6zel\u00edt\u00e9s, ismern\u00fcnk kell a k\u00f6zel\u00edt\u00e9s pontoss\u00e1g\u00e1t. \u00c1ltal\u00e1noss\u00e1gban, hogy jellemezz\u00fck e k\u00f6zel\u00edt\u00e9s pontoss\u00e1g\u00e1t, kisz\u00e1m\u00edtjuk a 95%-n\u00e9l l\u00e9v\u0151 intervallumot. Ez az intervallum megfelel :&nbsp;<\/p>\n\n\n\n<p><strong>95% intervallum = olyan intervallum, amelyben 95% es\u00e9ly van arra, hogy az eloszl\u00e1s \u00e1tlag\u00e1nak val\u00f3di \u00e9rt\u00e9ke ezen bel\u00fcl van.<\/strong>&nbsp;<\/p>\n\n\n\n<p>A statisztik\u00e1ban 95% a bizalom (1- \u03b1), amely kieg\u00e9sz\u00edti az \u03b1=5% els\u0151faj\u00fa kock\u00e1zatot. Ez a kock\u00e1zat annak az es\u00e9ly\u00e9t jelenti, hogy az eloszl\u00e1s \u00e1tlag\u00e1nak \u00e9rt\u00e9ke a konfidenciaintervallumon k\u00edv\u00fcl esik.&nbsp;<\/p>\n\n\n\n<p>Az al\u00e1bbiakban bemutatjuk, hogyan kell konfidenciaintervallumot konstru\u00e1lni az \u00e1tlaghoz:&nbsp;<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Az \"n\" m\u00e9ret\u0171 minta \u00e1tlag\u00e1nak \u00e9s sz\u00f3r\u00e1s\u00e1nak kisz\u00e1m\u00edt\u00e1sa: A minta adatainak felhaszn\u00e1l\u00e1s\u00e1val sz\u00e1m\u00edtsa ki a minta \u00e1tlag\u00e1t \u00e9s sz\u00f3r\u00e1s\u00e1t. <span class=\"wp-katex-eq\" data-display=\"false\">\\overline{X}<\/span>&nbsp;\u00e9s S.&nbsp;<\/li>\n\n\n\n<li>A megb\u00edzhat\u00f3s\u00e1gi szint kiv\u00e1laszt\u00e1sa (1- \u03b1): V\u00e1lasszon ki egy megb\u00edzhat\u00f3s\u00e1gi szintet, gyakran sz\u00e1zal\u00e9kban kifejezve, p\u00e9ld\u00e1ul 95% vagy 99%. A 95% megb\u00edzhat\u00f3s\u00e1gi szint azt jelenti, hogy 95% biztosak vagyunk abban, hogy az \u00e1ltalunk konstru\u00e1lt intervallum tartalmazza a val\u00f3di popul\u00e1ci\u00f3s \u00e1tlagot.&nbsp;<\/li>\n\n\n\n<li>Az intervallum meghat\u00e1roz\u00e1sa: Haszn\u00e1lja a megfelel\u0151 eloszl\u00e1s (Student vagy norm\u00e1l eloszl\u00e1s) szerinti k\u00f6z\u00e9p\u00e9rt\u00e9k bizalmi intervallum\u00e1nak k\u00e9plet\u00e9t:&nbsp;\n<ul class=\"wp-block-list\">\n<li>Ha ismeri a popul\u00e1ci\u00f3 \ud835\udf0e sz\u00f3r\u00e1s\u00e1t, haszn\u00e1lja a norm\u00e1lis eloszl\u00e1st:  \n<ul class=\"wp-block-list\">\n<li><span class=\"wp-katex-eq\" data-display=\"false\">IC = \\overline{X} \\underline{+}Z_{\\frac{a}{2}}\\ast \\frac{S}{\\sqrt{n}}<\/span> ahol:\n<ul class=\"wp-block-list\">\n<li><span class=\"wp-katex-eq\" data-display=\"false\">Z\\frac{2}{a}<\/span> a bizalmi szintnek megfelel\u0151 z-pontsz\u00e1m. (K\u00e9toldal\u00fa)<\/li>\n\n\n\n<li>n: a minta m\u00e9rete.&nbsp;<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Ha nem ismeri a popul\u00e1ci\u00f3 \ud835\udf0e sz\u00f3r\u00e1s\u00e1t, haszn\u00e1lja a Student-eloszl\u00e1st:&nbsp;\n<ul class=\"wp-block-list\">\n<li><span class=\"wp-katex-eq\" data-display=\"false\">IC = \\overline{X} \\underline{+}t_{\\frac{a}{2}n-1}\\ast \\frac{S}{\\sqrt{n}}<\/span> ahol:\n<ul class=\"wp-block-list\">\n<li><span class=\"wp-katex-eq\" data-display=\"false\">t_{rac{a}{2}n-1}<\/span>a megb\u00edzhat\u00f3s\u00e1gi szintnek megfelel\u0151 t-pontsz\u00e1m n-1 szabads\u00e1gi fok eset\u00e9n.&nbsp;&nbsp;<\/li>\n\n\n\n<li>n: a minta m\u00e9rete.&nbsp;<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>Az \u00e1tlag konfidenciaintervalluma teh\u00e1t azoknak az \u00e9rt\u00e9keknek a tartom\u00e1ny\u00e1t adja meg, amelyeken bel\u00fcl egy bizonyos megb\u00edzhat\u00f3s\u00e1gi szinten (1-\ud835\udefc) biztosak vagyunk abban, hogy a \u00b5 popul\u00e1ci\u00f3 val\u00f3di \u00e1tlaga van. Min\u00e9l magasabb a megb\u00edzhat\u00f3s\u00e1gi szint, ann\u00e1l sz\u00e9lesebb lesz az intervallum, ami a becsl\u00e9sbe vetett nagyobb fok\u00fa bizalmat t\u00fckr\u00f6zi.&nbsp;<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>&nbsp;<\/td><td>Standard elt\u00e9r\u00e9s S&nbsp;<\/td><td>Minta m\u00e9rete n&nbsp;<\/td><td>Bizalom (1-\u03b1)&nbsp;<\/td><\/tr><tr><td>Az \u00e1tlagos IC konfidenciaintervallum\u00e1nak sz\u00e9less\u00e9ge.&nbsp;<\/td><td>Az IC sz\u00e9less\u00e9ge n\u0151, ha a standard elt\u00e9r\u00e9s n\u0151&nbsp;<\/td><td>Az IC sz\u00e9less\u00e9ge cs\u00f6kken a minta m\u00e9ret\u00e9nek n\u00f6veked\u00e9s\u00e9vel&nbsp;<\/td><td>A CI sz\u00e9less\u00e9ge a bizalom n\u00f6veked\u00e9s\u00e9vel n\u0151&nbsp;<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>P\u00e9lda: Szeretn\u00e9nk tudni, hogyan lehet kisz\u00e1m\u00edtani a csal\u00e1donk\u00e9nti \u00e1tlagos cukorfogyaszt\u00e1s konfidenciaintervallum\u00e1t 95% konfidencia mellett. Egy 18 csal\u00e1db\u00f3l \u00e1ll\u00f3 mint\u00e1t vett\u00fcnk. Az al\u00e1bbi t\u00e1bl\u00e1zatban az eredm\u00e9nyek szerepelnek:&nbsp;<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>5&nbsp;<\/td><td>13&nbsp;<\/td><td>11&nbsp;<\/td><td>5&nbsp;<\/td><td>2&nbsp;<\/td><td>3&nbsp;<\/td><td>2&nbsp;<\/td><td>1&nbsp;<\/td><td>6&nbsp;<\/td><td>14&nbsp;<\/td><td>6&nbsp;<\/td><td>8&nbsp;<\/td><td>2&nbsp;<\/td><td>13&nbsp;<\/td><td>9&nbsp;<\/td><td>5&nbsp;<\/td><td>12&nbsp;<\/td><td>7&nbsp;<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>Megold\u00e1s:&nbsp;<\/p>\n\n\n\n<p>Sz\u00e1m\u00edtsuk ki az \u00e1tlagot, a sz\u00f3r\u00e1st \u00e9s a szabads\u00e1gfokok sz\u00e1m\u00e1t.&nbsp;<\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">\\overline{X} = \\frac{5+13+11+5+2+3+2+1+6+14+6+8+2+13+9+5+12+7}{18} = 6.88<\/span><\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">S = \\sqrt{\\frac{\\sum_{1}^{N}(xi-\\overline{x})^{2}}{17}}} = 4.25<\/span><\/p>\n\n\n\n<p class=\"has-text-align-center\">n-1 =17<\/p>\n\n\n\n<p>A Student's law t\u00e1bl\u00e1zatb\u00f3l, vagy a szoftver seg\u00edts\u00e9g\u00e9vel <a href=\"https:\/\/clooma.ai\/hu\/adatelemzesi-megoldasok\/\">Ellistat adatelemz\u00e9s<\/a>a t=2.110 \u00e9rt\u00e9ket tal\u00e1ljuk<\/p>\n\n\n\n<p class=\"has-text-align-center\"><img decoding=\"async\" alt=\"Sz\u00f6veget, diagramot, k\u00e9perny\u0151k\u00e9pet, automatikusan gener\u00e1lt PlotDescription k\u00e9pet tartalmaz\u00f3 k\u00e9p\" src=\"blob:https:\/\/clooma.ai\/5d36ab4a-a8d2-4d67-a3ed-af71b2dca181\"><\/p>\n\n\n\n<p>Ez\u00e9rt levezethetj\u00fck a k\u00f6vetkez\u0151 konfidenciaintervallumot:&nbsp;<\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">\\overline{X}-t_{\\frac{a}{2};n-1}\\ast \\frac{S}{\\sqrt{n}}}\\le \\mu\\le \\overline{X}+t_{\\frac{a}{2}n-1}\\ast \\frac{S}{\\sqrt{n}}<\/span><\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">6.88-2.110\\ast \\frac{4.25}{\\sqrt{18}}<\/span><\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">4.773 \\mu\\le 9.005<\/span><\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Sz\u00f3r\u00e1s \/ sz\u00f3r\u00e1s:<\/h2>\n\n\n\n<p>A popul\u00e1ci\u00f3 sz\u00f3r\u00e1s\u00e1ra vonatkoz\u00f3 konfidenciaintervallum megalkot\u00e1s\u00e1hoz a chi-2 eloszl\u00e1st haszn\u00e1ljuk (<span class=\"wp-katex-eq\" data-display=\"false\">x^{2}<\/span>). Tudjuk, hogy a variancia becsl\u00e9se a k\u00f6vetkez\u0151 k\u00e9plettel t\u00f6rt\u00e9nik:&nbsp;<\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">&lt;\/p&gt;\n\n\n\n&lt;p&gt;A chi-2 k&eacute;plet ([latex]x^{2}<\/span> a variancia a k\u00f6vetkez\u0151k\u00e9ppen \u00edrhat\u00f3 fel :&nbsp;<\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">X^{2} = \\frac{(n-1)S^{2}}}{sigma^{2}}<\/span><\/p>\n\n\n\n<p>A chi-2 s\u0171r\u0171s\u00e9gf\u00fcggv\u00e9ny g\u00f6rb\u00e9je (<span class=\"wp-katex-eq\" data-display=\"false\">x^{2}<\/span>) hasonl\u00edt a norm\u00e1lis eloszl\u00e1sra, de nem szimmetrikus. Alakja mindenekel\u0151tt a szabads\u00e1gfokok sz\u00e1m\u00e1t\u00f3l f\u00fcgg. Az al\u00e1bbi grafikon a chi-2 s\u0171r\u0171s\u00e9gf\u00fcggv\u00e9ny (<span class=\"wp-katex-eq\" data-display=\"false\">x^{2}<\/span>)n=4 szabads\u00e1gi fok eset\u00e9n.&nbsp;<\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large\"><img decoding=\"async\" src=\"https:\/\/clooma.ai\/wp-content\/uploads\/image-9-1024x838.png\" alt=\"\" class=\"wp-image-2728\"\/><\/figure><\/div>\n\n\n<p>A <span class=\"wp-katex-eq\" data-display=\"false\">x^{2}<\/span> haszn\u00e1lhat\u00f3 a variancia \ud835\udf0e\u00b2 konfidenciaintervallum\u00e1nak levezet\u00e9s\u00e9re, n mintam\u00e9ret \u00e9s 1-\u03b1 konfidencia eset\u00e9n.&nbsp;<\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">\\frac{(n-1)S^{2}}{X^{2}_{n-1;\\frac{a}{2}}}\\le \\sigma^{2}\\le \\frac{(n-1)S^{2}}{X^{2}n-1;1-\\frac{a}{2}}<\/span><\/p>\n\n\n\n<p>A variancia konfidenciaintervallum\u00e1nak kisz\u00e1m\u00edt\u00e1si elj\u00e1r\u00e1sa :&nbsp;<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>A sz\u00f3r\u00e1s \u00e9s a szabads\u00e1gfokok kisz\u00e1m\u00edt\u00e1sa: A mintaadatokb\u00f3l sz\u00e1m\u00edtsa ki az S\u00b2 sz\u00f3r\u00e1st \u00e9s a szabads\u00e1gfokokat (n-1).&nbsp;<\/li>\n\n\n\n<li>A kritikus chi-n\u00e9gyzet \u00e9rt\u00e9kek meghat\u00e1roz\u00e1sa: A kritikus chi-n\u00e9gyzet \u00e9rt\u00e9kek meghat\u00e1roz\u00e1sa <span class=\"wp-katex-eq\" data-display=\"false\">X^{2}n-1;\\frac{a}{2}\\text{et}X^{2}n-1;1-\\frac{a}{2}<\/span> a k\u00edv\u00e1nt megb\u00edzhat\u00f3s\u00e1gi szint \u00e9s szabads\u00e1gfok eset\u00e9n. Ezeket az \u00e9rt\u00e9keket a \ud835\udf122 eloszl\u00e1si t\u00e1bl\u00e1zatokban vagy az Ellistat seg\u00edts\u00e9g\u00e9vel tal\u00e1lhatja meg.&nbsp;<\/li>\n\n\n\n<li>A variancia konfidenciaintervallum\u00e1nak meghat\u00e1roz\u00e1s\u00e1hoz haszn\u00e1lja a k\u00f6vetkez\u0151 k\u00e9pleteket:&nbsp;<\/li>\n<\/ol>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">\\frac{(n-1)S^{2}}{X^{2}_{n-1;\\frac{a}{2}}}\\le \\sigma^{2}\\le \\frac{(n-1)S^{2}}{X^{2}n-1;1-\\frac{a}{2}}<\/span><\/p>\n\n\n\n<p>Megjegyz\u00e9s: a sz\u00f3r\u00e1s konfidenciaintervallum\u00e1t \u00fagy lehet levezetni, hogy a gy\u00f6keret mindk\u00e9t oldalra helyezz\u00fck.&nbsp;<\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">\\sqrt{\\frac{(n-1)S^{2}}{X^{2}n-1;\\frac{a}{2}}}\\le \\sigma\\le \\sqrt{\\frac{(n-1)S^{2}}{X^{2}n-1;1-\\frac{a}{2}}}<\/span><\/p>\n\n\n\n<p>P\u00e9lda: 10 palackb\u00f3l \u00e1ll\u00f3 mint\u00e1t vettek a gy\u00e1rt\u00e1sb\u00f3l. Szeretn\u00e9nk k\u00e9pet kapni a folyamat v\u00e1ltoz\u00e9konys\u00e1g\u00e1r\u00f3l. Hat\u00e1rozza meg az \ud835\udf0e2 variancia konfidenciaintervallum\u00e1t 95% konfidencia mellett:&nbsp;<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td>10&nbsp;<\/td><td>10&nbsp;<\/td><td>12&nbsp;<\/td><td>10&nbsp;<\/td><td>11&nbsp;<\/td><\/tr><tr><td>10&nbsp;<\/td><td>11&nbsp;<\/td><td>11&nbsp;<\/td><td>10&nbsp;<\/td><td>11&nbsp;<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>Megold\u00e1s:&nbsp;&nbsp;<\/p>\n\n\n\n<p>Sz\u00e1m\u00edtsuk ki az S sz\u00f3r\u00e1st \u00e9s a szabads\u00e1gfokok sz\u00e1m\u00e1t :&nbsp;<\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">S^{2} = \\frac{\\sum_{1}^{N}(xi-\\overline{x})^{2}}{9} = 0.489<\/span><\/p>\n\n\n\n<p class=\"has-text-align-center\">n-1=9&nbsp;<\/p>\n\n\n\n<p>A 95% (1-\u03b1) konfidenciaszint eset\u00e9n levezethetj\u00fck a variancia konfidenciaintervallum\u00e1nak kisz\u00e1m\u00edt\u00e1s\u00e1hoz haszn\u00e1lt kvantilisek \u00e9rt\u00e9keit:&nbsp;<\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">\\frac{\\alpha}{2}=0.025\\text{ \u00e9s } 1-\\frac{\\alpha}{2} = 0.975<\/span><\/p>\n\n\n\n<p>A \ud835\udf122 t\u00f6rv\u00e9ny t\u00e1bl\u00e1zat\u00e1b\u00f3l, vagy az Ellistat szoftverrel meg lehet tal\u00e1lni az \u00e9rt\u00e9ket.&nbsp;&nbsp;<span class=\"wp-katex-eq\" data-display=\"false\"> X^{2}<em>{9;\\frac{a}{2}}=19.02\\text{ et }X^{2}<\/em>{9;1-\\frac{a}{2}}=2.70<\/span><\/p>\n\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large\"><img decoding=\"async\" src=\"https:\/\/clooma.ai\/wp-content\/uploads\/image-8-1024x627.png\" alt=\"\" class=\"wp-image-2726\"\/><\/figure><\/div>\n\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large\"><img decoding=\"async\" src=\"https:\/\/clooma.ai\/wp-content\/uploads\/image-8-1024x627.png\" alt=\"\" class=\"wp-image-2727\"\/><\/figure><\/div>\n\n\n<p>Ez\u00e9rt kisz\u00e1m\u00edthatjuk a variancia konfidenciaintervallum\u00e1t 95% konfidenciaszinten.&nbsp;<\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">\\frac{(n-1)S^{2}}{X^{2}_{n-1;\\frac{a}{2}}}\\le \\sigma^{2}\\le \\frac{(n-1)S^{2}}{X^{2}n-1;1-\\frac{a}{2}}<\/span><\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">\\frac{9<em>0.489}{19.02}\\le \\sigma^{2}\\le \\frac{9<\/em>0.489}{2.70}<\/span><\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">0.231 \\sigma^{2}\\le 1.629<\/span><\/p>\n\n\n\n<p class=\"has-text-align-center\"><span class=\"wp-katex-eq\" data-display=\"false\">0,480 \\sigma\\le 1,276<\/span><\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Ar\u00e1ny<\/h2>\n\n\n\n<p>Egy ar\u00e1ny konfidenciaintervalluma olyan \u00e9rt\u00e9ktartom\u00e1ny, amelyen bel\u00fcl egy adott sokas\u00e1g egy adott ar\u00e1ny\u00e1nak egy bizonyos val\u00f3sz\u00edn\u0171s\u00e9ggel val\u00f3 becsl\u00e9se van. M\u00e1s sz\u00f3val, ez egy olyan, mintaadatokb\u00f3l konstru\u00e1lt \u00e9rt\u00e9kintervallum, amelyen bel\u00fcl a sokas\u00e1g val\u00f3di ar\u00e1nya becsl\u00e9sek szerint egy adott megb\u00edzhat\u00f3s\u00e1gi szinttel van.&nbsp;<\/p>\n\n\n\n<p>A statisztik\u00e1ban k\u00fcl\u00f6nb\u00f6z\u0151 m\u00f3dszerek l\u00e9teznek egy ar\u00e1ny konfidenciaintervallum\u00e1nak kisz\u00e1m\u00edt\u00e1s\u00e1ra, de a k\u00e9t leggyakrabban haszn\u00e1lt m\u00f3dszer a k\u00f6vetkez\u0151:&nbsp;<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Pontos m\u00f3dszer (kis mintanagys\u00e1g eset\u00e9n).\u00a0<\/li>\n<\/ul>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Megk\u00f6zel\u00edt\u0151 m\u00f3dszer (norm\u00e1lis eloszl\u00e1ssal)\u00a0<\/li>\n<\/ul>\n\n\n\n<p><strong>Pontos m\u00f3dszer<\/strong>(sz\u00e1m\u00edt\u00e1s binomi\u00e1lis eloszl\u00e1ssal)&nbsp;<\/p>\n\n\n\n<p>Az ar\u00e1nyok konfidenciaintervallum\u00e1nak kisz\u00e1m\u00edt\u00e1s\u00e1ra szolg\u00e1l\u00f3 pontos m\u00f3dszer a binomi\u00e1lis eloszl\u00e1son alapul, \u00e9s pontos megold\u00e1st ny\u00fajt az aszimptotikus m\u00f3dszerek \u00e1ltal alkalmazott k\u00f6zel\u00edt\u00e9sek n\u00e9lk\u00fcl. Ez k\u00fcl\u00f6n\u00f6sen hasznos kis mintanagys\u00e1g eset\u00e9n, vagy ha a megfigyelt ar\u00e1ny (<\/p>\n\n\n\n<p>\ud835\udc5d\u02c6p^<\/p>\n\n\n\n<p>) k\u00f6zel 0 vagy 1.&nbsp;<\/p>\n\n\n\n<p>Az ar\u00e1ny pontos konfidenciaintervallum\u00e1nak kisz\u00e1m\u00edt\u00e1s\u00e1hoz a k\u00f6vetkez\u0151 l\u00e9p\u00e9sek sz\u00fcks\u00e9gesek:&nbsp;<\/p>\n\n\n\n<p><strong>1. l\u00e9p\u00e9s:<\/strong> Sz\u00e1m\u00edtsa ki az n mint\u00e1n megfigyelt ar\u00e1nyt k sikerrel.\ud835\udc5d\u02c6=\ud835\udc58\ud835\udc58\ud835\udc5bp^=kn Hat\u00e1rozzuk meg a konfidenciaintervallum hat\u00e1rait .<\/p>\n\n\n\n<p><strong>2. szakasz<\/strong>Sz\u00e1m\u00edtsa ki a binomi\u00e1lis eloszl\u00e1s kvantiliseit. Ezek a kvantilisek hat\u00e1rolj\u00e1k a konfidenciaintervallumot. 1-\u03b1 konfidenciaszint eset\u00e9n a binomi\u00e1lis eloszl\u00e1s t\u00e1bl\u00e1zat\u00e1b\u00f3l meg kell tal\u00e1lnia a Q1 kvantilis\u00e9t a \ud835\udefc2\ud835\udefc2 percentilisn\u00e9l, majd a Q2 kvantilis\u00e9t az 1-\ud835\udefc21-\ud835\udefc2 percentilisn\u00e9l. Ezeket a kvantiliseket a binomi\u00e1lis eloszl\u00e1s t\u00e1bl\u00e1zataival vagy az Ellistat szoftverben lehet megtal\u00e1lni.\u00a0<\/p>\n\n\n\n<p><strong>3. l\u00e9p\u00e9s: <\/strong>A konfidenciaintervallum kisz\u00e1m\u00edt\u00e1sa: A konfidenciaintervallum kisz\u00e1m\u00edt\u00e1sa a k\u00f6vetkez\u0151 k\u00e9plet seg\u00edts\u00e9g\u00e9vel t\u00f6rt\u00e9nik: Ezut\u00e1n sz\u00e1m\u00edtsa ki a konfidenciaintervallumot [\ud835\udc441\ud835\udc5b;\ud835\udc442\ud835\udc5b][Q1n;Q2n].\u00a0\u00a0<\/p>\n\n\n\n<p>P\u00e9lda: Tegy\u00fck fel, hogy egy n=20 m\u00e9ret\u0171 minta felv\u00e9tele ut\u00e1n k=15 megfelel\u0151 alkatr\u00e9szt figyelt meg. Sz\u00e1m\u00edtsa ki a megfelel\u0151 alkatr\u00e9szek ar\u00e1ny\u00e1nak pontos konfidenciaintervallum\u00e1t 95-\u00f6s konfidenciaszint mellett)?&nbsp;<\/p>\n\n\n\n<p>Megold\u00e1s:&nbsp;<\/p>\n\n\n\n<p>A megfigyelt megfelel\u0151 alkatr\u00e9szek ar\u00e1nya: \ud835\udc5d\u02c6=1520=0,75p^=1520=0,75<\/p>\n\n\n\n<p>A Q1 \u00e9s Q2 meghat\u00e1roz\u00e1sa p=0,75 ar\u00e1ny \u00e9s 20 f\u0151s mintanagys\u00e1g eset\u00e9n.&nbsp;<\/p>\n\n\n\n<p>Az Ellistat szoftver seg\u00edts\u00e9g\u00e9vel a k\u00f6vetkez\u0151ket tal\u00e1ljuk: Q1=11 (11 a 0,025-hez legk\u00f6zelebbi \ud835\udefc\/2 \u00e9rt\u00e9ket adja) \u00e9s Q2=18.\u00a0<\/p>\n\n\n\n<p><img decoding=\"async\" alt=\"Sz\u00f6veget tartalmaz\u00f3 k\u00e9p, k\u00e9perny\u0151k\u00e9p, diagram, automatikusan gener\u00e1lt vonalle\u00edr\u00e1s\" src=\"blob:https:\/\/clooma.ai\/452a7f55-c11c-4fc1-a123-de2b51f2efa0\"><\/p>\n\n\n\n<p><img decoding=\"async\" alt=\"Sz\u00f6veget, diagramot, vonalat vagy automatikusan gener\u00e1lt PlotDescription k\u00e9pet tartalmaz\u00f3 k\u00e9p.\" src=\"blob:https:\/\/clooma.ai\/ffe282d7-519f-4138-9eff-8e17c09328ca\"><\/p>\n\n\n\n<p>Az ar\u00e1ny konfidenciaintervalluma 95% konfidencia eset\u00e9n: [\ud835\udc441\ud835\udc5b;\ud835\udc442\ud835\udc5b]= [1120;1820]Q1n;Q2n= [1120;1820]<\/p>\n\n\n\n<p>Fontos megjegyezni, hogy ez a m\u00f3dszer pontos megold\u00e1st ad, de sz\u00e1m\u00edt\u00e1sig\u00e9nyesebb lehet, k\u00fcl\u00f6n\u00f6sen nagy mintanagys\u00e1g eset\u00e9n, \u00e9s gyakran statisztikai szoftver haszn\u00e1lata sz\u00fcks\u00e9ges a sz\u00e1m\u00edt\u00e1sok elv\u00e9gz\u00e9s\u00e9hez.&nbsp;<\/p>\n\n\n\n<p><strong>Megk\u00f6zel\u00edt\u0151 m\u00f3dszer (norm\u00e1lis eloszl\u00e1ssal)<\/strong>:&nbsp;&nbsp;<\/p>\n\n\n\n<p>Egy popul\u00e1ci\u00f3ban l\u00e9v\u0151 ar\u00e1nyra vonatkoz\u00f3 konfidenciaintervallum megalkot\u00e1s\u00e1hoz a norm\u00e1lis eloszl\u00e1st haszn\u00e1ljuk, ha a k\u00f6zponti hat\u00e1r\u00e9rt\u00e9kt\u00e9tel felt\u00e9telei teljes\u00fclnek. Ha egy n m\u00e9ret\u0171 mint\u00e1t vesz\u00fcnk egy olyan sokas\u00e1gb\u00f3l, amely a p param\u00e9ter\u0171 binomi\u00e1lis eloszl\u00e1st k\u00f6veti, akkor az ebb\u0151l a mint\u00e1b\u00f3l sz\u00e1m\u00edtott ar\u00e1ny p^, felt\u00e9ve, hogy: \ud835\udc5d\u02c6=\ud835\udc65\ud835\udc5bp^=xn<\/p>\n\n\n\n<p>A :&nbsp;&nbsp;<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>x: a sikerek sz\u00e1ma.\u00a0<\/li>\n\n\n\n<li>n: a minta m\u00e9rete.\u00a0<\/li>\n<\/ul>\n\n\n\n<p>A popul\u00e1ci\u00f3 \u00e1tlaga \u00e9s sz\u00f3r\u00e1sa \ud835\udf07\ud835\udc5d\u02c6=\ud835\udc5d<\/p>\n\n\n\n<p class=\"has-text-align-center\">\ud835\udf0e\ud835\udc5d\u02c6=\ud835\udc5d (1-\ud835\udc5d)\ud835\udc5b\u203e\u203e\u203e\u203e\u203e\u203e\u203e\u203e\u203e\u221a\ud835\udf0ep^=p (1-p)n<\/p>\n\n\n\n<p>A korl\u00e1tozott centr\u00e1lis t\u00e9tel alkalmazhat\u00f3 a mint\u00e1k ar\u00e1ny\u00e1ra, ha \ud835\udc5b\u2217\ud835\udc5d\u22655n\u2217p\u22655 \u00e9s \ud835\udc5b\u2217\u2217(1-\ud835\udc5d)\u22655n\u2217(1-p)\u22655. Ez val\u00f3ban k\u00fcl\u00f6n\u00f6sen hasznos nagy mintanagys\u00e1gok eset\u00e9n, vagy amikor a megfigyelt ar\u00e1nyok nem k\u00f6zel\u00edtik meg az 1-et \u00e9s a 0-t.\u00a0<\/p>\n\n\n\n<p>A Z-pontsz\u00e1m k\u00e9plet teh\u00e1t alkalmazhat\u00f3: \ud835\udf0e\ud835\udc5d\u02c6=\ud835\udc5d (1-\ud835\udc5d)\ud835\udc5b\u203e\u203e\u203e\u203e\u203e\u203e\u203e\u203e\u203e\u221a\ud835\udf0ep^=p (1-p)n<\/p>\n\n\n\n<p>Ha0\u2264\ud835\udf07\ud835\udc5d\u02c6\u00b12\ud835\udf0e\ud835\udc5d\u02c6\u226410\u2264\ud835\udf07p^\u00b12\ud835\udf0ep^\u22641 , akkor \u00fagy tekinthetj\u00fck, hogy \ud835\udc5d\u02c6p^ megk\u00f6zel\u00edt\u0151leg norm\u00e1lis eloszl\u00e1st k\u00f6vet.\u00a0\u00a0<\/p>","protected":false},"featured_media":0,"template":"","meta":{"_acf_changed":false},"menu-ressource-pedagogique":[27],"class_list":["post-2723","ressource-pedagogiqu","type-ressource-pedagogiqu","status-publish","hentry","menu-ressource-pedagogique-3-statistiques-descriptives"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v25.9 - 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