MATHHX B
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15.3 Beviser til konfidensintervaller (A)
Vi vil udlede formlen:
-
Sætning 15.2.1
Hvis standardafvigelsen er kendt, bestemmes et konfidensinterval \(I\) for middelværdien i en normalfordeling ved formlen:
\[I=\left [\bar {x}-z_{1-\frac {\alpha }{2}}\cdot \frac {\sigma }{\sqrt {n}};\bar {x}+z_{1-\frac {\alpha }{2}}\cdot \frac {\sigma }{\sqrt {n}}\right ]\]
Her er:
- \(n\):
-
Stikprøvens størrelse
- \(\bar {x}\):
-
Gennemsnittet af stikprøven
- \(\sigma \):
-
Standardafvigelsen
- \(\alpha \):
-
Signifikansniveauet som decimaltal
- \(z_{1-\frac {\alpha }{2}}\):
-
\(\left (1-\frac {\alpha }{2}\right )\)-fraktilen i standardnormalfordelingen.
For at gøre det mere konkret, kigger vi på den situation hvor \(\alpha \) er \(5\%\), og vi har derfor at \(z_{1-\frac {\alpha }{2}}=1{,}96\) (se tabel 15.2, hvis du forvirret). Vi skal altså vise, at \(95\%\)-konfidensintervallet er givet ved:
\[I=\left [\overline {x}-1{,}96\cdot \frac {\sigma }{\sqrt {n}};\overline {x}+1{,}96\cdot \frac {\sigma }{\sqrt {n}}\right ]\]
Før vi går i gang med selve udledningen, er der lige nogle ting, vi skal have på plads.
Forudsætning 1
Vi husker:
-
Sætning 14.2.1 (68-95-99,7-reglen)
For en normalfordelt stokastisk variabel \(X\sim N(\mu ,\sigma )\) gælder følgende:
\(P(\mu -\sigma \leq X\leq \mu +\sigma )\approx 68\%\)
\(P(\mu -2\sigma \leq X\leq \mu +2\sigma )\approx 95\%\)
\(P(\mu -3\sigma \leq X\leq \mu +3\sigma )\approx 99{,}7\%\)
Sandsynlighederne i reglen er afrundede til hele tal, så når der står, at
\[P(\mu -2\sigma \leq X\leq \mu +2\sigma )\approx 95\%\]
så er den rigtige sandsynlighed faktisk \(95{,}4499736\ldots \%\). Vil man have \(95\%\) mere præcist, skal man kun gå \(1{,}96\) standardafvigelser ud (hvilket heller ikke giver præcis \(95\%\), men tættere på). Altså gælder:
\[P(\mu -1{,}96\sigma \leq X\leq \mu +1{,}96\sigma )= 95\%\]
Forudsætning 2
Hvis \(X\sim N(\mu ,\sigma )\) og vi vil lave en stikprøve, så er stikprøvens gennemsnit \(\overline {X}\) en stokastisk variabel som er normalfordelt \(\overline {X}\sim N(\mu ,\frac {\sigma }{\sqrt {n}})\). Vi vil
ikke bevise dette, men vi vil se på et eksempel for bedre at forstå, hvad det betyder:
-
Eksempel 15.3.1
Højden (målt i cm) af voksne mænd i Danmark er tilnærmelsesvis normalfordelt med \(X\sim N(180,7)\). Hvis vi udtager flere stikprøver, hver på 100 voksne mænd, vil vi få forskelligt
gennemsnit for hver stikprøve. Det kan være at gennemsnitshøjden, i den første stikprøve, er 182, den næste 179, den næste 181 osv. Gennemsnittet vil ligge tæt på de \(180\), men indtil stikprøven er udtaget, kan vi ikke sige
præcis hvad det vil blive. Gennemsnittet er altså en stokastisk variabel \(\overline {X}\). Påstanden er så, at denne stokastiske variable \(\overline {X}\) er normalfordelt:
\[\overline {X}\sim N(180,\frac {7}{\sqrt {100}})\]
Det vil sige
\[\overline {X}\sim N(180;{0{,}7})\]
Vi konkluderer, at gennemsnitshøjden i en stikprøve med \(100\) voksne mænd er normalfordelt med middelværdi \(180\) og standardafvigelse \(0{,}7\).
Vi ser af formlen \(\overline {X}\sim N(\mu ,\frac {\sigma }{\sqrt {n}})\), at jo større stikprøven er jo mindre bliver standardafvigelsen for stikprøvens gennemsnit. Det giver god mening, fordi en stor stikprøve altid vil
have et gennemsnit tæt på populationens gennemsnit.
Øvelse 15.3.1
Antag at vi køber en æbleplantage og at æblernes vægt er normalfordeling \(X\sim N(100,20)\).
-
a) Bestem \(\mu \) og \(\sigma \).
-
b) Vi vil nu udtage en stikprøve på 5 æbler og beregne gennemsnitsvægten i stikprøven. Gennemsnitsvægten i stikprøven vil da være en stokastisk
variabel. Hvordan er den fordelt?
-
c) Vi udtager nu stikprøven. Den består af æbler med vægten: \(79,105,67,104,128\). Bestem \(\overline {x}\).
-
d) Forklar forskellen på \(\overline {x}\) og \(\overline {X}\) i forbindelse med eksemplet med æblerne.
Løsning 15.3.1
-
a) \(\mu =100\) og \(\sigma =20\)
-
b) \(\overline {X}\sim N(100,\frac {20}{\sqrt {5}})\). Dvs. \(\overline {X}\sim N(100;8{,}94)\)
-
c) \(\overline {x}=96{,}6\)
-
d) Betegnelsen \(\overline {x}\) blev introduceret i afsnit 11.2. Den betyder bare gennemsnittet af en stikprøven, vi har udtaget.
Betegnelsen \(\overline {X}\) betyder gennemsnittet som stokastisk variabel. Man kan tænke på \(\overline {X}\) som gennemsnittet af stikprøven inden den udtages. Det er klart at gennemsnittet af æblernes vægt,
ikke er fastlagt før stikprøven udtages. Det afhænger jo af, hvilke æbler vi lige præcis får fat i. Men det er oplagt, at vi kan udtale os om sandsynligheden for at få et bestemt gennemsnit, og derfor er \(\overline {X}\) en
stokastisk variabel, som har en sandsynlighedsfordeling.
Udledning
Fra forudsætning 1 ved vi at, hvis vi har en normal fordelt stokastisk variabel \(X\sim N(\mu ,\sigma )\) så er
\[P(\mu -1{,}96\sigma \leq X\leq \mu +1{,}96\sigma )= 95\%\]
Fra forudsætning 2 ved vi at gennemsnittet for en stikprøve \(\overline {X}\) er normalfordelt \(\overline {X}\sim N(\mu ,\frac {\sigma }{\sqrt {n}})\). Vi bruger nu forudsætning 1 på \(\overline {X}\) og får:
\[P(\mu -1{,}96\frac {\sigma }{\sqrt {n}}\leq \overline {X}\leq \mu +1{,}96\frac {\sigma }{\sqrt {n}})= 95\%\]
Vi omskriver nu uligheden. Vi starter med at trække \(\overline {X}\) og \(\mu \) fra på begge sider af ulighedstegnet:
\[P(-\overline {X}-1{,}96\frac {\sigma }{\sqrt {n}}\leq -\mu \leq -\overline {X}+1{,}96\frac {\sigma }{\sqrt {n}})= 95\%\]
Nu ganger vi uligheden med \(-1\) (og husker at vi så skal vende ulighedstegnene)
\[P(\overline {X}+1{,}96\frac {\sigma }{\sqrt {n}}\geq \mu \geq \overline {X}-1{,}96\frac {\sigma }{\sqrt {n}})= 95\%\]
Hvilket jo er det samme som:
\[P(\overline {X}-1{,}96\frac {\sigma }{\sqrt {n}}\leq \mu \leq \overline {X}+1{,}96\frac {\sigma }{\sqrt {n}})= 95\%\]
Dette betyder, at hvis vi udtager stikprøve og regner dens gennemsnit \(\overline {x}\), så er der \(95\%\) sandsynlighed for at \(\mu \) vil ligge i intervallet:
\[\left [\overline {x}-1{,}96\frac {\sigma }{\sqrt {n}};\overline {x}+1{,}96\frac {\sigma }{\sqrt {n}}\right ]\]
At der er \(95\%\) sandsynlighed for at \(\mu \) vil ligge i intervallet betyder, at intervallet er et \(95\%\)-konfidensinterval.