Power testing and a Poisson Distribution












1












$begingroup$


Let $X_1, ..., X_n$ be a random sample from a Poisson distribution with parameter $lambda$.



If $X$ is Poisson with parameter $lambda$, then $text{E}[X] = text{Var}[X] = lambda$; hence $text{E}[X] = lambda$ and $text{Var}[X] = lambda/n$.



(a) Test $H_0 :lambda=4$ versus $H_A : lambda>4$ and that our test procedure is to reject $H_0$ if $x ge k$. Given $n = 100$, what value of $k$ gives a test with significance level approximately equal to $0.05$?



(b) What is the approximate power of the test at $lambda = 5$?



Hi all, I'm new to this so any help is appreciated. For (a), I tried setting up a $95%$ confidence interval because I thought that $95%$ confidence and $5%$ significance were about the same.






statistical-inference hypothesis-testing







share|cite|improve this question











$endgroup$












  • $begingroup$
    Would the downvoter care to explain?
    $endgroup$
    – M. Vinay
    Jun 30 '14 at 16:13










  • $begingroup$
    When you say x do you mean mean (x bar)
    $endgroup$
    – Kamster
    Jun 30 '14 at 16:26










  • $begingroup$
    no, not the sample mean. X is a random variable with the characteristics stated above. I am seeking helping with power tests.
    $endgroup$
    – Alina P
    Jun 30 '14 at 16:30










  • $begingroup$
    Oh I meant what do you mean by $xgeq k$ since you have 100 observations I don't know what x is
    $endgroup$
    – Kamster
    Jun 30 '14 at 16:33










  • $begingroup$
    From my class notes I'm getting that I want to reject the simple null hypothesis if x, the random variable, is greater than or equal to k. How do I set up a test?
    $endgroup$
    – Alina P
    Jun 30 '14 at 16:43
















1












$begingroup$


Let $X_1, ..., X_n$ be a random sample from a Poisson distribution with parameter $lambda$.



If $X$ is Poisson with parameter $lambda$, then $text{E}[X] = text{Var}[X] = lambda$; hence $text{E}[X] = lambda$ and $text{Var}[X] = lambda/n$.



(a) Test $H_0 :lambda=4$ versus $H_A : lambda>4$ and that our test procedure is to reject $H_0$ if $x ge k$. Given $n = 100$, what value of $k$ gives a test with significance level approximately equal to $0.05$?



(b) What is the approximate power of the test at $lambda = 5$?



Hi all, I'm new to this so any help is appreciated. For (a), I tried setting up a $95%$ confidence interval because I thought that $95%$ confidence and $5%$ significance were about the same.






statistical-inference hypothesis-testing







share|cite|improve this question











$endgroup$












  • $begingroup$
    Would the downvoter care to explain?
    $endgroup$
    – M. Vinay
    Jun 30 '14 at 16:13










  • $begingroup$
    When you say x do you mean mean (x bar)
    $endgroup$
    – Kamster
    Jun 30 '14 at 16:26










  • $begingroup$
    no, not the sample mean. X is a random variable with the characteristics stated above. I am seeking helping with power tests.
    $endgroup$
    – Alina P
    Jun 30 '14 at 16:30










  • $begingroup$
    Oh I meant what do you mean by $xgeq k$ since you have 100 observations I don't know what x is
    $endgroup$
    – Kamster
    Jun 30 '14 at 16:33










  • $begingroup$
    From my class notes I'm getting that I want to reject the simple null hypothesis if x, the random variable, is greater than or equal to k. How do I set up a test?
    $endgroup$
    – Alina P
    Jun 30 '14 at 16:43














1












1








1





$begingroup$


Let $X_1, ..., X_n$ be a random sample from a Poisson distribution with parameter $lambda$.



If $X$ is Poisson with parameter $lambda$, then $text{E}[X] = text{Var}[X] = lambda$; hence $text{E}[X] = lambda$ and $text{Var}[X] = lambda/n$.



(a) Test $H_0 :lambda=4$ versus $H_A : lambda>4$ and that our test procedure is to reject $H_0$ if $x ge k$. Given $n = 100$, what value of $k$ gives a test with significance level approximately equal to $0.05$?



(b) What is the approximate power of the test at $lambda = 5$?



Hi all, I'm new to this so any help is appreciated. For (a), I tried setting up a $95%$ confidence interval because I thought that $95%$ confidence and $5%$ significance were about the same.






statistical-inference hypothesis-testing







share|cite|improve this question











$endgroup$




Let $X_1, ..., X_n$ be a random sample from a Poisson distribution with parameter $lambda$.



If $X$ is Poisson with parameter $lambda$, then $text{E}[X] = text{Var}[X] = lambda$; hence $text{E}[X] = lambda$ and $text{Var}[X] = lambda/n$.



(a) Test $H_0 :lambda=4$ versus $H_A : lambda>4$ and that our test procedure is to reject $H_0$ if $x ge k$. Given $n = 100$, what value of $k$ gives a test with significance level approximately equal to $0.05$?



(b) What is the approximate power of the test at $lambda = 5$?



Hi all, I'm new to this so any help is appreciated. For (a), I tried setting up a $95%$ confidence interval because I thought that $95%$ confidence and $5%$ significance were about the same.






statistical-inference hypothesis-testing




statistical-inference hypothesis-testing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jun 30 '14 at 16:12









M. Vinay

7,33322136




7,33322136










asked Jun 30 '14 at 15:44









Alina PAlina P

164




164












  • $begingroup$
    Would the downvoter care to explain?
    $endgroup$
    – M. Vinay
    Jun 30 '14 at 16:13










  • $begingroup$
    When you say x do you mean mean (x bar)
    $endgroup$
    – Kamster
    Jun 30 '14 at 16:26










  • $begingroup$
    no, not the sample mean. X is a random variable with the characteristics stated above. I am seeking helping with power tests.
    $endgroup$
    – Alina P
    Jun 30 '14 at 16:30










  • $begingroup$
    Oh I meant what do you mean by $xgeq k$ since you have 100 observations I don't know what x is
    $endgroup$
    – Kamster
    Jun 30 '14 at 16:33










  • $begingroup$
    From my class notes I'm getting that I want to reject the simple null hypothesis if x, the random variable, is greater than or equal to k. How do I set up a test?
    $endgroup$
    – Alina P
    Jun 30 '14 at 16:43


















  • $begingroup$
    Would the downvoter care to explain?
    $endgroup$
    – M. Vinay
    Jun 30 '14 at 16:13










  • $begingroup$
    When you say x do you mean mean (x bar)
    $endgroup$
    – Kamster
    Jun 30 '14 at 16:26










  • $begingroup$
    no, not the sample mean. X is a random variable with the characteristics stated above. I am seeking helping with power tests.
    $endgroup$
    – Alina P
    Jun 30 '14 at 16:30










  • $begingroup$
    Oh I meant what do you mean by $xgeq k$ since you have 100 observations I don't know what x is
    $endgroup$
    – Kamster
    Jun 30 '14 at 16:33










  • $begingroup$
    From my class notes I'm getting that I want to reject the simple null hypothesis if x, the random variable, is greater than or equal to k. How do I set up a test?
    $endgroup$
    – Alina P
    Jun 30 '14 at 16:43
















$begingroup$
Would the downvoter care to explain?
$endgroup$
– M. Vinay
Jun 30 '14 at 16:13




$begingroup$
Would the downvoter care to explain?
$endgroup$
– M. Vinay
Jun 30 '14 at 16:13












$begingroup$
When you say x do you mean mean (x bar)
$endgroup$
– Kamster
Jun 30 '14 at 16:26




$begingroup$
When you say x do you mean mean (x bar)
$endgroup$
– Kamster
Jun 30 '14 at 16:26












$begingroup$
no, not the sample mean. X is a random variable with the characteristics stated above. I am seeking helping with power tests.
$endgroup$
– Alina P
Jun 30 '14 at 16:30




$begingroup$
no, not the sample mean. X is a random variable with the characteristics stated above. I am seeking helping with power tests.
$endgroup$
– Alina P
Jun 30 '14 at 16:30












$begingroup$
Oh I meant what do you mean by $xgeq k$ since you have 100 observations I don't know what x is
$endgroup$
– Kamster
Jun 30 '14 at 16:33




$begingroup$
Oh I meant what do you mean by $xgeq k$ since you have 100 observations I don't know what x is
$endgroup$
– Kamster
Jun 30 '14 at 16:33












$begingroup$
From my class notes I'm getting that I want to reject the simple null hypothesis if x, the random variable, is greater than or equal to k. How do I set up a test?
$endgroup$
– Alina P
Jun 30 '14 at 16:43




$begingroup$
From my class notes I'm getting that I want to reject the simple null hypothesis if x, the random variable, is greater than or equal to k. How do I set up a test?
$endgroup$
– Alina P
Jun 30 '14 at 16:43










1 Answer
1






active

oldest

votes


















0












$begingroup$

So first to find test with significance level of at approximately 0.05 we see that we want
$$P_{lambdainTheta_{Ho}}(bar{X}geq k)=P_{lambda=4}left(sum Xgeq 100k right) = 1-P_{lambda=4}left(sum X< 100kright)=0.05$$
Where we see that $sum Xsim operatorname{Pois}(100cdot4)=operatorname{Pois}(400)$ Thus in order to find when this is true it can be a little tedious but you are going to have to calculate a list of possible $100k=t$'s and calculate $1-P(sum X< t)$ (remember for $operatorname{Pois}(400)$ and trying doing on excel) and see which is closest to $0.05$ (not above though). after you get $t^{*}$ that does this call it $t^{*}$ then $k=frac{t^{*}}{100}$.



Now to calculate Power we just calculate
$$operatorname{power}(lambda=5)=P_{lambda=5}left(sum Xgeq t^{*}right) = 1-P_{lambda=5}left(sum X< t^{*}right)$$
where now $sum Xsim operatorname{Pois}(100cdot5)=operatorname{Pois}(500)$



also note $sum X=sum_{i=1}^{100} X_i$



Also to your question about confidence interval, it is true you could calculate a $95%$ confidence interval and that would correspond to $5%$ significance level but the problem with trying to do this is that I don't really know what CI interval is for mean in poisson and also even if you could derive it, CI intervals still rely on some critical value and since you can't really standardize or have table to look at for that it becomes situation above where you just have calculate probabilities for list of values and see which ones closets to significance level






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I've never used excel to do calculations before, so could you help me with what I need to input? I entered poisson distribution with mean 4, x=100 and cumulative 0. My answer was 3.154 and I don't think that's right.
    $endgroup$
    – Alina P
    Jun 30 '14 at 20:42












  • $begingroup$
    Ok so on excel when you have cumulative=TRUE that means it calculated Pr(X<=x) thus to get => you must subrtact that one from 1 then and Pr(X=x)
    $endgroup$
    – Kamster
    Jun 30 '14 at 23:53










  • $begingroup$
    You can write $100times4$ or $100cdot4.$ Using an asterisk for that purpose is a workaround for occasions when you are limited to the symbols on the keyboard.
    $endgroup$
    – Michael Hardy
    May 17 '18 at 17:41














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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

So first to find test with significance level of at approximately 0.05 we see that we want
$$P_{lambdainTheta_{Ho}}(bar{X}geq k)=P_{lambda=4}left(sum Xgeq 100k right) = 1-P_{lambda=4}left(sum X< 100kright)=0.05$$
Where we see that $sum Xsim operatorname{Pois}(100cdot4)=operatorname{Pois}(400)$ Thus in order to find when this is true it can be a little tedious but you are going to have to calculate a list of possible $100k=t$'s and calculate $1-P(sum X< t)$ (remember for $operatorname{Pois}(400)$ and trying doing on excel) and see which is closest to $0.05$ (not above though). after you get $t^{*}$ that does this call it $t^{*}$ then $k=frac{t^{*}}{100}$.



Now to calculate Power we just calculate
$$operatorname{power}(lambda=5)=P_{lambda=5}left(sum Xgeq t^{*}right) = 1-P_{lambda=5}left(sum X< t^{*}right)$$
where now $sum Xsim operatorname{Pois}(100cdot5)=operatorname{Pois}(500)$



also note $sum X=sum_{i=1}^{100} X_i$



Also to your question about confidence interval, it is true you could calculate a $95%$ confidence interval and that would correspond to $5%$ significance level but the problem with trying to do this is that I don't really know what CI interval is for mean in poisson and also even if you could derive it, CI intervals still rely on some critical value and since you can't really standardize or have table to look at for that it becomes situation above where you just have calculate probabilities for list of values and see which ones closets to significance level






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I've never used excel to do calculations before, so could you help me with what I need to input? I entered poisson distribution with mean 4, x=100 and cumulative 0. My answer was 3.154 and I don't think that's right.
    $endgroup$
    – Alina P
    Jun 30 '14 at 20:42












  • $begingroup$
    Ok so on excel when you have cumulative=TRUE that means it calculated Pr(X<=x) thus to get => you must subrtact that one from 1 then and Pr(X=x)
    $endgroup$
    – Kamster
    Jun 30 '14 at 23:53










  • $begingroup$
    You can write $100times4$ or $100cdot4.$ Using an asterisk for that purpose is a workaround for occasions when you are limited to the symbols on the keyboard.
    $endgroup$
    – Michael Hardy
    May 17 '18 at 17:41


















0












$begingroup$

So first to find test with significance level of at approximately 0.05 we see that we want
$$P_{lambdainTheta_{Ho}}(bar{X}geq k)=P_{lambda=4}left(sum Xgeq 100k right) = 1-P_{lambda=4}left(sum X< 100kright)=0.05$$
Where we see that $sum Xsim operatorname{Pois}(100cdot4)=operatorname{Pois}(400)$ Thus in order to find when this is true it can be a little tedious but you are going to have to calculate a list of possible $100k=t$'s and calculate $1-P(sum X< t)$ (remember for $operatorname{Pois}(400)$ and trying doing on excel) and see which is closest to $0.05$ (not above though). after you get $t^{*}$ that does this call it $t^{*}$ then $k=frac{t^{*}}{100}$.



Now to calculate Power we just calculate
$$operatorname{power}(lambda=5)=P_{lambda=5}left(sum Xgeq t^{*}right) = 1-P_{lambda=5}left(sum X< t^{*}right)$$
where now $sum Xsim operatorname{Pois}(100cdot5)=operatorname{Pois}(500)$



also note $sum X=sum_{i=1}^{100} X_i$



Also to your question about confidence interval, it is true you could calculate a $95%$ confidence interval and that would correspond to $5%$ significance level but the problem with trying to do this is that I don't really know what CI interval is for mean in poisson and also even if you could derive it, CI intervals still rely on some critical value and since you can't really standardize or have table to look at for that it becomes situation above where you just have calculate probabilities for list of values and see which ones closets to significance level






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I've never used excel to do calculations before, so could you help me with what I need to input? I entered poisson distribution with mean 4, x=100 and cumulative 0. My answer was 3.154 and I don't think that's right.
    $endgroup$
    – Alina P
    Jun 30 '14 at 20:42












  • $begingroup$
    Ok so on excel when you have cumulative=TRUE that means it calculated Pr(X<=x) thus to get => you must subrtact that one from 1 then and Pr(X=x)
    $endgroup$
    – Kamster
    Jun 30 '14 at 23:53










  • $begingroup$
    You can write $100times4$ or $100cdot4.$ Using an asterisk for that purpose is a workaround for occasions when you are limited to the symbols on the keyboard.
    $endgroup$
    – Michael Hardy
    May 17 '18 at 17:41
















0












0








0





$begingroup$

So first to find test with significance level of at approximately 0.05 we see that we want
$$P_{lambdainTheta_{Ho}}(bar{X}geq k)=P_{lambda=4}left(sum Xgeq 100k right) = 1-P_{lambda=4}left(sum X< 100kright)=0.05$$
Where we see that $sum Xsim operatorname{Pois}(100cdot4)=operatorname{Pois}(400)$ Thus in order to find when this is true it can be a little tedious but you are going to have to calculate a list of possible $100k=t$'s and calculate $1-P(sum X< t)$ (remember for $operatorname{Pois}(400)$ and trying doing on excel) and see which is closest to $0.05$ (not above though). after you get $t^{*}$ that does this call it $t^{*}$ then $k=frac{t^{*}}{100}$.



Now to calculate Power we just calculate
$$operatorname{power}(lambda=5)=P_{lambda=5}left(sum Xgeq t^{*}right) = 1-P_{lambda=5}left(sum X< t^{*}right)$$
where now $sum Xsim operatorname{Pois}(100cdot5)=operatorname{Pois}(500)$



also note $sum X=sum_{i=1}^{100} X_i$



Also to your question about confidence interval, it is true you could calculate a $95%$ confidence interval and that would correspond to $5%$ significance level but the problem with trying to do this is that I don't really know what CI interval is for mean in poisson and also even if you could derive it, CI intervals still rely on some critical value and since you can't really standardize or have table to look at for that it becomes situation above where you just have calculate probabilities for list of values and see which ones closets to significance level






share|cite|improve this answer











$endgroup$



So first to find test with significance level of at approximately 0.05 we see that we want
$$P_{lambdainTheta_{Ho}}(bar{X}geq k)=P_{lambda=4}left(sum Xgeq 100k right) = 1-P_{lambda=4}left(sum X< 100kright)=0.05$$
Where we see that $sum Xsim operatorname{Pois}(100cdot4)=operatorname{Pois}(400)$ Thus in order to find when this is true it can be a little tedious but you are going to have to calculate a list of possible $100k=t$'s and calculate $1-P(sum X< t)$ (remember for $operatorname{Pois}(400)$ and trying doing on excel) and see which is closest to $0.05$ (not above though). after you get $t^{*}$ that does this call it $t^{*}$ then $k=frac{t^{*}}{100}$.



Now to calculate Power we just calculate
$$operatorname{power}(lambda=5)=P_{lambda=5}left(sum Xgeq t^{*}right) = 1-P_{lambda=5}left(sum X< t^{*}right)$$
where now $sum Xsim operatorname{Pois}(100cdot5)=operatorname{Pois}(500)$



also note $sum X=sum_{i=1}^{100} X_i$



Also to your question about confidence interval, it is true you could calculate a $95%$ confidence interval and that would correspond to $5%$ significance level but the problem with trying to do this is that I don't really know what CI interval is for mean in poisson and also even if you could derive it, CI intervals still rely on some critical value and since you can't really standardize or have table to look at for that it becomes situation above where you just have calculate probabilities for list of values and see which ones closets to significance level







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited May 17 '18 at 17:41









Michael Hardy

1




1










answered Jun 30 '14 at 16:59









KamsterKamster

1,687917




1,687917












  • $begingroup$
    I've never used excel to do calculations before, so could you help me with what I need to input? I entered poisson distribution with mean 4, x=100 and cumulative 0. My answer was 3.154 and I don't think that's right.
    $endgroup$
    – Alina P
    Jun 30 '14 at 20:42












  • $begingroup$
    Ok so on excel when you have cumulative=TRUE that means it calculated Pr(X<=x) thus to get => you must subrtact that one from 1 then and Pr(X=x)
    $endgroup$
    – Kamster
    Jun 30 '14 at 23:53










  • $begingroup$
    You can write $100times4$ or $100cdot4.$ Using an asterisk for that purpose is a workaround for occasions when you are limited to the symbols on the keyboard.
    $endgroup$
    – Michael Hardy
    May 17 '18 at 17:41




















  • $begingroup$
    I've never used excel to do calculations before, so could you help me with what I need to input? I entered poisson distribution with mean 4, x=100 and cumulative 0. My answer was 3.154 and I don't think that's right.
    $endgroup$
    – Alina P
    Jun 30 '14 at 20:42












  • $begingroup$
    Ok so on excel when you have cumulative=TRUE that means it calculated Pr(X<=x) thus to get => you must subrtact that one from 1 then and Pr(X=x)
    $endgroup$
    – Kamster
    Jun 30 '14 at 23:53










  • $begingroup$
    You can write $100times4$ or $100cdot4.$ Using an asterisk for that purpose is a workaround for occasions when you are limited to the symbols on the keyboard.
    $endgroup$
    – Michael Hardy
    May 17 '18 at 17:41


















$begingroup$
I've never used excel to do calculations before, so could you help me with what I need to input? I entered poisson distribution with mean 4, x=100 and cumulative 0. My answer was 3.154 and I don't think that's right.
$endgroup$
– Alina P
Jun 30 '14 at 20:42






$begingroup$
I've never used excel to do calculations before, so could you help me with what I need to input? I entered poisson distribution with mean 4, x=100 and cumulative 0. My answer was 3.154 and I don't think that's right.
$endgroup$
– Alina P
Jun 30 '14 at 20:42














$begingroup$
Ok so on excel when you have cumulative=TRUE that means it calculated Pr(X<=x) thus to get => you must subrtact that one from 1 then and Pr(X=x)
$endgroup$
– Kamster
Jun 30 '14 at 23:53




$begingroup$
Ok so on excel when you have cumulative=TRUE that means it calculated Pr(X<=x) thus to get => you must subrtact that one from 1 then and Pr(X=x)
$endgroup$
– Kamster
Jun 30 '14 at 23:53












$begingroup$
You can write $100times4$ or $100cdot4.$ Using an asterisk for that purpose is a workaround for occasions when you are limited to the symbols on the keyboard.
$endgroup$
– Michael Hardy
May 17 '18 at 17:41






$begingroup$
You can write $100times4$ or $100cdot4.$ Using an asterisk for that purpose is a workaround for occasions when you are limited to the symbols on the keyboard.
$endgroup$
– Michael Hardy
May 17 '18 at 17:41




















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Wiesbaden

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Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
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Marschland

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Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
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Dieringhausen

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Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
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