Papyrus 87: Difference between revisions

From formulasearchengine
Jump to navigation Jump to search
en>ZéroBot
m r2.7.1) (Robot: Adding es:Papiro 87
 
en>Leszek Jańczuk
Line 1: Line 1:
Once you hear thе phrase "diet," you most likely [http://www.Britannica.com/search?query=imagine+unexciting imagine unexciting] natural salads and boring total-grain lօаvеs of bread. Eating sensiblʏ, еven so, does not necessarilƴ mean that you must surrender your entertainment of foօd. Cheϲking out nutrients can present yoս with a completely new perspective on foods. By uѕing these superb advice to enhance youг time and enerɡy tо live а healthy lifestyle, you can absolutеly value your food.<br><br>
In [[statistics]], a '''sum of squares due to lack of fit''', or more tersely a '''lack-of-fit sum of squares''', is one of the components of a partition of the [[Sum of squares (statistics)|sum of squares]] in an [[analysis of variance]], used in the [[numerator]] in an [[F-test]] of the [[null hypothesis]] that says that a proposed model fits well.


A good way to boost general nutrіtion without abandoning your exіstіng seleсtions is always to add a very little anything healtҺy to the diѕhes. This is particularly helpful for picky eɑters or finicky children, but sneaking in healthy substances is perfect for you, way too. Examples of sneaking nutrients into every single day food items incorporate adding health proteins powder to milk, combining cɑuliflower intο mac and cheese, or using apple inc marinade іn place of essential oil in thе favored baƙing menu. You are going to all trу to eat a lot more nutritiously, and no one muѕt know.<br><br>Coach yourself to eat till yօu arе pleased, not jammed complete. You'll then remain from eating too muсh, and your physique can absorb the meals in correct ways. As soon as you commence to really feel complete, quit and prߋѵide your number of occasions to ascertаin if you are total.<br><br>Correct nourishment is key to your greater functioning body. A multi-supplement will ɦelp make certain уou are finding the proper quantities of essеntial vitamins and nutrients each day. Үou may try looking in a supplеment retail stoге to diѕcover what ѡorks well with you. For example, a female in her own fifties will neeԀ a multiple-vitamin supplement that is for miԁsection-aged  [http://city-Wiz.com/node/208851 vigrx Plus Deals] ladies. Consider these dietary supplements with Һ2o every single day.<br><br>To preserve a healthy protein intake although ѕlicing rear the quantity of red lean meats you taкe in, attempt to add Quinoa in your diet. Quinoa can be a rare foߋds that may be not beef but still consіsts ߋf amino acids. It has no gluten and iѕ also pасked with nutrіtional vіtamins. It preferences fantastic as well.<br><br>Ginger herb will help significantly in case ƴou aгe bоthered by motiоn dіsease. Supplements is one type ginger comes in. One hour аhead of the vacation, take in ginger herb, around 1,000 mg. Perform repeatedly in a few 60 minutеs timе intervals. Thіs should hеlp yߋu tгuly feel much bettеr and maintain ƴou notify during your journeʏ. Try out some ginger candy or some [http://www.Wikipedia.org/wiki/ginger+herb ginger herb] green tea.<br><br>Ingesting а balanced diet will give you a far healthier physical appearance and make you sense much more energetic. A terrific way to take care of oneself is as simple as lowering your enhanced sսgar intaҡe. Keep an eye out for poor food products and refreshments like juice and soda put. They are contribսtors in terms of becoming loaded with sugar. Avoid them. Eradicate sugar from your meals that you simply consume and you may see a maʝor ɗifference. Your health, power and looқ will manage to benefit from lowering or getting rid of these all kinds оf sugar.<br><br>Those who are affected by diabetes may have рroblems getting all of thеir nutritiouѕ requires achieved. Theѕe challеnges might be addressed by eating over a schedule as it helps to keep glucosе at standɑrd ranges. They might require a lot of vegetables and fresh fruits, products աith wholegгain and products from dairy fooԀ with reduced excеѕs fat. People suffering from diabetes have to eat at [http://thesaurus.com/browse/consistent+times consistent times] on a daily basis.<br><br>Zinc is a good supplemеnt to assist support a good entire body. Zinc can increase the fitness of your immune system, helping you feel much better rapidly and protecting уou upcoming аilments. Wonderful sources of zinc involve ƅerriеs, ѡheat or grain germ, peaches, and pumpκin plant sеeds. These food types also contain some anti-oxidants.<br><br>If you chooѕe sensibly, nuts may bе ɑn extremеly nutritious and whοlesome snaϲk selection. Normal walnuts havе a ѕuperіor fiber articleѕ and make a fantastic-flavored, crunchy snack food.<br><br>Mushrooms are a spесific thing wɦich sҺould invariably be made properly. Preparing food the mushгοoms nicely may help disintegгatе cаrcinogens which might be in the mushrooms. When you are conscious of your health and comply with general suggestions, you make sure you will hɑve a much simpler time shedding pounds and burning fat.<br><br>Refreshing beets certainly are a great meɑls to include in your diet plan, just be sure they aren't those that are deriνed from a can. Beets are loaded with fibег content and minerals, but if they are processed theѕe are loaded wіth sodium. A wise idea is usually to steam overcome gгeen veggies or put beetroots with your salad.<br><br>The value of nutrients іn suitable health ought not to ƅe overlooқed. If you ԝish to feel great and look very good, a healthy diet plan is vital. One wondeгful healtɦ insurance and nutrition suggestion would be to reduce refined ѕugars. Soft drinks are notoriously bad. These beverages are loaded witɦ sugar, that ought to be significantly restrictеd on your diet progrɑm. Oncе you reduce the ɑll kinds of sugar in уour diet, you will loose weіgҺt գuicker. The body will look better, and yоu'll feel much bettеr too.<br><br>Beets are an excellent supplement, but only new ones. New beets are high in fiber and vitamins, but the canned assortment is usually loaded with sea salt. You can water vapor clean beet plants and add these peߋple to salads.<br><br>Have an acquaintance or family membeг diet regime togetɦeг with you to рresent you with enthusiasm. Thiѕ  [http://clan.gentlemen-soldiers.net/index.php?mod=users&action=view&id=18149 Vigrx plus website] will either be a advisοr who may have already changed thіngs  [http://City-Wiz.com/node/210254 Vigrx plus retail Stores] and can present you with guidance, or a person who is in the identical level whеn you. Possessing somebody around that one could speak with will be verʏ convenient.<br><br>You may monitor everything you try to eat with an Online going on a diet monitor. Appear more carefully at what you're havіng if you gained weight. High cаrbs meals and high fat foods coսld force yߋu to gain pounds in case ƴou are not training. Οbѕerving your food consumption ԝill allow you to make better selections about diet program.<br><br>Nutrition is a very sορhisticateԀ topic. Mаking wholesome, conscious selections consistently can help you adhere to a healthy dіet strategy. With а little luck, the rеcommendations yоu pօssess gotten in this article wіll help you move ahead.
== Sketch of the idea ==
 
In order for the lack-of-fit sum of squares to differ from the [[Residual sum of squares|sum of squares of residuals]], there must be [[replication (statistics)|more than one]] value of the [[response variable]] for at least one of the values of the set of predictor variables. For example, consider fitting a line
 
: <math> y = \alpha x + \beta \, </math>
 
by the method of [[least squares]]. One takes as estimates of ''α'' and ''β'' the values that minimize the sum of squares of residuals, i.e., the sum of squares of the differences between the observed ''y''-value and the fitted ''y''-value.  To have a lack-of-fit sum of squares that differs from the residual sum of squares, one must observe more than one ''y''-value for each of one or more of the ''x''-values. One then partitions the "sum of squares due to error", i.e., the sum of squares of residuals, into two components:
 
: sum of squares due to error = (sum of squares due to "pure" error) + (sum of squares due to lack of fit).
 
The sum of squares due to "pure" error is the sum of squares of the differences between each observed ''y''-value and the average of all ''y''-values corresponding to the same ''x''-value.
 
The sum of squares due to lack of fit is the ''weighted'' sum of squares of differences between each average of ''y''-values corresponding to the same ''x''-value and the corresponding fitted ''y''-value, the weight in each case being simply the number of observed ''y''-values for that ''x''-value.<ref>{{cite book |first=Richard J. |last=Brook |first2=Gregory C. |last2=Arnold |title=Applied Regression Analysis and Experimental Design |publisher=[[CRC Press]] |location= |year=1985 |pages=48–49 |isbn=0824772520 }}</ref><ref>{{cite book |first=John |last=Neter |first2=Michael H. |last2=Kutner |first3=Christopher J. |last3=Nachstheim |first4=William |last4=Wasserman |title=Applied Linear Statistical Models |edition=Fourth |publisher=Irwin |location=Chicago |year=1996 |pages=121–122 |isbn=0256117365 }}</ref> Because it is a property of least squares regression that the vector whose components are "pure errors" and the vector of lack-of-fit components are orthogonal to each other, the following equality holds:
 
: <math>
\begin{align}
&\sum (\text{observed value} - \text{fitted value})^2 & & \text{(error)} \\
&\qquad = \sum (\text{observed value} - \text{local average})^2 & & \text{(pure error)} \\
& {} \qquad\qquad {} + \sum \text{weight}\times (\text{local average} - \text{fitted value})^2. & & \text{(lack of fit)}
\end{align}
</math>
 
Hence the residual sum of squares has been completely decomposed into two components.
 
== Mathematical details ==
Consider fitting a line with one predictor variable. Define ''i'' as an index of each of the ''n'' distinct ''x'' values, ''j'' as an index of the response variable observations for a given ''x'' value, and ''n''<sub>''i''</sub> as the number of ''y'' values associated with the ''i'' <sup>th</sup> ''x'' value. The value of each response variable observation can be represented by
 
: <math> Y_{ij} = \alpha x_i + \beta + \varepsilon_{ij},\qquad i = 1,\dots, n,\quad j = 1,\dots,n_i.</math>
 
Let
 
: <math> \widehat\alpha, \widehat\beta \,</math>
 
be the [[least squares]] estimates of the unobservable parameters ''α'' and ''β'' based on the observed values of ''x''<sub>&nbsp;''i''</sub> and ''Y''<sub>&nbsp;''i&nbsp;j''</sub>.
 
Let
 
: <math> \widehat Y_i = \widehat\alpha x_i + \widehat\beta \,</math>
 
be the fitted values of the response variable. Then
 
: <math> \widehat\varepsilon_{ij} = Y_{ij} - \widehat Y_i \,</math>
 
are the [[errors and residuals in statistics|residuals]], which are observable estimates of the unobservable values of the error term&nbsp;''ε''<sub>&nbsp;''ij''</sub>.  Because of the nature of the method of least squares, the whole vector of residuals, with  
 
:<math> N = \sum_{i=1}^n n_i </math>
 
scalar components, necessarily satisfies the two constraints
 
: <math> \sum_{i=1}^n \sum_{j=1}^{n_i} \widehat\varepsilon_{ij} = 0 \,</math>
 
: <math> \sum_{i=1}^n \left(x_i \sum_{j=1}^{n_i} \widehat\varepsilon_{ij} \right) = 0. \,</math>
 
It is thus constrained to lie in an (''N''&nbsp;&minus;&nbsp;2)-dimensional subspace of '''R'''<sup>&nbsp;''N''</sup>, i.e. there are ''N''&nbsp;&minus;&nbsp;2 "[[degrees of freedom (statistics)|degrees of freedom]] for error".
 
Now let
 
: <math> \overline{Y}_{i\bullet} = \frac{1}{n_i} \sum_{j=1}^{n_i} Y_{ij} </math>
 
be the average of all ''Y''-values associated with the ''i'' <sup>th</sup> ''x''-value.
 
We partition the sum of squares due to error into two components:
 
:<math>
\begin{align}
& \sum_{i=1}^n \sum_{j=1}^{n_i} \widehat\varepsilon_{ij}^{\,2}
= \sum_{i=1}^n \sum_{j=1}^{n_i} \left( Y_{ij} - \widehat Y_i \right)^2 \\
& = \underbrace{ \sum_{i=1}^n \sum_{j=1}^{n_i} \left(Y_{ij} - \overline Y_{i\bullet}\right)^2 }_\text{(sum of squares due to pure error)}
+ \underbrace{ \sum_{i=1}^n n_i \left( \overline Y_{i\bullet} - \widehat Y_i \right)^2. }_\text{(sum of squares due to lack of fit)}
\end{align}
</math>
 
== Probability distributions ==
 
=== Sums of squares ===
 
Suppose the [[errors and residuals in statistics|error terms]] ''ε''<sub>&nbsp;''i&nbsp;j''</sub> are [[statistical independence|independent]] and [[normal distribution|normally distributed]] with [[expected value]]&nbsp;0 and [[variance]]&nbsp;''σ''<sup>2</sup>. We treat ''x''<sub>&nbsp;''i''</sub> as constant rather than random. Then the response variables ''Y''<sub>&nbsp;''i&nbsp;j''</sub> are random only because the errors ''ε''<sub>&nbsp;''i&nbsp;j''</sub> are random.
 
It can be shown to follow that if the straight-line model is correct, then the '''sum of squares due to error''' divided by the error variance,
 
: <math> \frac{1}{\sigma^2}\sum_{i=1}^n \sum_{j=1}^{n_i} \widehat\varepsilon_{ij}^{\,2} </math>
 
has a [[chi-squared distribution]] with ''N''&nbsp;&minus;&nbsp;2 degrees of freedom.
 
Moreover, given the total number of observations ''N'', the number of levels of the independent variable ''n,'' and the number of parameters in the model ''p'':
 
* The sum of squares due to pure error, divided by the error variance ''σ''<sup>2</sup>, has a chi-squared distribution with ''N''&nbsp;&minus;&nbsp;''n'' degrees of freedom;
* The sum of squares due to lack of fit, divided by the error variance ''σ''<sup>2</sup>, has a chi-squared distribution with ''n''&nbsp;&minus;&nbsp;''p'' degrees of freedom (here ''p''&nbsp;=&nbsp;2 as there are two parameters in the straight-line model);
* The two sums of squares are probabilistically independent.
 
=== The test statistic ===
 
It then follows that the statistic
 
: <math>
\begin{align}
F & = \frac{ \text{lack-of-fit sum of squares} /\text{degrees of freedom} }{\text{pure-error sum of squares} / \text{degrees of freedom} } \\[8pt]
& = \frac{\left.\sum_{i=1}^n n_i \left( \overline Y_{i\bullet} - \widehat Y_i \right)^2\right/ (n-p)}{\left.\sum_{i=1}^n \sum_{j=1}^{n_i} \left(Y_{ij} - \overline Y_{i\bullet}\right)^2 \right/ (N - n)}
\end{align}
</math>
 
has an [[F-distribution]] with the corresponding number of degrees of freedom in the numerator and the denominator, provided that the model is correct. If the model is wrong, then the probability distribution of the denominator is still as stated above, and the numerator and denominator are still independent.  But the numerator then has a [[noncentral chi-squared distribution]], and consequently the quotient as a whole has a [[non-central F-distribution]].
 
One uses this F-statistic to test the [[null hypothesis]] that there is no lack of linear fit. Since the non-central F-distribution is [[stochastic order|stochastically larger]] than the (central) F-distribution, one rejects the null hypothesis if the F-statistic is larger than the critical F value. The critical value corresponds to the [[cumulative distribution function]] of the [[F distribution]] with ''x'' equal to the desired [[confidence level]], and degrees of freedom ''d''<sub>1</sub>&nbsp;=&nbsp;(''n''&nbsp;&minus;&nbsp;''p'') and ''d''<sub>2</sub>&nbsp;=&nbsp;(''N''&nbsp;&minus;&nbsp;''n''). This critical value can be calculated using online tools<ref name=soperds>{{cite web|last=Soper|first=D.S.|title=Critical F-value Calculator (Online Software)|url=http://www.danielsoper.com/statcalc3|work=Statistics Calculators|accessdate=19 April 2012}}</ref> or found in tables of statistical values.<ref name=lowryr>{{cite web|last=Lowry|first=Richard|title=VassarStats|url=http://vassarstats.net/textbook/apx_d.html|work=Concepts and Applications of Inferential Statistics|accessdate=19 April 2012}}</ref>
 
The assumptions of [[normal distribution]] of errors and [[independence (probability theory)|independence]] can be shown to entail that this [[lack-of-fit test]] is the [[likelihood-ratio test]] of this null hypothesis.
 
== See also ==
* [[Linear regression]]
 
== Notes ==
{{reflist}}
 
[[Category:Analysis of variance]]
[[Category:Regression analysis]]
[[Category:Design of experiments]]
[[Category:Least squares]]

Revision as of 16:04, 26 February 2013

In statistics, a sum of squares due to lack of fit, or more tersely a lack-of-fit sum of squares, is one of the components of a partition of the sum of squares in an analysis of variance, used in the numerator in an F-test of the null hypothesis that says that a proposed model fits well.

Sketch of the idea

In order for the lack-of-fit sum of squares to differ from the sum of squares of residuals, there must be more than one value of the response variable for at least one of the values of the set of predictor variables. For example, consider fitting a line

y=αx+β

by the method of least squares. One takes as estimates of α and β the values that minimize the sum of squares of residuals, i.e., the sum of squares of the differences between the observed y-value and the fitted y-value. To have a lack-of-fit sum of squares that differs from the residual sum of squares, one must observe more than one y-value for each of one or more of the x-values. One then partitions the "sum of squares due to error", i.e., the sum of squares of residuals, into two components:

sum of squares due to error = (sum of squares due to "pure" error) + (sum of squares due to lack of fit).

The sum of squares due to "pure" error is the sum of squares of the differences between each observed y-value and the average of all y-values corresponding to the same x-value.

The sum of squares due to lack of fit is the weighted sum of squares of differences between each average of y-values corresponding to the same x-value and the corresponding fitted y-value, the weight in each case being simply the number of observed y-values for that x-value.[1][2] Because it is a property of least squares regression that the vector whose components are "pure errors" and the vector of lack-of-fit components are orthogonal to each other, the following equality holds:

(observed valuefitted value)2(error)=(observed valuelocal average)2(pure error)+weight×(local averagefitted value)2.(lack of fit)

Hence the residual sum of squares has been completely decomposed into two components.

Mathematical details

Consider fitting a line with one predictor variable. Define i as an index of each of the n distinct x values, j as an index of the response variable observations for a given x value, and ni as the number of y values associated with the i th x value. The value of each response variable observation can be represented by

Yij=αxi+β+εij,i=1,,n,j=1,,ni.

Let

α^,β^

be the least squares estimates of the unobservable parameters α and β based on the observed values of x i and Y i j.

Let

Y^i=α^xi+β^

be the fitted values of the response variable. Then

ε^ij=YijY^i

are the residuals, which are observable estimates of the unobservable values of the error term ε ij. Because of the nature of the method of least squares, the whole vector of residuals, with

N=i=1nni

scalar components, necessarily satisfies the two constraints

i=1nj=1niε^ij=0
i=1n(xij=1niε^ij)=0.

It is thus constrained to lie in an (N − 2)-dimensional subspace of R N, i.e. there are N − 2 "degrees of freedom for error".

Now let

Yi=1nij=1niYij

be the average of all Y-values associated with the i th x-value.

We partition the sum of squares due to error into two components:

i=1nj=1niε^ij2=i=1nj=1ni(YijY^i)2=i=1nj=1ni(YijYi)2(sum of squares due to pure error)+i=1nni(YiY^i)2.(sum of squares due to lack of fit)

Probability distributions

Sums of squares

Suppose the error terms ε i j are independent and normally distributed with expected value 0 and variance σ2. We treat x i as constant rather than random. Then the response variables Y i j are random only because the errors ε i j are random.

It can be shown to follow that if the straight-line model is correct, then the sum of squares due to error divided by the error variance,

1σ2i=1nj=1niε^ij2

has a chi-squared distribution with N − 2 degrees of freedom.

Moreover, given the total number of observations N, the number of levels of the independent variable n, and the number of parameters in the model p:

  • The sum of squares due to pure error, divided by the error variance σ2, has a chi-squared distribution with N − n degrees of freedom;
  • The sum of squares due to lack of fit, divided by the error variance σ2, has a chi-squared distribution with n − p degrees of freedom (here p = 2 as there are two parameters in the straight-line model);
  • The two sums of squares are probabilistically independent.

The test statistic

It then follows that the statistic

F=lack-of-fit sum of squares/degrees of freedompure-error sum of squares/degrees of freedom=i=1nni(YiY^i)2/(np)i=1nj=1ni(YijYi)2/(Nn)

has an F-distribution with the corresponding number of degrees of freedom in the numerator and the denominator, provided that the model is correct. If the model is wrong, then the probability distribution of the denominator is still as stated above, and the numerator and denominator are still independent. But the numerator then has a noncentral chi-squared distribution, and consequently the quotient as a whole has a non-central F-distribution.

One uses this F-statistic to test the null hypothesis that there is no lack of linear fit. Since the non-central F-distribution is stochastically larger than the (central) F-distribution, one rejects the null hypothesis if the F-statistic is larger than the critical F value. The critical value corresponds to the cumulative distribution function of the F distribution with x equal to the desired confidence level, and degrees of freedom d1 = (n − p) and d2 = (N − n). This critical value can be calculated using online tools[3] or found in tables of statistical values.[4]

The assumptions of normal distribution of errors and independence can be shown to entail that this lack-of-fit test is the likelihood-ratio test of this null hypothesis.

See also

Notes

43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.

  1. 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.

    My blog: http://www.primaboinca.com/view_profile.php?userid=5889534
  2. 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.

    My blog: http://www.primaboinca.com/view_profile.php?userid=5889534
  3. Template:Cite web
  4. Template:Cite web