The test null hypothesis is:
H _{ 0 } : Âµ1 = Âµ2 = Âµ3,
and the alternative:
H _{ a } : at least two of the means are not equal.
Using the
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JavaScripts, we obtain the needed information in constructing the following ANOVA table:
Conclusion: The p-value is P= 0.006, indicating a strong evidence against the null hypothesis. The means of the populations are not equal. Here, one may conclude that person who has consumed more than certain level of alcohol commits more driving errors.
A"block design sampling" implies studying more than two dependent populations. For testing the equality of means of more than two populations based on block design sampling, you may use Two-Way ANOVA Test JavaScript. In the case of having block design data with replications, use Two-Way ANOVA with Replications JavaScript to obtain the needed information for constructing the ANOVA tables.
An Application: Suppose we wish to test the null hypothesis
That is, all three population proportions are almost identical. The sample data from each of the three populations are given in the following table:
The Chi-square statistic is 8.95 with d.f. = (3-1)(3-1) = 4. The p-value is equal to 0.062, indicating that there is moderate evidence against the null hypothesis that the three populations are statistically identical.
You might like to use Testing Proportions to perform this test.
Prior to applying the K-S test it is necessary to arrange each of the two sample observations in a frequency table. The frequency table must have a common classification. Therefore the test is based on the frequency table, which belongs to the family of distribution-free tests.
The K-S Test process is as follows:
K-S statistic = D = Maximum | F1 _{ i } - F2 _{ i } |, for all i = 1, 2, .., k.
The above process is depicted in the following figure.
The critical values of K-S statistic can be found at Computers and Computational Statistics with Applications
An Application: The daily sales of the two subsidiaries of The PC Accessories Company are shown in the following table, with n1 = 44, and n2 = 54:
has a Chi-square distribution with d.f. = k-1. Where i is the count's number, N is its counts, and N = N/k.
One may extend this useful test to where the duration of obtaining the i count is t. Then the above test statistic becomes:
and has a Chi-square distribution with d.f. = k-1, where i is the count's number, N is its counts, and N = N/t.
You might like to use the Compatibility of Multi-Counts JavaScript to check your computations, and to perform some numerical experimentation for a deeper understanding of the concepts.
Like any statistical test procedures, the Chi-square based testing must meet certain necessary conditions to apply; otherwise, any obtained conclusion might be wrong or misleading. This is true in particular for using the Chi-square-based test for cross-tabulated data.Necessary conditions for the Chi-square based tests for crosstable data are:
Suppose the monthly number of accidents reported in a factory in three eight-hour shifts is 1, 7, and 7, respectively. Are the working conditions and the exposure to risk similar for all shifts? Clearly, the answer must be, No they are not. However, applying the goodness-of-fit, at 0.05, under the null hypothesis that there are no differences in the number of accidents in three shifts, one expects 5, 5, and 5 accidents in each shift. The Chi-square test statistic is:
However, since = 5.99, there is no reason to reject that there is no difference, which is a very strange conclusion. What is wrong with this application?
You might like to use this JavaScript to verify your computation.
Suppose a population has a normal distribution. The manager is to test a specific claim made about the quality of the population by way of testing its variance . Among three possible scenarios, the interesting case is in testing the following null hypothesis based on a set of n random sample observations:: Variation is about the claimed value. : The variation is more than what is claimed, indicating the quality is much lower than expected.
Upon computing the estimated variance S based on n observations, then the statistic:
has a Chi-square distribution with degree of freedom = n - 1. This statistic is then used for testing the above null hypothesis.
You might like to use
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JavaScript to check your computations.
is used to test if k samples have equal variances. It compares the Geometric Mean of the group variances to the arithmetic mean; therefore, it is a Chi-square statistic with (k-1) degrees of freedom, where k is the number of categories in the independent variable. The test is sensitive to departures from normality. The sample sizes do not have to be equal but each must be at least 6. Just like the two population t-test, ANOVA can go wrong when the equality of variances condition is not met.
The Bartlett test statistic is designed to test for equality of variances across groups against the alternative that variances are unequal for at least two groups. Formally,
: All variances are almost equal.
The test statistic:
In the above, S is the variance of the ith group, n is the sample size of the i group, k is the number of groups, and S is the pooled variance. The pooled variance is a weighted average of the group variances and is defined as:
and
You might like to use the
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JavaScript to check your computations, and to perform some numerical experimentation for a deeper understanding of the concepts.
For 3 or more populations, there is a practical rule known as the"Rule of 2". According to this rule, one divides the highest variance of a sample by the lowest variance of the other sample. Given that the sample sizes are almost the same, and the value of this division is less than 2, then, the variations of the populations are almost the same.
Consider the following three random samples from three populations, P1, P2, P3:
With an F = 4.38 and a p-value of 0.023, we reject the null at = 0.05. This is not good news, since ANOVA, like the two-sample t-test, can go wrong when the equality of variances condition is not met.
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Where the Fisher Z-transformation is
Under the null hypothesis:
: All correlation coefficients are almost equal.
The test statistic has (k-1) degrees of freedom, where k is the number of populations.
Consider the following correlation coefficients obtained by random sampling form ten independent populations.
Using the above formula -statistic = 19.916, that has a p-value of 0.02. Therefore, there is moderate evidence against the null hypothesis.
In such a case, one may omit a few outliers from the group, then use the Test for Equality of Several Correlation Coefficients JavaScript. Repeat this process until a possible homogeneous sub-group may emerge.
You might need to use
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JavaScript at the design stage of your statistical investigation in decision making with specific subjective requirements.
The following contains the main essential steps during modeling and analysis of regression model building, presented in the context of an applied numerical example.
A taxicab company manager believes that the monthly repair costs (Y) of cabs are related to age (X) of the cabs. Five cabs are selected randomly and from their records we obtained the following data: (x, y) = {(2, 2), (3, 5), (4, 7), (5, 10), (6, 11)}.
The first step in constructing a simple linear regression model is to draw a scattered diagram, as shown in the following figure for our numerical example:
The manager of the first branch is claiming that"since the daily sales are random phenomena, my overall performance is as good as the other manager's performance." In other words:
In the five species with 2n = 24, pairs 1, 2, and 8 were metacentric, and pair 3 submetacentric. Pair 4 was subtelocentric in B . calcarata , B . cf. lanciformis and B . raniceps , and submetacentric in B . almendarizae and B . heilprini . In B . almendarizae , B . calcarata and B . raniceps pairs 5 and 6 were submetacentric, whereas these pairs were respectively metacentric and submetacentric in B . heilprini and, submetacentric and subtelocentric in B . cf. lanciformis . In B . calcarata and B . raniceps , pair 7 was metacentric, and in the other species, B . almendarizae , B . heilprini and B . cf. lanciformis were submetacentric. Moreover, pair 9 was metacentric in B . calcarata and B . cf. lanciformis , but submetacentric in B . almendarizae , B . heilprini , and B . raniceps . Pairs 10 and 11 were metacentric in B . almendarizae , B . calcarata , and B . heilprini and B . raniceps , and submetacentric in B . cf. lanciformis , the latter being a distinctive feature of this species. Finally, the smallest pair 12 was metacentric in B . almendarizae , B . calcarata and B . cf. lanciformis , and submetacentric in B . heilprini and B . raniceps .
NORs were located terminally on pair 8p in
B
. cf.
alfaroi
,
B
.
leucocheila
, and
B
.
multifasciata
, on pair 11q in
B
. cf.
lanciformis
and
B
.
raniceps
, and on pair 12 in
B
.
almendarizae
(
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).
In
Boana multifasciata
C-bands occured only on pairs 1, 7, 8, and 9; interstitially on 1p and 7p7q, and pericentromerically on 8p and 9q (
Fig 6A
).
Boana almendarizae
and
B
.
calcarata
showed a similar heterochromatic pattern restricted exclusively to all centromeres (
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), whereas in
B
.
raniceps
C+ heterochromatin was pericentromeric and interstitial on pairs 7p and 11q, respectively (
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).
A. B . multifasciata . B. B . almendarizae . C . B . calcarata . D. B . raniceps . E. B . cinerascens . F. B . punctata . G. B . boans . H . B . cf. semilineata . I. B . wavrini . J. B . pellucens .
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Cryptography can be a hard subject to understand. It’s full of mathematical proofs. But unless you are actually developing cryptographic systems, much of that complexity is not necessary to understand what is going on at a high level.
If you opened this article hoping to create the next HTTPS protocol, I’m sorry to say that pigeons won’t be enough. Otherwise, brew some coffee and enjoy the article.
Any activity you do on the Internet (reading this article, buying stuff on Amazon, uploading cat pictures) comes down to sending and receiving messages to and from a server.
This can be a bit abstract so let’s imagine that those messages were delivered by carrier pigeons . I know that this may seem very arbitrary, but trust me HTTPS works the same way, albeit a lot faster.
Also instead of talking about servers, clients and hackers, we will talk about Alice, Bob and Mallory. If this isn’t your first time trying to understand cryptographic concepts you will recognize those names, because they are widely used in technical literature.
If Alice wants to send a message to Bob, she attaches the message on the carrier pigeon’s leg and sends it to Bob. Bob receives the message, reads it and it’s all is good.
But what if Mallory intercepted Alice’s pigeon in flight and changed the message? Bob would have no way of knowing that the message that was sent by Alice was modified in transit.
This is how HTTP works. Pretty scary right? I wouldn’t send my bank credentials over HTTP and neither should you.
Now what if Alice and Bob are very crafty. They agree that they will write their messages using a secret code. They will shift each letter by 3 positions in the alphabet. For example D → A, E → B, F → C. The plain text message “secret message” would be “pbzobq jbppxdb”.
Now if Mallory intercepts the pigeon she won’t be able to change the message into something meaningful nor understand what it says, because she doesn’t know the code. But Bob can simply apply the code in reverse and decrypt the message where A → D, B → E, C → F. The cipher text “pbzobq jbppxdb” would be decrypted back to “secret message”.
Success!
This is called symmetric key cryptography , because if you know how to encrypt a message you also know how to decrypt it.
The code I described above is commonly known as the Caesar cipher . In real life, we use fancier and more complex codes, but the main idea is the same.
Symmetric key cryptography is very secure if no one apart from the sender and receiver know what key was used. In the Caesar cipher, the key is an offset of how many letters we shift each letter by. In our example we used an offset of 3, but could have also used 4 or 12.
The issue is that if Alice and Bob don’t meet before starting to send messages with the pigeon, they would have no way to establish a key securely. If they send the key in the message itself, Mallory would intercept the message and discover the key. This would allow Mallory to then read or change the message as she wishes before and after Alice and Bob start to encrypt their messages.
This is the typical example of a Man in the Middle Attack and the only way to avoid it is to change the encryption system all together.
So Alice and Bob come up with an even better system. When Bob wants to send Alice a message she will follow the procedure below:
This way Mallory can’t change the message by intercepting the pigeon, because she doesn’t have the key. The same process is followed when Alice wants to send Bob a message.
Alice and Bob just used what is commonly known as asymmetric key cryptography .It’s called asymmetric, because even if you can encrypt a message (lock the box) you can’t decrypt it (open a closed box).In technical speech the box is known as the public key and the key to open it is known as the private key .
If you paid attention you may have noticed that we still have a problem. When Bob receives that open box how can he be sure that it came from Alice and that Mallory didn’t intercept the pigeon and changed the box with one she has the key to?
Alice decides that she will sign the box, this way when Bob receives the box he checks the signature and knows that it was Alice who sent the box.
Some of you may be thinking, how would Bob identify Alice’s signature in the first place? Good question. Alice and Bob had this problem too, so they decided that, instead of Alice signing the box, Ted will sign the box.
Who is Ted? Ted is a very famous, well known and trustworthy guy. Ted gave his signature to everyone and everybody trusts that he will only sign boxes for legitimate people.
Ted will only sign an Alice box if he’s sure that the one asking for the signature is Alice. So Mallory cannot get an Alice box signed by Ted on behalf of her as Bob will know that the box is a fraud because Ted only signs boxes for people after verifying their identity.
Ted in technical terms is commonly referred to as a Certification Authority and the browser you are reading this article with comes packaged with the signatures of various Certification Authorities.
So when you connect to a website for the first time you trust its box because you trust Ted and Ted tells you that the box is legitimate.
Alice and Bob now have a reliable system to communicate, but they realize that pigeons carrying boxes are slower than the ones carrying only the message.
They decide that they will use the box method (asymmetric cryptography) only to choose a key to encrypt the message using symmetric cryptography with (remember the Caesar cipher?).
This way they get the best of both worlds. The reliability of asymmetric cryptography and the efficiency of symmetric cryptography.
In the real world there aren’t slow pigeons, but nonetheless encrypting messages using asymmetric cryptography is slower than using symmetric cryptography, so we only use it to exchange the encryption keys.
Now you know how HTTPS works and your coffee should also be ready. Go drink it you deserved it 😉
From a quick cheer to a standing ovation, clap to show how much you enjoyed this story.
Until now, successful gene identification by homozygosity mapping has been mostly based on consanguineous kindred that have multiple affected individuals. Close consanguinity of such kindred generate homozygous segments broad enough to be detected by SNP marker sets of low density (e.g., 50 K marker set). The presence of multiple siblings or cousins helps refine the candidate region to fewer cZLR peaks that overlap between affected siblings or cousins. For example, the number of 9, 12 and 10 cZLR peaks in siblings A1538-1, -2, and -3, respectively is reduced to 3, 4, and 5 cZLR peaks if each pair of siblings is evaluated together and to 2 cZLR peaks if all three sibs are evaluated together ( Figure 3 ). However, a severe limitation to successful positional cloning of recessive disease genes by homozygosity mapping lies in the fact that sufficient numbers of consanguineous kindred with multiple affecteds and mutation of the same gene are very hard to ascertain. In our 16-year experience in worldwide recruitment of 1,069 patients with NPHP, non-consanguineous single cases (n = 918) were 6-times as frequent as consanguineous cases (n = 151). In 595 patients with SRNS and mutations in recessive disease-causing genes the ascertainment of non-consanguineous single cases (n = 512) was also about 6-times as high as consanguineous cases (n = 83). Therefore, the ability to employ individuals from outbred populations in homozygosity mapping should greatly accelerate gene discovery.
Another strong limitation to gene identification is imposed by the experience that studies of consanguineous kindred result in segments of homozygosity that are too numerous and too large and contain too many positional candidates for efficient gene identification by mutation analysis. For instance, in our worldwide cohort individuals from consanguineous background had 4–60 cZLR peaks ( Figure 3 ). These homozygous segments contained hundreds of positional candidate genes. Because of both limitations we wanted to evaluate whether gene identification by homozygosity mapping is possible in single individuals from outbred populations. Using a medium resolution SNP marker set (250 K) we were able to detect homozygous segments that contain the disease-causing mutations in single individuals from outbred populations as shown in Figure 3 :
With the exception of 4 families (A1218, A646, F1158, and A825), all other 23 outbred families exhibited 4 or less cZLR peaks ( Figure 3 ). In 2 of the 3 outbred families with 4 cZLR peaks, calculation of both affected individuals together reduced the number of cZLR peak from 4 and 3 cZLR peaks to 2 cZLR peaks (A237) and from 4 and 2 peaks to 1 cZLR peak (F60-61). Four outbred families had 3 cZLR peaks (F138, F456, F50, and F1183). Seven outbred individuals exhibited only 2 cZLR peaks (A762, F53, A1298, A567, A7, F54, and A1338) ( Figure 3 ). In these individuals the cZLR peak that contained the homozygous gene mutation had a median size of 8.3 Mb (range 2.4 - 40.1 Mb) ( Figure 3, insert ).