Drosophila is a model organism for genetics experiments because it reproduces quickly and when it reproduces it reproduces in large quantities. Also it is a model organism because so much is already known about it. The Drosophila egg is about half a millimeter long. One day after fertilization the embryo develops and hatches worm like larvae. The larva continuously eats and grows, and moults four times. The fourth time it moults it forms an immobile pupa and turns into the winged form. It hatches in about 4 days and is fertile in 12 hours.

Statistical analysis can be used to determine if there is a significant difference between two of group data sets. One way to do this is to use a Chi square. The Chi square test produces a number which you compare to a statistical Chi square number. Each of these statistical numbers has a significance level. Significance levels show you how likely a result is due to chance. The most common level, which is also used in this lab, is .95 which makes something good enough to be believed. This means that 95% of the time the findings will be true, and 5% of the time they will not. If the Chi square produces a number which is less then the statistical value, you accept your null hypothesis, meaning that there is no significant difference between the data. If the Chi square test produces a number higher than the statistical value then you must refute your null hypothesis, meaning that there is a significant difference in your data. The null hypothesis used is that the pattern for inheritance is autosomal. The expected phenotype ratios for mutant to wild type flies were 1:4. The expected genotype ratios were 1:4 homozygous dominant (ant+/ant+), 2:4 heterozygous (ant+/ant), and 1:4 homozygous recessive (ant/ant). The null hypothesis is that there is no significant difference between this expected data and the data we collected.
II. Results

...Chisquare test for independence of two attributes. Suppose N observations are considered and classified according two characteristics say A and B. We may be interested to test whether the two characteristics are independent. In such a case, we can use Chisquare test for independence of two attributes.
The example considered above testing for independence of success in the English test vis a vis immigrant status is a case fit for analysis using this test.
This lesson explains how to conduct a chi-square test for independence. The test is applied when you have two categorical variables from a single population. It is used to determine whether there is a significant association between the two variables.
For example, in an election survey, voters might be classified by gender (male or female) and voting preference (Democrat, Republican, or Independent). We could use a chi-square test for independence to determine whether gender is related to voting preference. The sample problem at the end of the lesson considers this example.
When to Use Chi-Square Test for Independence
The test procedure described in this lesson is appropriate when the following conditions are met:
* The sampling method is simple random sampling.
* Each population is at least 10 times as large as its respective sample.
* The variables...

...2.3. The Chi-Square Distribution
One of the most important special cases of the gamma distribution is the chi-square distribution because the sum of the squares of independent normal random variables with mean zero and standard deviation one has a chi-square distribution. This section collects some basic properties of chi-square random variables, all of which are well known; see Hogg and Tanis [6].
A random variable X has a chi-square distribution with n degrees of freedom if it is a gamma random variable with parameters m = n/2 and = 2, i.e X ~ (n/2,2). Therefore, its probability density function (pdf) has the form
(1) f(t) = f(t; n) =
In this case we shall say X is a chi-square random variable with n degrees of freedom and write X ~ (n). Usually n is assumed to be an integer, but we only assume n > 0.
Proposition 1. If X has a gamma distribution with parameters m and then 2X/ has a chi-square distribution with 2m degrees of freedom.
Proof. By Proposition 5 in section 2.2 the random variable X has a gamma distribution with parameters m and 2, i.e X ~ (m,2) = ((2m)/2,2). The proposition follows from this.
Proposition 2. If X has a chi-square distribution with n degrees of freedom, then the...

...Chi-square requires that you use numerical values, not percentages or ratios.
Then calculate 2 using this formula, as shown in Table B.1. Note that we get a value of 2.668 for 2. But what does this number mean? Here's how to interpret the 2 value:
1. Determine degrees of freedom (df). Degrees of freedom can be calculated as the number of categories in the problem minus 1. In our example, there are two categories (green and yellow); therefore, there is I degree of freedom.
2. Determine a relative standard to serve as the basis for accepting or rejecting the hypothesis. The relative standard commonly used in biological research is p >0.05. The p value is the probability that the deviation of the observed from that expected is due to chance alone (no other forces acting). In this case, using p >0.05, you would expect any deviation to be due to chance alone 5% of the time or less.
3. Refer to a chi-square distribution table (Table B.2). Using the appropriate degrees of 'freedom, locate the value closest to your calculated chi-square in the table. Determine the closestp (probability) value associated with your chi-square and degrees of freedom. In this case (2=2.668), the p value is about 0.10, which means that there is a 10% probability that any deviation from expected results is due to chance only. Based on our standard p > 0.05, this is within the...

...A chi-squared test, also referred to as chi-square test or χw² test, is any statistical hypothesis test in which the sampling distribution of the test statistic is a chi-squared distribution when the null hypothesis is true. Also considered a chi-squared test is a test in which this is asymptotically true, meaning that the sampling distribution (if the null hypothesis is true) can be made to approximate achi-squared distribution as closely as desired by making the sample size large enough.
Some examples of chi-squared tests where the chi-squared distribution is only approximately valid:
Pearson's chi-squared test, also known as the chi-squared goodness-of-fit test or chi-squared test for independence. When the chi-squared test is mentioned without any modifiers or without other precluding context, this test is usually meant (for an exact test used in place of χ², see Fisher's exact test).
Yates's correction for continuity, also known as Yates' chi-squared test.
Cochran–Mantel–Haenszel chi-squared test.
McNemar's test, used in certain 2 × 2 tables with pairing
Tukey's test of additivity
The portmanteau test in time-series analysis, testing for the presence of autocorrelation
Likelihood-ratio tests in general...

...CHI-SQUARE TEST (χ²):
Chi-square is a statistical test commonly used to compare observed data with data we would expect to obtain according to a specific hypothesis. For example, if, according to Mendel's laws, you expected 10 of 20 offspring from a cross to be male and the actual observed number was 8 males, then you might want to know about the "goodness to fit" between the observed and expected. Were the deviations (differences between observed and expected) the result of chance, or were they due to other factors. How much deviation can occur before you, the investigator, must conclude that something other than chance is at work, causing the observed to differ from the expected. The chi-square test is always testing what scientists call the null hypothesis, which states that there is no significant difference between the expected and observed result.
The formula for calculating chi-square (χ²) is:
2= (o-e) ²/e
That is, chi-square is the sum of the squared difference between observed (o) and the expected (e) data (or the deviation, d), divided by the expected data in all possible categories.
INTERPRETATION OF CHI-SQUARE TEST
1. Determine degrees of freedom (DF). Degrees of freedom can be calculated as the number of categories in the problem minus 1.
2. Determine a relative...

...Testing statisticalsignificance is an excellent way to identify probably relevance between a total data set mean/sigma and a smaller sample data set mean/sigma, otherwise known as a population mean/sigma and sample data set mean/sigma. This classification of testing is also very useful in proving probable relevance between data samples. Although testing statisticalsignificance is not a 100% fool proof, if testing to the 95% probability on two data sets the statistical probability is .25% chance that the results of the two samplings was due to chance. When testing at this level of probability and with a data set size that is big enough, a level of certainty can be created to help determine if further investigation is warranted. The following is a problem is used to illustrate how testing statisticalsignificance paints a more descriptive picture of data set relationships.
Sam Sleep researcher hypothesizes that people who are allowed to sleep for only four hours will score significantly lower than people who are allowed to sleep for eight hours on a management ability test. He brings sixteen participants into his sleep lab and randomly assigns them to one of two groups. In one group he has participants sleep for eight hours and in the other group he has them sleep for four. The next morning he administers the SMAT (Sam's Management Ability Test) to all participants....

...Counseling Psychology 1983, Vol. 30, No, 3,459-463
Copyright 1983 by the American Psychological Association, Inc.
StatisticalSignificance, Power, and Effect Size: A Response to the Reexamination of Reviewer Bias
Bruce E. Wampold
Department of Educational Psychology University of Utah
Michael J. Furlong and Donald R. Atkinson
Graduate School of Education University of California, Santa Barbara
In responding to our study of the influence thatstatisticalsignificance has on reviewers' recommendations for the acceptance or rejection of a manuscript for publication (Atkinson, Furlong, & Wampold, 1982), Fagley and McKinney (1983) argue that reviewers were justified in rejecting the bogus study when nonsignificant results were reported due to what Fagley and McKinney indicate is the low power of the bogus study. The concept of power is discussed in the present article to show that the bogus study actually had adequate power to detect a large experimental effect and that attempts to design studies sensitive to small experimental effects are typically impractical for counseling research when complex designs are used. In addition, it is argued that the power of the bogus study compares favorably to that of research published in the Journal of Counseling Psychology at the time our study was completed. Finally, the importance of considering statisticalsignificance, power, and effect...

...Effects of Alcohol on Drosophila Behavior
Your Name
Bio1100
Instructor: Dr. Gary Bulla
Section 6
11/14/02
Notes: Title should be descriptive but concise, containing some key words
Title page should include only the information shown here
Use no smaller than 12 pt font
Capitalize words in title (except “of”, “the”, “in”, ……..)
ABSTRACT (Capitalized with bold face, underline or all caps)
A simple summary of the important points of the paper. Usually a 1-2 paragraphs. It
includes the hypotheses, a brief description of the types of experiments done, the major
results found and conclusions.
Example
Drosophila (fruit fly) is a well-studied organism which has been used as model to
understand higher-order organisms both at the behavior and m
olecular levels. Because
this organism reproduces rapidly a number of mutants phenotypes exit, Drosophila has
proven to be very useful tool. Because high alcohol levels has been shown to have an
effect of development of the human fetus, we asked whether similar effects are observed
and to what extent in the Drosophila. In order to test this hypothesis, Drosophila were
treated with varying levels of alcohol at various times during embryogenesis. Our results
demonstrate that as little as 10% alcohol treatme nt has profound effects on Drosophila
development. In addition, embryogenesis was delayed and abnormal wing development
was...

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