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From Wikipedia

Contingency table

In statistics, a contingency table (also referred to as cross tabulationor cross tab) is often used to record and analyze the relation between two or morecategorical variables. It displays the (multivariate) frequency distribution of the variables in a matrix format.

The term contingency table was first used by Karl Pearson in "On the Theory of Contingency and Its Relation to Association and Normal Correlation", part of the Drapers' Company Research Memoirs Biometric Series I published in 1904.

Example

Suppose that we have two variables, sex (male or female) and handedness (right- or left-handed). Further suppose that 100 individuals are randomly sampled from a very large population as part of a study of sex differences in handedness. A contingency table can be created to display the numbers of individuals who are male and right-handed, male and left-handed, female and right-handed, and female and left-handed. Such a contingency table is shown below.

The numbers of the males, females, and right- and left-handed individuals are called marginal totals. The grand total, i.e., the total number of individuals represented in the contingency table, is the number in the bottom right corner.

The table allows us to see at a glance that the proportion of men who are right-handed is about the same as the proportion of women who are right-handed although the proportions are not identical. The significance of the difference between the two proportions can be assessed with a variety of statistical tests including Pearson's chi-square test, the G-test, Fisher's exact test, and Barnard's test, provided the entries in the table represent individuals randomly sampled from the population about which we want to draw a conclusion. If the proportions of individuals in the different columns vary significantly between rows (or vice versa), we say that there is a contingency between the two variables. In other words, the two variables are not independent. If there is no contingency, we say that the two variables are independent.

The example above is the simplest kind of contingency table, a table in which each variable has only two levels; this is called a 2 x 2 contingency table. In principle, any number of rows and columns may be used. There may also be more than two variables, but higher order contingency tables are difficult to represent on paper. The relation between ordinal variables, or between ordinal and categorical variables, may also be represented in contingency tables, although such a practice is rare.

Measures of association

The degree of association between the two variables can be assessed by a number of coefficients: the simplest is the phi coefficient defined by

\phi=\sqrt{\frac{\chi^2}{N}},

where χ2 is derived from Pearson's chi-square test, and N is the grand total of observations. φ varies from 0 (corresponding to no association between the variables) to 1 or -1 (complete association or complete inverse association). This coefficient can only be calculated for frequency data represented in 2 x 2 tables. φ can reach a minimum value -1.00 and a maximum value of 1.00 only when every marginal proportion is equal to .50 (and two diagonal cells are empty). Otherwise, the phi coefficient cannot reach those minimal and maximal values.

Alternatives include the tetrachoric correlation coefficient (also only applicable to 2 × 2 tables), the contingency coefficientC, and Cramér'sV.

C suffers from the disadvantage that it does not reach a maximum of 1 or the minimum of -1; the highest it can reach in a 2 x 2 table is .707; the maximum it can reach in a 4 x 4 table is 0.870. It can reach values closer to 1 in contingency tables with more categories. It should, therefore, not be used to compare associations among tables with different numbers of categories. Moreover, it does not apply to asymmetrical tables (those where the numbers of row and columns are not equal).

The formulae for the C and V coefficients are:

C=\sqrt{\frac{\chi^2}{N+\chi^2}} and
V=\sqrt{\frac{\chi^2}{N(k-1)}},

k being the number of rows or the number of columns, whichever is less.

C can be adjusted so it reaches a maximum of 1 when there is complete association in a table of any number of rows and columns by dividing C by \sqrt{\frac{k-1}{k}} (recall that C only applies to tables in which the number of rows is equal to the number of columns and therefore equal to k).

The tetrachoric correlation coefficient assumes that the variable underlying each dichotomous measure is normally distributed. The tetrachoric correlation coefficient provides "a convenient measure of [the Pearson product-moment] correlation when graduated measurements have been reduced to two categories." The tetrachoric correlation should not be confused with the Pearson product-moment correlation coefficient computed by assigning, say, values 0 and 1 to represent the two levels of each variable (which is mathematically equivalent to the phi coefficient). An extension of the tetrachoric correlation to tables involving variables with more than two levels is the polychoric correlation coefficient.

The Lambda coefficientis a measure the strength of association of the cross tabulations when the variables are measured at thenominal level. Values range from 0 (no association) to 1 (the theoretical maximum possible association). Asymmetric lambdameasures the percentage improvement in predicting the dependent variable.

Inverse trigonometric functions

In mathematics, the inverse trigonometric functions or cyclometric functions are the inverse functions of the trigonometric functions, though they do not meet the official definition for inverse functions as their ranges are subsets of the domains of the original functions. Since none of the six trigonometric functions are one-to-one (by failing the horizontal line test), they must be restricted in order to have inverse functions.

For example, just as the square root function y = \sqrt{x} is defined such that y2 = x, the function y = arcsin(x) is defined so that sin(y) = x. There are multiple numbers y such that sin(y) = x; for example, sin(0) = 0, but also sin(π) = 0, sin(2π) = 0, etc. It follows that the arcsine function is multivalued: arcsin(0) = 0, but also arcsin(0) = π, arcsin(0) = 2π, etc. When only one value is desired, the function may be restricted to its principal branch. With this restriction, for each x in the domain the expression arcsin(x) will evaluate only to a single value, called its principal value. These properties apply to all the inverse trigonometric functions.

The principal inverses are listed in the following table.

If x is allowed to be a complex number, then the range of y applies only to its real part.

The notations sin−1, cos−1, etc. are often used for arcsin, arccos, etc., but this convention logically conflicts with the common semantics for expressions like sin2(x), which refer to numeric power rather than function composition, and therefore may result in confusion between multiplicative inverse and compositional inverse.

In computer programming languages the functions arcsin, arccos, arctan, are usually called asin, acos, atan. Many programming languages also provide the two-argument atan2 function, which computes the arctangent of y / x given y and x, but with a range of (−Ï€, Ï€].

Relationships among the inverse trigonometric functions

Complementary angles:

\arccos x = \frac{\pi}{2} - \arcsin x
\arccot x = \frac{\pi}{2} - \arctan x
\arccsc x = \frac{\pi}{2} - \arcsec x

Negative arguments:

\arcsin (-x) = - \arcsin x \!
\arccos (-x) = \pi - \arccos x \!
\arctan (-x) = - \arctan x \!
\arccot (-x) = \pi - \arccot x \!
\arcsec (-x) = \pi - \arcsec x \!
\arccsc (-x) = - \arccsc x \!

Reciprocal arguments:

\arccos x^{-1} \,= \arcsec x \,
\arcsin x^{-1} \,= \arccsc x \,
\arctan x^{-1} = \tfrac{1}{2}\pi - \arctan x =\arccot x,\text{ if }x > 0 \,
\arctan x^{-1} = -\tfrac{1}{2}\pi - \arctan x = -\pi + \arccot x,\text{ if }x < 0 \,
\arccot x^{-1} = \tfrac{1}{2}\pi - \arccot x =\arctan x,\text{ if }x > 0 \,
\arccot x^{-1} = \tfrac{3}{2}\pi - \arccot x = \pi + \arctan x\text{ if }x < 0 \,
\arcsec x^{-1} = \arccos x \,
\arccsc x^{-1} = \arcsin x \,

If you only have a fragment of a sine table:

\arccos x = \arcsin \sqrt{1-x^2},\text{ if }0 \leq x \leq 1
\arctan x = \arcsin \frac{x}{\sqrt{x^2+1}}

Whenever the square root of a complex number is used here, we choose the root with the positive real part (or positive imaginary part if the square was negative real).

From the half-angle formula \tan \frac{\theta}{2} = \frac{\sin \theta}{1+\cos \theta} , we get:

\arcsin x = 2 \arctan \frac{x}{1+\sqrt{1-x^2}}
\arccos x = 2 \arctan \frac{\sqrt{1-x^2}}{1+x},\text{ if }-1 < x \leq +1
\arctan x = 2 \arctan \frac{x}{1+\sqrt{1+x^2}}

Relationships between trigonometric functions and inverse trigonometric functions

\sin (\arccos x) = \cos(\arcsin x) = \sqrt{1-x^2}
\sin (\arctan x) = \frac{x}{\sqrt{1+x^2}}
\cos (\arctan x) = \frac{1}{\sqrt{1+x^2}}
\tan (\arcsin x) = \frac{x}{\sqrt{1-x^2}}
\tan (\arccos x) = \frac{\sqrt{1-x^2}}{x}

General solutions

Each of the trigonometric functions is periodic in the real part of its argument, running through all its values twice in each interval of 2π. Sine and cosecant begin their period at 2πk− π/2 (where k is an integer), finish it at 2πk + π/2, and then reverse themselves over 2πk + π/2 to 2πk + 3π/2. Cosine and secant begin their period at 2πk, finish it at 2πk + π, and then reverse themselves over 2πk + π to 2πk + 2π. Tangent begins its period at 2πk− π/2, finishes it at 2πk + π/2, and then repeats it (forward) over 2πk + π/2 to 2πk + 3π/2. Cotangent begins its period at 2πk, finishes it at 2πk + π, and then repeats it (forward) over 2πk + π to 2πk + 2π.

This periodicity is reflected in the general inverses where k is some integer:

\sin(y) = x \ \Leftrightarrow\ y = \arcsin(x) + 2k\pi \text{ or } y = \pi - \arcsin(x) + 2k\pi
\cos(y) = x \ \Leftrightarrow\ y = \arccos(x) + 2k\pi \text{ or } y = 2\pi - \arccos(x) + 2k\pi
\tan(y) = x \ \Leftrightarrow\ y = \arctan(x) + k\pi
\cot(y) = x \ \Leftrightarrow\ y = \arccot(x) + k\pi
\sec(y) = x \ \Leftrightarrow\ y = \arcsec(x) + 2k\pi \text{ or } y = 2\pi - \arcsec (x) + 2k\pi
\csc(y) = x \ \Leftrightarrow\ y = \arccsc(x) + 2k\pi \text{ or } y = \pi - \arccsc(x) + 2k\pi

Derivatives of inverse trigonometric functions

Simple derivatives for real and complex values of x are as follows:

\begin{align} \frac{d}{dx} \arcsin x & {}= \frac{1}{\sqrt{1-x^2}}\\ \frac{d}{dx} \arccos x & {}= \frac{-1}{\sqrt{1-x^2}}\\ \frac{d}{dx} \arctan x & {}= \frac{1}{1+x^2}\\ \frac{d}{dx} \arccot x & {}= \frac{-1}{1+x^2}\\ \frac{d}{dx} \arcsec x & {}= \frac{1}{x\,\sqrt{x^2-1}}\\ \frac{d}{dx} \arccsc x & {}= \frac{-1}{x\,\sqrt{x^2-1}} \end{align} Only for real values of x:



From Yahoo Answers

Question:I'm 14 years old and I'm in 8th grade. I am currently taking 9th grade level math, but the thing is I don't understand what these things are: square roots, GCF, LCF, perfect square binomial, perfect square trinomial, ect. And I don't know how to do these: FOIL, subtracting and adding fractions, solving proportions, factoring, pythagorean theorem, and I can't memorize most of the basic multiplication tables. Ever since 6th grade I've failed every single test and quiz in math. But the thing is I'm doing excellently in all my other classes and electives while my classmates struggle in them. I've gotten nearly perfect scores on my reading, writing and science FCATs, but I just barely passed the math FCAT. No matter how hard I study, no matter how many tutoring sessions I go to and no matter who I ask for help, I can't keep my grades up. My test and quiz grades have been 50% and below. My most recent ones were 36, 8, and 0%. Is there anything wrong with me?

Answers:It's normal. I'm the same way, I just don't get math.

Question:I am looking for the equation to generate the normal distribution table in order to calculate probability based on the critical value (z) in the right tail of the curve. I know that the equation for P(x) is 1/(SIGMA * SQRT(2 * PI)) * EXP(-(x - MU)^2/(2 * SIGMA^2)) , but I am looking for the representation using critical value (z). I am writing an application that requires the equation instead of the distribution table. I know that there has to be an algorithm that is commonly used. I just need to find the equation.

Answers:The standard normal distribution, which is what I think you suggest by z, has a mean of 0 and a standard deviation of 1. Therefore just put these figures (mu = 0, sigma = 1) into the general normal distribution formula that you have above and you get the formula you want.

Question:the mean life of a certain kind of light bulb is 900 hours with a standard deviation of 30 hours. Assuming the lives of the light bulbs are normally distributed, find the percent of the light bulbs that will last for the given interval. use the standard normal table if necessary. 1. more than 984 hours like i know the normal curve and all but 984 is not on the normal curve. how do you find the answer using the standard normal table or if you don't need it, please explain im so confused.

Answers:Hi, On a TI-83, press [2nd][VARS] to get to DISTR. Choose #2 normalcdf(. It needs normalcdf(lower bound, upper bound,mean,standard deviation). If you enter normalcdf(984,99999,900,30), it equals .002555 which is .2555% of the bulbs will last more than 984 hours. (99999 is just a very large number to approximate infinity.) I hope that helps!! :-)

Question:Let x be a continuous random variable that follows a normal distribution with a mean of 550 and a standard deviation of 75. a) Find the value of x so that the area under the normal curve to the left of x is .0250 b) Find the value of x so that the area under the normal curve to the right of x is .9345 c) Find the value of x so that the area under the normal curve to the right of x is approximately .0275 Thanks!

Answers:a) ANSWER: x = -1.96 Why??? NORMAL DISTRIBUTION, STANDARDIZED VARIABLE z, PROBABILITY "LOOK-UP" P = 0.025 (2.5%) probability to the left of 0.250 inches Table "LOOK-UP" Inverse Cumulative Distribution Function Normal with mean = 0 and standard deviation = 1 P( X <= x ) x 0.025 -1.96 b) ANSWER: x = 1.51 Inverse Cumulative Distribution Function Normal with mean = 0 and standard deviation = 1 P( X <= x ) x 0.9345 1.51 c) ANSWER: x = -1.92 Inverse Cumulative Distribution Function Normal with mean = 0 and standard deviation = 1 P( X <= x ) x 0.0275 -1.92

From Youtube

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