zscore
(noun)Definition of zscore
The standardized value of observation x from a distribution that has mean μ and standard deviation σ.
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Examples of zscore in the following topics:

ZScores and Location in a Distribution
 A zscore is the signed number of standard deviations an observation is above the mean of a distribution.
 Thus, a positive zscore represents an observation above the mean, while a negative zscore represents an observation below the mean.
 We obtain a zscore through a conversion process known as standardizing or normalizing.Zscores are also called standard scores, zvalues, normal scores or standardized variables.
 The zscore, in turn, provides an assessment of how offtarget a process is operating.The conversion of a raw score, x, to a zscore can be performed using the following equation:$z=\frac { x\mu }{ \sigma }$where μ is the mean of the population and σ is the standard deviation of the population.
 Z is negative when the raw score is below the mean and positive when the raw score is above the mean.A key point is that calculating z requires the population mean and the population standard deviation, not the sample mean nor sample deviation.
 A zscore is the signed number of standard deviations an observation is above the mean of a distribution.

Calculating a Normal Approximation
 This can be done by finding zscores and using the zscore table, which can be seen in .
 If we did not have the normal area calculator, we could find the solution using a table of the standard normal distribution (a ztable) as follows:Find a Z score for 7.5 using the formula $Z=\frac { 7.55 }{ 1.5811 } =1.5811$Find the area below a Z of 1.58 = 0.943.Find a Z score for 8.5 using the formula $Z=\frac { 8.55 }{ 1.5811 } =2.21$Find the area below a Z of 2.21 = .987.Subtract the value in step 2 from the value in step 4 to get 0.044.The same logic applies when calculating the probability of a range of outcomes.

The Standard Normal Curve
 Luckily, one can transform any normal distribution with a certain mean μ and standard deviation σ into a standard normal distribution, by the zscore conversion formula:$z=\frac { x\mu }{ \sigma }$ Therefore, a zscore is the standardized value of observation x from a distribution that has mean μ and standard deviation σ (how many standard deviations you are away from zero).
 The zscore gets its name because of the denomination of the standard normal distribution as the "Z" distribution.
 Since x = 70.4, μ = 64 inches and σ = 2.5 inches, we need to calculate z: $z=\frac { 70.464 }{ 2.5 } =\frac { 6.4 }{ 2.5 } =2.56$ Therefore, the probability P(X>70.4) is equal to P(Z>2.56), where X is the normally distributed height with mean μ = 64 inches and standard deviation σ = 2.5 inches {X~N(64,2.5) for short}, and Z is a standard normal distribution {Z~N(0,1)}.
 The next step requires that we use what is known as the zscore table to calculate probabilities for the standard normal distribution.
 Therefore: P(60.3<X<65) = P(1.48<Z<0.40) = P(Z<0.40)P(Z<1.48) = (0.5+0.1554)(0.50694) = 0.65540.0694 = 0.5860.

Finding the Area Under the Normal Curve
 In order to do this, we use a zscore table, as shown in (same as in the process of standardization discussed in the previous section).
 The following is another simple example: find P(Z≥1.17).
 Since all the probabilities must sum to 1: P(Z>1.17) = 1 − P(Z<1.17) = 0.121 As a final example: find P(−1.16≤Z≤1.32).
 P(−1.16≥Z≤1.32) = P(Z≤1.32) − P(Z≤−1.16) The difficulty arrises from the fact that our table of values does not allow us to directly calculate P(Z≤−1.16).
 However, we can use the symmetry of the distribution, as follows: P(Z≤−1.16) = 1 − P(Z≤1.16) = 0.1230 So, we can say that: P(−1.16≤Z≤1.32) = 0.9066 − 0.1230 = 0.7836.

Change of Scale
 Thus, a positive standard score represents a datum above the mean, while a negative standard score represents a datum below the mean.
 The use of "Z" is because the normal distribution is also known as the "Z distribution".
 They are most frequently used to compare a sample to a standard normal deviate (standard normal distribution, with μ = 0 and σ = 1).The zscore is only defined if one knows the population parameters.
 If one only has a sample set, then the analogous computation with sample mean and sample standard deviation yields the Student's tstatistic.The standard score of a raw score x is:$z=\frac { x\mu }{ \sigma }$,where μ is the mean of the population, and σ is the standard deviation of the population.
 The absolute value of z represents the distance between the raw score and the population mean in units of the standard deviation. z is negative when the raw score is below the mean, positive when above.A key point is that calculating z requires the population mean and the population standard deviation, not the sample mean or sample deviation.

Wilcoxon tTest
 Thus,For ${ N }_{ r }\ge 10$, a zscore can be calculated as ).If z > zcritical then reject H0.For Nr < 10, W is compared to a critical value from a reference table.If $W\ge { W }_{ critical,{ N }_{ r } }$ then reject H0.Alternatively, a pvalue can be calculated from enumeration of all possible combinations of W given Nr.

Using the Normal Curve
 ZValueThe functional form for a normal distribution is a bit complicated.
 The table only gives the probabilities to the left of the zvalue.
 For example, if we wanted to find out the probability that a variable is more than 0.51 sigmas above the mean, P(z>0.51), we just need to calculate 1P(z<0.51)=10.6950=0.3050, or 30.5%.There is another note of caution to take into consideration when using the table: The table provided only gives values for positive zvalues, which correspond to values above the mean.
 What if we wished instead to find out the probability that a value falls below a zvalue of 0.51, or 0.51 standard deviations below the mean?
 We must remember that the standard normal curve is symmetrical , meaning that P(z<0.51)=P(z>0.51), which we calculate above to be 30.5%.We may even wish to find the probability that a variable is between two zvalues, such as between 0.50 and 1.50, or P(0.50).689599.7 RuleAlthough we can always use the zscore table to find probabilities, the 689599.7 rule helps for quick calculations.

Comparing Two Independent Population Proportions
 To conduct the test, we use a pooled proportion, pc.The pooled proportion is calculated as follows:${ p }_{ c }=\frac { { x }_{ A }+{ x }_{ B } }{ { n }_{ A }+{ n }_{ B } }$The distribution for the differences is:${ P }_{ A }^{ ' }{ P }_{ B }^{ ' }\sim N\left[ 0,\sqrt { { p }_{ c }\cdot (1{ p }_{ c })\cdot \left( \frac { 1 }{ { n }_{ A } } +\frac { 1 }{ { n }_{ B } } \right) } \right]$.The test statistic (zscore) is:$z=\frac { { (p }_{ A }^{ ' }{ p }_{ B }^{ ' })({ p }_{ A }{ p }_{ B }) }{ \sqrt { { p }_{ c }\cdot (1{ p }_{ c })\cdot \left( \frac { 1 }{ { n }_{ A } } +\frac { 1 }{ { n }_{ B } } \right) } }$.ExampleTwo types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. 20 out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. 12 out of another random sample of 200 adults given medication B still had hives 30 minutes after taking the medication.

Estimates and Sample Size
 To calculate E, we need to know the desired confidence level (${ Z }_{ \frac { \alpha }{ 2 } }$) and the population standard deviation, $\sigma$.
 $E={ Z }_{ \frac { \alpha }{ 2 } }\frac { \sigma }{ \sqrt { n } }$To change the size of the error (E), two variables in the formula could be changed: the level of confidence (${ Z }_{ \frac { \alpha }{ 2 } }$) or the sample size (n).
 To determine this, begin by solving the equation for the E in terms of n:$n={ \left( \frac { { Z }_{ \frac { \alpha }{ 2 } }\sigma }{ E } \right) }^{ 2 }$where ${ Z }_{ \frac { \alpha }{ 2 } }$ is the critical z score based on the desired confidence level, E is the desired margin of error, and $\sigma$ is the population standard deviation.Since the population standard deviation is often unknown, the sample standard deviation from a previous sample of size n ≥ 30 may be used as an approximation to s.
 In this case, the margin of error, E, is found using the formula:$E={ Z }_{ \frac { \alpha }{ 2 } }\sqrt { \frac { p'q' }{ n } }$where:p' = x/n is the point estimate for the population proportionx is the number of successes in the samplen is the number in the sample; andq' = 1  p'Then, solving for the minimum sample size n needed to estimate p:$n=p'q'\left( \frac { { Z }_{ \frac { \alpha }{ 2 } } }{ E } \right) ^{ 2 }$ExampleThe Mesa College mathematics department has noticed that a number of students place in a nontransfer level course and only need a 6 week refresher rather than an entire semester long course.