If a random variable X has a Normal distribution with mean M and variance S2, then
Z = (X - M)/S
They should be.
Another term for z-scores is standard scores. Z-scores indicate how many standard deviations a data point is from the mean of its distribution, allowing for comparison between different datasets. They are commonly used in statistics to standardize scores and facilitate further analysis.
Yes, although the z-scores associated with p-values of 0.01 and 0.05 have special significance, perhaps mostly for historical reasons, all possible z-scores from negative infinity to positive infinity have meaning in statistical theory and practice.
A Z score of 300 is an extremely large number as the z scores very rarely fall above 4 or below -4. About 0 percent of the scores fall above a z score of 300.
z = 1.281551
z-scores are distributed according to the standard normal distribution. That is, with the parameters: mean 0 and variance 1.
If the distribution is Gaussian (or Normal) use z-scores. If it is Student's t, then use t-scores.
They should be.
True or False, One major advantage of transforming X values into z-scores is that the z-scores always form a normal distribution
You either look it up in a table of z scores or you can use a calculator such as the TI8 and use normalcdf.
No, the Z-test is not the same as a Z-score. The Z-test is where you take the Z-score and compare it to a critical value to determine if the null hypothesis will be rejected or fail to be rejected.
Yes, although the z-scores associated with p-values of 0.01 and 0.05 have special significance, perhaps mostly for historical reasons, all possible z-scores from negative infinity to positive infinity have meaning in statistical theory and practice.
Z = (x minus mu) divided by sigma.
A Z score of 300 is an extremely large number as the z scores very rarely fall above 4 or below -4. About 0 percent of the scores fall above a z score of 300.
z = 1.281551
No, they do not. They are pure numbers.
X = 50 => Z = (50 - 70)/12 = -20/12 = -1.33 So prob(X < 50) = Prob(Z < -1.33...) = 0.091