They should be.
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
Best to use a histogram i think! z scores can probably be used too however they seem more a method of how to transform outliers in workable scores.
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.
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 = (x minus mu) divided by sigma.
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