Calculator UseThe z-score is the number of standard deviations a data point is from the population mean. You can calculate a z-score for any raw data value on a normal distribution. Show When you calculate a z-score you are converting a raw data value to a standardized score on a standardized normal distribution. The z-score allows you to compare data from different samples because z-scores are in terms of standard deviations. A positive z-score means the data value is higher than average. A negative z-score means it's lower than average. You can also determine the percentage of the population that lies above or below any z-score using a z-score table. Using the Z-Score CalculatorThis calculator can find the z-score given:
With the first method above, enter one or more data points separated by commas or spaces and the calculator will calculate the z-score for each data point provided from the same population. With the last method above enter a sample set of values. Enter values separated by commas or spaces. You can also copy and paste lines of data from spreadsheets or text documents. See all allowable formats below. Z-Score FormulaWhen calculating the z-score of a single data point x; the formula to calculate the z-score is the difference of the raw data score minus the population mean, divided by the population standard deviation. \[ z = \dfrac{x - \mu}{\sigma} \]
When calculating the z-score of a sample with known population standard deviation; the formula to calculate the z-score is the difference of the sample mean minus the population mean, divided by the Standard Error of the Mean for a Population which is the population standard deviation divided by the square root of the sample size. \[ z = \dfrac{\overline{x} - \mu}{\dfrac{\sigma}{\sqrt{n}}} \]
Acceptable Data Formats Column (New Lines) 42 42, 54, 65, 47, 59, 40, 53 Comma Separated 42, or 42, 54, 65, 47, 59, 40, 53 42, 54, 65, 47, 59, 40, 53 Spaces 42 54 or 42 54 65 47 59 40 53 42, 54, 65, 47, 59, 40, 53 Mixed Delimiters 42 42, 54, 65, 47, 59, 40, 53 Follow CalculatorSoup:  The statistic used to estimate the mean of a population, μ, is the sample mean, If X has a distribution with mean μ, and standard deviation σ, and is approximately normally distributed or n is large, then When σ Is Known If the standard deviation, σ, is known, we can transform
Example: From the previous example, μ=20, and σ=5. Suppose we draw a sample of size n=16 from this population and want to know how likely we are to see a sample average greater than 22, that is P( So the probability that the sample mean will be >22 is the probability that Z is > 1.6 We use the Z table to determine this: P( > 22) = P(Z > 1.6) = 0.0548. Exercise: Suppose we were to select a sample of size 49 in the example above instead of n=16. How will this affect the standard error of the mean? How do you think this will affect the probability that the sample mean will be >22? Use the Z table to determine the probability. Answer When σ Is Unknown If the standard deviation, σ, is unknown, we cannot transform
If X is approximately normally distributed, then has a t distribution with (n-1) degrees of freedom (df) Using the t-tableNote: If n is large, then t is approximately normally distributed.
The z table gives detailed correspondences of P(Z>z) for values of z from 0 to 3, by .01 (0.00, 0.01, 0.02, 0.03,…2.99. 3.00). The (one-tailed) probabilities are inside the table, and the critical values of z are in the first column and top row. The t-table is presented differently, with separate rows for each df, with columns representing the two-tailed probability, and with the critical value in the inside of the table. The t-table also provides much less detail; all the information in the z-table is summarized in the last row of the t-table, indexed by df = ∞. So, if we look at the last row for z=1.96 and follow up to the top row, we find that P(|Z| > 1.96) = 0.05 Exercise: What is the critical value associated with a two-tailed probability of 0.01? Answer Now, suppose that we want to know the probability that Z is more extreme than 2.00. The t-table gives us P(|Z| > 1.96) = 0.05 And P(|Z| > 2.326) = 0.02 So, all we can say is that P(|Z| > 2.00) is between 2% and 5%, probably closer to 5%! Using the z-table, we found that it was exactly 4.56%. Example: In the previous example we drew a sample of n=16 from a population with μ=20 and σ=5. We found that the probability that the sample mean is greater than 22 is P( From the tables we see that the two-tailed probability is between 0.01 and 0.05. P(|T| > 1.711) = 0.05 And P(|T| > 2.064) = 0.01 To obtain the one-tailed probability, divide the two-tailed probability by 2. P(T > 1.711) = ½ P(|T| > 1.711) = ½(0.05) = 0.025 And P(T > 2.064) = ½ P(|T| > 2.064) = ½(0.01) = 0.005 So the probability that the sample mean is greater than 22 is between 0.005 and 0.025 (or between 0.5% and 2.5%) Exercise: . If μ=15, s=6, and n=16, what is the probability that Answer return to top | previous page | next page What is the variance for the following population of scores scores 5 2 5 4?Summary: The variance for the following population of scores: 5, 2, 5, 4 is 1.5.
What are the values for SS and variance for the following sample of N 4 scores?What are the values for SS and variance for the following sample of n = 4 scores? Scores: 1, 3, 0, 4a. ss=10 and variance=2.5b.
What is the value of SS for the following sample scores 1/2 6?Scores: 1, 2, 6 0 SS = 17 0 ss = (9)2 SS = 14 5S = 13/2. Video Answer: Video Player is loading.
What zData that is two standard deviations below the mean will have a z-score of -2, data that is two standard deviations above the mean will have a z-score of +2. Data beyond two standard deviations away from the mean will have z-scores beyond -2 or 2.
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