When to use z-score and t score

A Z-score compares your bone density to the average bone density of people your own age and gender. For example, if you are a 60-year-old female, a Z-score compares your bone density to the average bone density of 60-year-old females.

NOTE: Any post menopausal woman should always request her T-score rather than just her Z-score

A Z-score is helpful in diagnosing secondary osteoporosis and is always used for children, young adults, women who are pre-menopausal, and men under age 50. If you have a very low Z-score (more than 2 standard deviations below other individuals your age), your doctor should consider whether other medical conditions or medications may be causing lower than expected bone density.

Although you may have low bone density when you have your first test, your doctor cannot tell if you have lost bone density or if you have always had lower bone density due to family or medical history. Your peak adult bone mass may have been below that of the average individual. For example, if you have a T-score of -2.5, it is not appropriate to say that you have lost 25% of your bone density unless you had a bone density test when you reached peak bone density. There are lab tests that can help your doctor determine if you are currently losing bone density.

The z-score and t-score (aka z-value and t-value) show how many standard deviations away from the mean of the distribution you are, assuming your data follow a z-distribution or a t-distribution.

These scores are used in statistical tests to show how far from the mean of the predicted distribution your statistical estimate is. If your test produces a z-score of 2.5, this means that your estimate is 2.5 standard deviations from the predicted mean.

The predicted mean and distribution of your estimate are generated by the null hypothesis of the statistical test you are using. The more standard deviations away from the predicted mean your estimate is, the less likely it is that the estimate could have occurred under the null hypothesis.

Both are used extensively when performing hypothesis tests or constructing confidence intervals, but they’re slightly different.

Here’s the formula for each:

t-score = (x – μ) / (s/√n)

where:

  • x: Sample mean
  • μ: Population mean
  • s: Sample standard deviation
  • n: Sample size

z-score = (x – μ) / σ

where:

  • x: Raw data value
  • μ: Population mean
  • σ: Population standard deviation

This flow chart shows when you should use each, depending on your data:

The following examples show how to calculate a t-score and z-score in practice.

Example 1: Calculating a T-Score

Suppose a restaurant makes burgers that claim to have a mean weight of μ = 0.25 pounds.

Suppose we take a random sample of n = 20 burgers and find that the sample mean weight is x = 0.22 pounds with a standard deviation of s = 0.05 pounds. Perform a hypothesis test to determine if the true mean weight of all burgers produced by this restaurant is equal to 0.25 pounds.

For this example, we would use a t-score to perform the hypothesis test because neither of the following two conditions are met.

  • The population standard deviation (σ) is known. (σ is not provided in this example)
  • The sample size is greater than 30. (n = 20 in this example)

Thus, we would calculate the t-score as:

  • t-score = (x – μ) / (s/√n)
  • t-score = (.22 – .25) / (.05 / √20)
  • t- score = -2.68

According to the T Score to P Value Calculator, the p-value that corresponds to this t-score is 0.01481.

Since this p-value is less than .05, we have sufficient evidence to say that the mean weight of burgers produced at this restaurant is not equal to 0.25 pounds.

Example 2: Calculating a Z-Score

Suppose a company manufactures batteries that are known to have a lifespan that follows a normal distribution with a mean of μ = 20 hours and a standard deviation of σ = 5 hours.

Suppose we take a random sample of n = 50 batteries and find that the sample mean is x = 21 hours. Perform a hypothesis test to determine if the true mean lifespan of all batteries manufactured by this company is equal to 20 hours.

For this example, we would use a z-score to perform the hypothesis test because the following two conditions are met:

  • The population standard deviation (σ) is known. (σ is equal to 5 in this example)
  • The sample size is greater than 30. (n = 50 in this example)

Thus, we would calculate the z-score as:

  • z-score = (x – μ) / σ
  • z-score = (21 – 20) / 5
  • z- score = 0.2

According to the Z Score to P Value Calculator, the p-value that corresponds to this z-score is 0.84184.

Since this p-value is not less than .05, we don’t have sufficient evidence to say that the mean lifespan of all batteries manufactured by this company is different than 20 hours.

What is the difference between z

Z score is the standardization from the population raw data or more than 30 sample data to standard score while T score is standardization from the sample data of less than 30 data to a standard score. Z score ranges from -3 to 3, while the T score ranges from 20 to 80.

How do you know if you should use Z or T?

When to use Z-test vs T-test?.
If the population standard deviation is known and the sample size is greater than 30, Z-test is recommended to be used..
If the population standard deviation is known, and the size of the sample is less than or equal to 30, T-test is recommended..

What are T scores and z scores used for?

T-scores compare bone density with that of a healthy person, whereas Z-scores use the average bone density of people of the same age, sex, and size as a comparator. Although both scores can be useful, most experts prefer using Z-scores for children, teenagers, premenopausal females, and younger males.

Why do we use t

As mentioned, a t-test is primarily used for research with limited sample sizes whereas a z-test is deployed for hypothesis testing that requires researchers to look at a population size that's larger than 30.