# T-Score vs. Z-Score: When to Use Each

Two terms that often confuse students in statistics classes are t-scores and z-scores.

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.