Z-Scores and T-Scores

Understanding Z-Scores and T-Scores

Solution & Explanation

Both Z-scores and T-scores are measures of standardization used in statistical analysis, particularly when comparing data points to a population or sample mean. They help determine how far or different a data point is from the mean.

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Z-Score

Definition

A Z-score, also known as a standard score, quantifies the number of standard deviations a data point is from the mean of a distribution. It is used when the population standard deviation is known and the data is assumed to follow a normal distribution.

Formula

$Z = \frac{(X - \mu)}{\sigma}$

• X: The value being standardized.
• $\mu$: The mean of the population.
• $\sigma$: The standard deviation of the population.

Application

• Comparison: Z-scores allow for comparison of data points from different distributions by transforming them to a common scale.
• Normal Distribution: Used when the data follows a normal distribution with known population parameters.

Example

If an IQ test has a mean ($\mu$) of 100 and a standard deviation ($\sigma$) of 15, and an individual scores 120, their Z-score is calculated as: $Z = \frac{(120 - 100)}{15} = 1.33$ This indicates the individual scored 1.33 standard deviations above the mean.

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T-Score

Definition

A T-score is used when the sample size is small (typically less than 30) and the population standard deviation is unknown. It measures how many standard errors the sample mean is from the population mean.

Formula

$T = \frac{(\bar{X} - \mu)}{\frac{s}{\sqrt{n}}}$

• $\bar{X}$: The sample mean.
• $\mu$: The population mean (or the value being compared against).
• s: The sample standard deviation.
• n: The sample size.

Application

• Hypothesis Testing: Used in t-tests to determine the significance of the difference between sample means and a hypothesized population mean.
• Small Samples: Suitable for small sample sizes or when the population standard deviation is unknown.

Example

If a sample of 10 students has a mean score ($\bar{X}$) of 82, the population mean ($\mu$) is 78, and the sample standard deviation (s) is 5, the T-score is: $T = \frac{(82 - 78)}{\frac{5}{\sqrt{10}}} = 2.53$ This means the sample mean is 2.53 standard errors above the population mean.

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Key Differences

• Population Standard Deviation: Z-scores require a known population standard deviation, whereas T-scores use sample standard deviation.
• Sample Size: Z-scores are used for large samples, while T-scores are appropriate for small samples.
• Distribution: Z-scores assume normal distribution, whereas T-scores are used when normality is not assured.

Understanding when and how to use Z-scores and T-scores is essential for accurate statistical analysis and hypothesis testing. They provide a means to assess the relative standing of data points and sample means in a distribution.