To calculate the z-score and t-score, we need to clarify which of the two sets of data you're referencing.

1. Z-Score:

The z-score measures how many standard deviations an observation is from the population mean. It is used when the population standard deviation is known.

The formula for the z-score is:

$z = \frac{X - \mu}{\sigma}$

Where:

• $X$ is the sample mean
• $\mu$ is the population mean
• $\sigma$ is the population standard deviation

In your case:

• $X = 119$ (sample mean)
• $\mu = 100$ (population mean)
• $\sigma = 6$ (population standard deviation)

Substitute the values:

$z = \frac{119 - 100}{6} = \frac{19}{6} \approx 3.17$

So, the z-score is approximately 3.17.

2. T-Score:

The t-score is used when the population standard deviation is unknown, and we only have the sample standard deviation. It is computed using:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

Where:

• $X = 119$ (sample mean)
• $\mu = 100$ (population mean)
• $s = 8$ (sample standard deviation)
• $n = 4$ (sample size)

Now substitute the values:

$t = \frac{119 - 100}{\frac{8}{\sqrt{4}}} = \frac{19}{\frac{8}{2}} = \frac{19}{4} = 4.75$

So, the t-score is 4.75.

Summary:

• z-score: $\approx 3.17$
• t-score: $4.75$

Let me know if you'd like more details or have any further questions!

5 related questions:

1. What is the difference between a z-score and a t-score?
2. When should you use a t-distribution instead of a normal distribution?
3. How does sample size affect the t-distribution?
4. What happens to the t-score if the sample size increases?
5. Can we use a z-score when the population standard deviation is unknown?

Tip:

For small sample sizes (typically $n \leq 30$), use the t-distribution since it accounts for more uncertainty in estimating the population parameters.