To determine the interval in which the additional measurement is expected to fall with 95% probability, we use the prediction interval formula:

$\text{Interval} = \bar{x} \pm t_{\alpha/2, n-1} \cdot s \cdot \sqrt{1 + \frac{1}{n}}$

Where:

• $\bar{x} = 49.9$ (sample mean)
• $s = 0.94$ (sample standard deviation)
• $n = 8$ (sample size)
• $t_{\alpha/2, n-1}$ is the critical t-value for a 95% confidence level with 7 degrees of freedom (n-1 = 7).

Step 1: Find $t_{\alpha/2,7}$

Using a t-table or calculator, for 95% confidence and 7 degrees of freedom: $t_{0.025, 7} \approx 2.447$

Step 2: Compute the Value

$\text{Value} = 2.447 \times 0.94 \times \sqrt{1 + \frac{1}{8}}$

$= 2.447 \times 0.94 \times \sqrt{1.125}$

$= 2.447 \times 0.94 \times 1.061$

$= 2.447 \times 0.9973$

$\approx 2.44$

Final Answer:

$\mathbf{2.44}$

Would you like a breakdown of any step?

Here are some related questions:

1. How does a confidence interval differ from a prediction interval?
2. Why do we use the t-distribution instead of the normal distribution?
3. What happens to the interval if the sample size increases?
4. How does the standard deviation affect the prediction interval?
5. What is the effect of changing the confidence level from 95% to 99%?

Tip: The prediction interval is always wider than a confidence interval because it accounts for both sample variability and individual measurement variability.