To find the 95% prediction interval for the average number of sick days an employee will take given an age of 34, we can use the prediction interval formula. The prediction interval for a value $\hat{y}$ at a specific $x$ is given by:

$\hat{y} \pm t_{\alpha/2, n-2} \times S_e \sqrt{1 + \frac{1}{n} + \frac{(x - \bar{x})^2}{\sum (x_i - \bar{x})^2}}$

where:

• $\hat{y}$ is the predicted value for the given $x$,
• $t_{\alpha/2, n-2}$ is the critical t-value for a 95% confidence level with $n - 2$ degrees of freedom,
• $S_e$ is the standard error,
• $x$ is the specific age for prediction (34),
• $\bar{x}$ is the mean of the sample ages,
• $n$ is the sample size.

Step 1: Calculate $\hat{y}$

Given the regression equation: $\text{Sick Days} = 14.310162 - 0.236900 \times \text{Age}$ we substitute Age = 34: $\hat{y} = 14.310162 - 0.236900 \times 34$

Step 2: Calculate the mean of the sample ages ($\bar{x}$)

The ages are: 30, 50, 40, 55, 30, 28, 60, 25, 30, 45.

Step 3: Calculate the sum of squared deviations from the mean ($\sum (x_i - \bar{x})^2$)

Step 4: Determine the t-value for a 95% prediction interval with $n-2 = 8$ degrees of freedom.

Let's calculate this.The 95% prediction interval for the average number of sick days an employee will take per year, given the employee is 34 years old, is approximately $2.15$ to $10.36$ days.

Would you like a detailed breakdown of each calculation step, or do you have further questions?

Here are some related questions:

1. How would the prediction interval change with different confidence levels?
2. What would be the interval if the standard error, $S_e$, were lower?
3. How does increasing the sample size impact the width of the prediction interval?
4. What is the interpretation of the regression slope $-0.236900$ in this context?
5. How does the variability in ages affect the accuracy of predictions?

Tip: Remember, wider intervals imply greater uncertainty; smaller intervals reflect more precise predictions.