Questions: The estimated regression line and the standard error are given. Defects per Countertop = 6.717822 - 1.004950 (Hours of Training) sc = 1.229787 Suppose a new employee has had 10 hours of training, What would be the 99% prediction interval for the number of defects per countertop? Round your answer to two decimal places.

Solution Steps

To estimate the number of defects per countertop for a new employee with 10 hours of training, we use the regression line given by:

\[ \text{Defects per Countertop} = 6.717822 - 1.004950 \times (\text{Hours of Training}) \]

Substituting \( \text{Hours of Training} = 10 \):

\[ \text{Predicted defects} = 6.717822 - 1.004950 \times 10 = -3.331678 \]

To calculate the margin of error for a 99% prediction interval, we first determine the Z-score corresponding to a 99% confidence level, which is \( Z = 2.5758 \).

The margin of error \( E \) is calculated using the formula:

\[ E = \frac{Z \times \sigma}{\sqrt{n}} \]

where:

• \( \sigma = 1.229787 \) (standard error)
• \( n = 10 \) (sample size)

Substituting the values:

\[ E = \frac{2.5758 \times 1.229787}{\sqrt{10}} = 1.0017 \]

The 99% prediction interval is calculated as follows:

\[ \text{Lower Bound} = \text{Predicted defects} - E = -3.331678 - 1.0017 = -4.333378 \]

\[ \text{Upper Bound} = \text{Predicted defects} + E = -3.331678 + 1.0017 = -2.329978 \]

Rounding to two decimal places, we have:

\[ \text{Lower Bound} \approx -4.33, \quad \text{Upper Bound} \approx -2.33 \]

Final Answer