**Title: Analyzing the Relationship Between Training Hours and Defects in Manufacturing**A manufacturing company that produces laminate for countertops is interested in studying the relationship between the number of hours of training that an employee receives and the number of defects per countertop produced. Ten employees are randomly selected. The number of hours of training each employee has received is recorded, and the number of defects on the most recent countertop produced is determined. The results are as follows:| Hours of Training | Defects per Countertop ||-------------------|------------------------|| 1 | 5 || 4 | 1 || 7 | 0 || 3 | 3 || 2 | 4 || 5 | 2 || 2 | 3 || 4 | 2 || 5 | 2 || 6 | 2 |**Regression Analysis:**The estimated regression equation and the standard error are given:\[ \text{Defects per Countertop} = 6.717822 - 1.004950 \times (\text{Hours of Training}) \]\[ S_e = 1.229787 \]**Prediction Interval:**Suppose a new employee has had 4 hours of training. What would be the 90% prediction interval for the number of defects per countertop? Round your answer to two decimal places.**Interactive Element:**- Input your calculations to find the prediction interval.- Use the regression equation to predict the defects for 4 hours of training.- The task involves interpreting standard error and constructing a prediction interval.**How to Use:**- Understand how training impacts production quality.- Apply regression analysis to practical scenarios.- Evaluate the importance of employee training in reducing defects.Submit your calculated answer and verify its accuracy. This exercise aids in comprehending effective workforce development strategies in manufacturing settings.**Note:**Knowledge of regression analysis, prediction intervals, and statistical significance is essential for this exercise. You may refer to supplementary materials on these topics for better understanding.

Prediction interval for an individual y for x = xp is given by, PI = y^ ±tα/2, n-1Se1+1n+xp- x¯2∑(xi -x¯)2 Where , y^ is predicted value of y when x = xp Se is standard Error x¯ is sample mean of variable x n is number of data pairs (x, y) tα/2, n-2 is critical value for given confidence level with df = n - 2 Given , y : Defect per countertop ( Dependent variable ) x : Hours of training ( independent variable ) Regression equation Defect per countertop = 6.717822 - 1.004950(Hours of training) i.e. y^ = 6.717822 - 1.004950*x Standard Error = Se = 1.229787 Suppose a new employee has had 4 hours of training i.e. xp = 4 The predicted value of y when x = 4 is, y^ = 6.717822 - 1.004950*4 = 2.698022 Let's find sample mean (x¯) x¯ = ∑xin=1+4+7+3+2+2+5+5+1+610 = 3610 = 3.6 (xp-x¯)2 = (4-3.6)2= 0.16 Now let's find ∑xi-x¯2 we got , ∑xi-x¯2 = 40.4 Now let's find critical value tα/2, n-2 confidence level = 90% = 0.9 Significance level = α = 1-0.9 = 0.1 degrees of freedom (df) = n - 2 = 10 - 2= 8 Using Excel function , "=TINV(α, df)" , this function returns two tailed inverse of t distribution. So, tα/2, n-2 = TINV(0.1, 8) = 1.85955 Now put all values in formula , we get PI = y^ ±tα/2, n-1Se1+1n+xp- x¯2∑(xi -x¯)2 PI = 2.698022 ±1.85955*1.2297871+110+0.1640.4 PI = 2.698022 ±2.402782747 PI = (2.698022 -2.402782747, 2.698022 +2.402782747) PI = (0.295239253, 5.100804747) PI ≈ (0.30 , 5.10) (rounded to 2 decimal places )