A manufacturing company that produces laminate for countertops is interested in

Question

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 4 4 The estimated regression line and the standard error are given. Defects per Countertop 6.717822-1.004950(Hours of Training) = 1.229787 Suppose a new employee has had 9 hours of training. What would be the 99% prediction interval for the number of defects per countertop? Round your answer to two decimal places

Explanation / Answer

The statistical software output for this problem is:

Simple linear regression results:
Dependent Variable: Defects per Countertop
Independent Variable: Hours of Training
Defects per Countertop = 6.7178218 - 1.0049505 Hours of Training
Sample size: 10
R (correlation coefficient) = -0.87822848
R-sq = 0.77128526
Estimate of error standard deviation: 1.2297871

Parameter estimates:

Analysis of variance table for regression model:

Predicted values:

Hence,

99% prediction interval of y for x = 9 is:

(-7.90, 3.24)

Parameter Estimate Std. Err. Alternative DF T-Stat P-value Intercept 6.7178218 0.79774428 0 8 8.4210215 <0.0001 Slope -1.0049505 0.19348141 0 8 -5.1940417 0.0008

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