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 1 4 7 3 2 2 5 5 1 6 Copy Data The estimated regression line and the standard error are given. Defects per Countertop= Defects per Countertop 5 1 0 3 5 4 1 2 8 2 6.717822-1.004950(Hours of Training) 5,= 1.229787 Suppose a new employee has had 5 hours of training. What would be the 99 % prediction interval for the number of defects per countertop?
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 1 4 7 3 2 2 5 5 1 6 Copy Data The estimated regression line and the standard error are given. Defects per Countertop= Defects per Countertop 5 1 0 3 5 4 1 2 8 2 6.717822-1.004950(Hours of Training) 5,= 1.229787 Suppose a new employee has had 5 hours of training. What would be the 99 % prediction interval for the number of defects per countertop?
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Defects per Countertop= 6.717822-1.004950(Hours of Training)
= 1.229787
Suppose a new employee has had 5 hours of training. What would be the 99 % prediction interval for the number of defects per countertop?
Round your answer to two decimal places
Answer
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Transcribed Image Text:Prev
Defects per Countertop= 6.717822-1.004950(Hours of Training)
= 1.229787
Suppose a new employee has had 5 hours of training. What would be the 99 % prediction interval for the number of defects per countertop?
Round your answer to two decimal places
Answer
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Keypad
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
Transcribed Image Text: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
1
4
7
3
2
2
5
5
1
6
Copy Data
The estimated regression line and the standard error are given.
Defects per Countertop
5
1
0
3
5
4
1
2
8
2
Defects per Countertop = 6.717822-1.004950(Hours of Training)
5, 1.229787
Suppose a new employee has had 5 hours of training. What would be the 99 % prediction interval for the number of defects per countertop?
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