An electronic device factory is studying the length of life of the electronic components they produced. The manager selects two assembly lines and takes all samples on those two lines. He got a sample of 500 electronic components and records the length of life in the life test. From the sample he found the average length of life was 200,000 hours and that the standard deviation was 1,000 hours. He wants to find the confidence interval for the average length of life of the electronic components they produced. Based on the information, what advice will you give to him?
Select one or more:
a. The distribution of the length of life of the electronic components is usually right skewed. Thus she should not compute the confidence interval.
b. He can calculate the confidence interval but should use a t-distribution because the population standard deviation is unknown.
c. The mean and standard deviation are large enough to compute the confidence interval.
d. The population is right skewed, but the sample size is large enough. Thus, he can compute the confidence interval.
e. He can calculate the confidence interval but should use a t-distribution because the population standard deviation is unknown.
Answer:
Given that:
He got a sample of 500 electronic components and records the length of life in the life test. From the sample he found the average length of life was 200,000 hours and that the standard deviation was 1,000 hours.
b) He can calculate the confidence interval but should use a t-distribution because the population standard deviation is unknown.
Point estimate = sample mean = = 200000
sample standard deviation = s = 1000
sample size = n = 500
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