A user reports his or her computer is slow to boot. You check the boot order and determine that the computer is checking the C d
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Answer:
1) 402.7 grams. This estimate is called the sample mean.
2) (399.11, 406.29)
3) The 99 percent confidence limits is between 399.11 grams and 406.29 grams.
I am 99% sure that the value lies between 399.11 grams and 406.29 grams.
Explanation:
sample size (n) = 40, the mean weight (x)= 402.7 grams and the standard deviation (σ)=8.8 grams
1) The point estimated mean weight of the population is 402.7 grams. This estimate is called the sample mean.
2) c = 99% = 0.99
α = 1 - 0.99 = 0.01
.
The z score of 0.005 corresponds with the z score of 0.495 (0.5 - 0.005).
.
The margin of error (e) =
The confidence interval = x ± e = 402.7 ± 3.59 = (399.11, 406.29)
3) The 99 percent confidence limits is between 399.11 grams and 406.29 grams.
I am 99% sure that the value lies between 399.11 grams and 406.29 grams.