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
[tex]\frac{\alpha }{2} =\frac{0.01}{2} =0.005[/tex].
The z score of 0.005 corresponds with the z score of 0.495 (0.5 - 0.005).
[tex]z_\frac{\alpha }{2} =2.58[/tex].
The margin of error (e) = [tex]z_\frac{\alpha }{2}*\frac{\sigma}{\sqrt{n} } =2.58*\frac{8.8}{\sqrt{40} } =3.59[/tex]
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.
