Answer to Question #346760 in Statistics and Probability for Bethsheba Kiap

Question #346760

A random sample is drawn from a normally distributed population of known standard deviation 5. Construct a 99.8% confidence interval for the population mean based on the information given (not all of the information given need be used).

a. n = 16, 𝑥̅= 98, 𝑠 = 5.6

b. n = 9, 𝑥̅= 98, 𝑠 = 5.6


1
Expert's answer
2022-06-03T12:55:25-0400

1. The critical value for α=0.002\alpha = 0.002 zc=z1α/2=3.0902z_c = z_{1-\alpha/2} = 3.0902

The corresponding confidence interval is computed as shown below:



CI=(Xˉzc×σn,Xˉ+zc×σn)CI=(\bar{X}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{X}+z_c\times\dfrac{\sigma}{\sqrt{n}})=(983.0902×516,=(98-3.0902\times\dfrac{5}{\sqrt{16}},98+3.0902×516)98+3.0902\times\dfrac{5}{\sqrt{16}})=(94.137,101.863)=(94.137,101.863)

Therefore, based on the data provided, the 95 confidence interval for the population mean is 94.137<μ<101.863,94.137 < \mu <101.863, which indicates that we are 99.8% confident that the true population mean μ\mu is contained by the interval (94.137,101.863).(94.137,101.863).


2. The critical value for α=0.002\alpha = 0.002 zc=z1α/2=3.0902z_c = z_{1-\alpha/2} = 3.0902

The corresponding confidence interval is computed as shown below:



CI=(Xˉzc×σn,Xˉ+zc×σn)CI=(\bar{X}-z_c\times\dfrac{\sigma}{\sqrt{n}}, \bar{X}+z_c\times\dfrac{\sigma}{\sqrt{n}})=(983.0902×59,=(98-3.0902\times\dfrac{5}{\sqrt{9}},98+3.0902×59)98+3.0902\times\dfrac{5}{\sqrt{9}})=(92.850,103.150)=(92.850,103.150)

Therefore, based on the data provided, the 95 confidence interval for the population mean is 92.850<μ<103.150,92.850 < \mu <103.150, which indicates that we are 99.8% confident that the true population mean μ\mu is contained by the interval (92.85,103.150).(92.85,103.150).

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog