Answer to Question #339430 in Statistics and Probability for Liam

Question #339430

survey of 31 randomly selected students finds that they save a mean of $82 per semester by

using a website. Assume the date comes from a normal distribution and the sample standard deviation is $18 per month.


Confidence Interval: What is the 99% confidence interval to estimate the population mean? (Round your

answers to two decimal places.)


____< u < _____


1
Expert's answer
2022-05-12T15:18:49-0400

The critical value for α=0.01\alpha = 0.01 and df=n1=30df = n-1 = 30 degrees of freedom is tc=z1α/2;n1=2.749996.t_c = z_{1-\alpha/2; n-1} = 2.749996.

The corresponding confidence interval is computed as shown below:


CI=(xˉtc×sn,xˉ+tc×sn)CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}},\bar{x}+t_c\times\dfrac{s}{\sqrt{n}})

=(822.749996×1831,=(82- 2.749996\times\dfrac{18}{\sqrt{31}},

82+2.749996×1831)82+ 2.749996\times\dfrac{18}{\sqrt{31}})

=(73.11,90.89)=(73.11, 90.89)

Therefore, based on the data provided, the 99% confidence interval for the population mean is 73.11<μ<90.89,73.11 < \mu < 90.89 , which indicates that we are 99% confident that the true population mean μ\mu  is contained by the interval (73.11,90.89).(73.11, 90.89).



73.11<μ<90.8973.11 < \mu < 90.89


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