Statistics and Confidence IntervalsGiven the following set of values:10,11,14,95,73,30,29,9,97,94,70How do I calculate a 99% confidence interval for the sample mean? I am assuming that the variance is 10Well, the idea I have is to assume that the distribution is normal, but after that i'm not completely sure what to do next. In particular, I am unable to find the z-score that corresponds to the z-score in the formula for the CI, i'm not sure what to input in R to find the CI. The formula for the confidence interval is: x − z a 2 σ 2 s q r t ( n ) and x + z a 2 σ 2 s q r t ( n ) where x denotes the mean. Here the significance level (a) is 0.01.

Question

Statistics and Confidence IntervalsGiven the following set of values:10,11,14,95,73,30,29,9,97,94,70How do I calculate a 99% confidence interval for the sample mean? I am assuming that the variance is 10Well, the idea I have is to assume that the distribution is normal, but after that i&#039;m not completely sure what to do next. In particular, I am unable to find the z-score that corresponds to the z-score in the formula for the CI, i&#039;m not sure what to input in R to find the CI. The formula for the confidence interval is:  x  −      z                  a        2                        σ      2              s      q      r      t      (      n      )       and   x  +      z                  a        2                        σ      2              s      q      r      t      (      n      )       where x denotes the mean. Here the significance level (a) is 0.01.

London Maldonado · Accepted Answer

Step 1A conventional setup for the problem is that you have a sample       X    =  (      X    1    ,      X    2    ,  …  ,      X    n    ) of size   n  =  11 from a   N  (  μ  ,      σ    0    2    ) population (by assumption) with       σ    0    2    =  10. You have to find a confidence interval for the mean   μ.A suitable pivotal quantity here is  Q  (      X    ,  μ  )  =                    n            (              X        ¯            −      μ      )              σ      0        ∼  N  (  0  ,  1  ), where       X    ¯    =      1    n        ∑          i      =      1        n        X    i   is the sample mean.So if       z          α              /            2       be such that   P  (  Z  &amp;gt;      z          α              /            2        )  =  α      /    2 where   Z  ∼  N  (  0  ,  1  ), you have      P          μ        (  −      z          α              /            2        ≤  Q  ≤      z          α              /            2        )  =  1  −  α      ∀    μYou have to use the above to arrive at the form      P          μ        (      c    1      ≤  μ  ≤          c    2    )  =  1  −  α      ∀    μ  100  (  1  −  α  )  % confidence interval for   μ is then   [      c    1    ,      c    2    ].

Statistics and Confidence Intervals. Given the following set of values: 10,11,14,95,73,30,29,9,97,94,70

Answered question

Answer & Explanation

New Questions in College Statistics