Find confidence levels from given intervalsA set of 40 data items, produces a confidence interval for the population mean of 94.93 &lt; μ &lt; 105.07. If ∑ x 2 = 424 375, find the confidence level.So the idea of confidence intervals is still rather new to me and one that isn't fully clear in my mind, would someone be able to give me some hints about solving this and explain what they are doing to solve it.I get that the confidence interval is given by x ¯ − z σ n &lt; μ &lt; x ¯ + z σ n and that z = Φ − 1 ( c ) where c is the confidence level, however this is as far as my knowledge goes.

Question

Find confidence levels from given intervalsA set of 40 data items, produces a confidence interval for the population mean of   94.93  &amp;lt;  μ  &amp;lt;  105.07. If   ∑      x    2    =  424  375, find the confidence level.So the idea of confidence intervals is still rather new to me and one that isn&#039;t fully clear in my mind, would someone be able to give me some hints about solving this and explain what they are doing to solve it.I get that the confidence interval is given by            x      ¯        −  z      σ          n        &amp;lt;  μ  &amp;lt;            x      ¯        +  z      σ          n      and that   z  =      Φ          −      1        (  c  ) where c is the confidence level, however this is as far as my knowledge goes.

altaryjny94 · Accepted Answer

Step 1To match the confidence interval with your numerical data, we get            x      ¯        −  z      σ          n        =  94.93and             x      ¯        +  z      σ          n        =  105.07You are interested in the confidence level, which you could know if you get z, i.e. via the normal tail. Subtracting both equations you get  2  z      σ          n        =  105.07  −  94.93  =  10.14so   z  =      10.14          2              σ                  n                    Since your sample size consists of 40 elements, then   n  =  40, hence  z  =      10.14          2              σ                  40                    and your   σ could be estimated using an unbiased estimator      σ    2    =                    ∑                  i          =          1                          40                            x        i        2            −              1        n                              (                      ∑                          i              =              1                                      40                                            x            i                    )                2                    n      −      1        =            424375      −              1        40                              (                      ∑                          i              =              1                                      40                                            x            i                    )                2                    40      −      1      Step 2The sum   ∑      x    i   is nothing other than       ∑          i      =      1              40            x    i    =  n            x      ¯       which could be obtained by adding the first two equations:  2            x      ¯        =  105.07  +  94.93  =  200hence            x      ¯        =  100so      ∑          i      =      1              40            x    i    =  n            x      ¯        =  40  (  100  )  =  4000Replacing we get  σ  =                    424375        −                  1          40                                      (            4000            )                    2                            40        −        1            So  z  =      10.14          2                                                  424375              −                              1                40                                                              (                  4000                  )                                2                                                    40              −              1                                                40

kennadiceKesezt · Accepted Answer

Step 1If   z  =      Φ          −      1        (  c  ), then let the   100  (  2  c  −  1  )  % confidence interval for   μ be       μ    1    &amp;lt;  μ  &amp;lt;      μ    2  .This is because the normal distribution (which is what we are assuming here) is two-tailed, so we have that  P  (  X  &amp;lt;  z  )  =  c  ,    P  (  X  &amp;gt;  z  )  =  1  −  cand we seek the region   P  (  −  z  &amp;lt;  X  &amp;lt;  z  )  =  c  −  (  1  −  c  )  =  2  c  −  1 due to symmetry. Here, X is just a random variable.Step 2Therefore, using the standard inequalities for confidence intervals, we obtain the following equalities                    (1)                                          x            ¯                          −        z                  σ                      n                          =                  μ          1                                        (2)                                          x            ¯                          +        z                  σ                      n                          =                  μ          2                .            Adding the two yields             x      ¯        =  (      μ    1    +      μ    2    )      /    2, or that                    (3)                    ∑                  x          i                =                  n          2                (                  μ          1                +                  μ          2                )        .            Similarly, performing   (  2  )  −  (  1  ) results in                    (4)                    z                  σ                      n                          =                                            μ              2                        −                          μ              1                                2                        ⟹                z        =                              (                          μ              2                        −                          μ              1                        )                          n                                            2            σ                              but recall that the unbiased estimator for the variance is                    (5)                              σ          2                =                              ∑                          x              i              2                        −                          1              n                                                      (                ∑                                  x                  i                                )                            2                                            n            −            1                          .            You now have enough information to calculate z, and thus   c  =  Φ  (  z  ).

Find confidence levels from given intervals. A set of 40 data items, produces a confidence interval for the population mean of 94.93<mu<105.07. If sum x^2=424375, find the confidence level.

Answered question

Answer & Explanation

New Questions in College Statistics