A coach has made a statement that his players have bigger lung capacity than the average of the population of the same age which is 3.4. (Normal distribution)The measurements yield the following data: 3.4, 3.6, 3.8, 3.3, 3.4, 3.5, 3.7, 3.6, 3.7, 3.4 and 3.6. n = 11 X ¯ = 3.545 S = 0.157Find the required sample size, which lung capacity should be measured, so coach can state his statement with 99% confidence. (assume σ 2 = 0.09)I don't even know how should I start. My initial thought was to use the U statistics U = X ¯ − μ σ n ~ N ( 0 , 1 ). But I don't know the U.

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A coach has made a statement that his players have bigger lung capacity than the average of the population of the same age which is 3.4. (Normal distribution)The measurements yield the following data: 3.4, 3.6, 3.8, 3.3, 3.4, 3.5, 3.7, 3.6, 3.7, 3.4 and 3.6.  n  =  11            X      ¯        =  3.545  S  =  0.157Find the required sample size, which lung capacity should be measured, so coach can state his statement with 99% confidence. (assume       σ    2    =  0.09)I don&#039;t even know how should I start. My initial thought was to use the U statistics   U  =                              X          ¯                    −      μ        σ        n  ~  N  (  0  ,  1  ). But I don&#039;t know the U.

mentest91k99 · Accepted Answer

Step 1The setup is                              H          0                                    :                              average lung capacity of players equals population average                                              H          1                                    :                              average lung capacity of players &amp;gt; population average                    Step 2So, since we can define   Z      ∼      N  (  0  ,  1  ) where   Z      :=                                              X            ¯                          −        μ                    σ                  /                          n                    , in order to reject       H    0   we need a sample-size n such that  P      (    Z            &amp;gt;                  Z              0.01              )        =      0.01    .      Z          0.01            ≈      2.33 can be determined from Standard-Normal tables. Thus, for             X      ¯            =      3.545,   μ      =      3.4 and       σ    2        =      0.09, n should be, at least,  n      =                (                                                  Z                              0.01                                      0.3                                3.545            −            3.4                              )        2        ≈      23.24    ,about twice the number of players the coach has at present! But, of course, if the sample variance is used (and, why not, since this should be indicative of how similar the team members are) instead of the population variance, we have, instead,  n      ≈      6.36which is half of the team-size; the coach has more than enough &quot;evidence&quot; in this regard. This is not surprising: the coach was selective in choosing his players, in that they display top physical ability and, therefore, might be more likely to have &quot;larger&quot; lungs, on average, than that of the population.

A coach has made a statement that his players have bigger lung capacity than the average of the population of the same age which is 3.4. (Normal distribution)

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