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)

Aryanna Fisher

Aryanna Fisher

Answered question

2022-11-15

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 = 0.157
Find 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.

Answer & Explanation

mentest91k99

mentest91k99

Beginner2022-11-16Added 17 answers

Step 1
The setup is
H 0 : average lung capacity of players equals population average H 1 : average lung capacity of players > population average
Step 2
So, 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 > 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 6.36
which is half of the team-size; the coach has more than enough "evidence" 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 "larger" lungs, on average, than that of the population.

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