An article in Medicine and Science in Sports and ExerciseSol

An article in Medicine and Science in Sports and Exercise

Solution

As the sessions were carried out 3 times per week for 3 weeks on 17 players, sample size

= 3x3x17 = 153

A sample of 153 units has a variance ?2 of 0.09.

First find degrees of freedom:
Degrees of Freedom = n - 1
153 - 1 = 152

Calculate ?:
? = 1 - confidence% = 0.05

Find low end confidence interval value:
?_{low end} = ?/2
?_{low end} = 0.05/2
?_{low end} = 0.025

Find low end ?² value for 0.025
?²_0.025 = 188.0262

Low End = ?(152)(0.09)/188.0262) = ?13.68/188.0262
= ?0.0727558180722= 0.2697

Find high end confidence interval value:
?_{high end} = 1 - ?/2
?_{high end} = 1 - 0.05/2
?_{high end} = 0.975

Find high end ?² value for 0.975
?²_0.975 = 119.7587

Calculate high end confidence interval total:
High End = Square Root((n - 1)s²/?²_{1 - ?/2})
= ?(152)(0.09)/119.7587) = ?13.68/119.7587
= ?0.114229696882
= 0.338

Thus conf interval for sigma is
0.2697 < ? < 0.338 <---- This is our 95% confidence interval
What this means is if we repeated experiments, the proportion of such intervals that contain ? would be 95%