Respuesta :
Answer:
95% confidence interval for the average height of all NCAA D-I players is [68.61 , 75.54].
Step-by-step explanation:
We are given that in a random sample of 10 players, the sample average is 72.075 inches with a standard deviation of 4.8507 inches.
So, the pivotal quantity for 95% confidence interval for the average height is given by;
      P.Q. = [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample average height = 72.075 inches
       s = sample standard deviation = 4.8507 inches
       n = sample of players = 10
       [tex]\mu[/tex] = population mean height
So, 95% confidence interval for the population mean height, [tex]\mu[/tex] is ;
P(-2.262 < [tex]t_9[/tex] < 2.262) = 0.95
P(-2.262 < [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] < 2.262) = 0.95
P( [tex]-2.262 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]{\bar X - \mu}[/tex] < [tex]2.262 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.95
P( [tex]\bar X - 2.262 \times {\frac{s}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]\bar X + 2.262 \times {\frac{s}{\sqrt{n} }[/tex] ) = 0.95
95% confidence interval for [tex]\mu[/tex] = [ [tex]\bar X - 2.262 \times {\frac{s}{\sqrt{n} }[/tex] , Â [tex]\bar X + 2.262 \times {\frac{s}{\sqrt{n} }[/tex] ]
                        = [ [tex]72.075 - 2.262 \times {\frac{4.8507}{\sqrt{10} }[/tex] , [tex]72.075 + 2.262 \times {\frac{4.8507}{\sqrt{10} }[/tex] ]
                         = [68.61 , 75.54]
Therefore, 95% confidence interval for the average height of all NCAA D-I players is [68.61 , 75.54].
