Question

The number of wins in a season for 32 randomly selected professional football teams arelisted below. Construct a 90% confidence interval for the true mean number of wins in a season.
Mean(x) = sum of 32 observations / 32 = 7.97



Standard deviation( sd )=2.62



Sample Size(n)=32



Confidence Interval = [ 7.97 ± Z a/2 ( 2.62/ Sqrt ( 32) ) ] = [ 7.97 - 1.64 * (0.4632) , 7.97 + 1.64 * (0.4632) ] = [ 7.2,8.7 . What are the steps used to solve this? How did they find their Standard Deviation? Also what does the Z stand for in the confidence interval?

Answer 1

Answer:

A)CI 90 % = [ 7,2103 ;  8,7297]

B) s = √ ∑(xi - xp)²/n-1

c) z is the score for the given significance level in this case is 1,64

Step-by-step explanation:

For a confidence Interval of 90%, the significance level is α = 10 %

α = 0,1     α/2 = 0,05  and from z-table we find  z(c) for 0,05   z(c) = 1,64

CI 90 % = [ x  ±  z(c) * s/√n ]

CI 90 % = [ 7,97 ±  1,64 * 2.62/√32]

CI 90 % = [ 7,97 ± 0,7597]

CI 90 % = [ 7,2103 ;  8,7297]

B) To calculate the standard deviation of the sample the equation is:

s = √ ∑(xi - xp)²/n-1

xi - xp  is the difference between each one of the elements of the sample

xp is the mean or average of the elements of the sample and n is the sample size