Question

Ivan surveyed 49 randomly selected players from the 134 players of the soccer club teams to see if they wanted the games to be played on Saturdays or Sundays. Twenty-two of the players said that they preferred that the games be played on Saturdays. Ivan correctly determined that the margin of error, E, of his survey using a 99% confidence interval (z*score 2.58) is approximately 18% What is the 99% confidence level for players who prefer the games be played on Saturdays?

Answer 1

Twenty-two of the players said that they preferred that the games be played on Saturdays. Ivan correctly determined that the margin of error, E, of his survey using a 99% confidence interval (z*score 2.58) is approximately 18%

Ivan surveyed 49 randomly select

Answer 2

`0.269,0.629`

Explanation:

Ivan surveyed randomly selected 49 players out of 134.

22 of the players said that they preferred that the games be played on Saturdays.

Here,

p = proportion =
`(22)/(49)`,

n = sample size = 49,

We know that,

`\text{M.E}=Z_(critical)\cdot \sqrt{(p(1-p))/(n)}`

It is given as 18% or 0.18

We also know that, confidence interval will be

`=p\pm \text{M.E}`

Putting the values,

`=(22)/(49)\pm 0.18`

`=0.449\pm 0.18`

`=0.269,0.629`