A random sample of 504 out of 1600 students in a school play soccer. How large should the sample of students be if a 90% confidence interval is desired with a margin of error of 2.2 percentage points for the proportion of students that play soccer in the school?

Ilnaus5

Ilnaus5

Answered question

2022-09-25

90% Confidence Interval
A random sample of 504 out of 1600 students in a school play soccer. How large should the sample of students be if a 90% confidence interval is desired with a margin of error of 2.2 percentage points for the proportion of students that play soccer in the school?
My main problem here is the interpretation of margin of error. The confidence interval I came up with was 504 16000 + 1.645 ( 8631 40000 ) n . The + and - back to back represent a "plus or minus". 1.645 is the respective z value for a 90% confidence interval and ( 8631 40000 ) = 504 1600 ( 1 504 1600 ) is the variance I calculated.
I thought then that I needed to find a value for 1.645 ( 8631 40000 ) n that was 2.2% of 504 1600 . This means that I got a final n value of approximately 12158, but I am not sure if this is what is required in this situation.

Answer & Explanation

rae2721

rae2721

Beginner2022-09-26Added 8 answers

Step 1
Your approach appears to be correct. You need to find estimates for the sample mean and sample variance and then set the right hand side of your equation ( 1.645 s n ) equal to .022 and solve. However you have made some calculation errors. Recheck your work and see if you can get the answer.
Step 2
Also note that you can check your answer. Does 1.645 8631 / 40000 12158 equal .022?

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