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# Find the sample size with a margin of error less than 0.04.?

Cora wants to determine a 99 percent confidence interval for the true proportion of high school students in the area who attend their home basketball games. How large of a sample must she have to get a margin of error less than 0.04? [Note that you don't have an estimate for p*!]

[Round to the smallest integer that works.] n =

### 1 Answer

- AlanLv 72 years agoFavorite Answer
Margin of Error will be the largest when

p*(1-p) is a maximum

This occurs when p = 1-p

2p =1

p =0.50

p= 0.50

Source of formula:

https://stattrek.com/estimation/confidence-interva...

standard deviation = sqrt( (p*(1-p)/ n) = sqrt(0.50*0.50/n) = sqrt(0.25)/sqrt(n) =1/(2*sqrt(N))

so for a 99 percent confidence interval

You go from 0.005 to 0.995 to cover 99 %

P(z< Z) = 0.995

From the following z-table

https://www.stat.tamu.edu/~lzhou/stat302/standardn...

P(z< 2.57) = .99492

P(z< 2.58) = .99506

so interpolating

=2.57 + ( 0.995-0.99492)*(0.01)/ (0.99506-0.99492) = 2.576 (when rounded to 3

digits after the decimal)

so

Margin of error = z_critical*standard deviation = 2.576*1/ (2*sqrt(N))

0.04 = 2.576/(2*sqrt(N) )

0.08*sqrt(N) = 2.576

sqrt(N) = 2.576 /0.08 = 32.2

N = 32.2^2 =1036.84

so the smallest integer is

N = 1037

=== checking

Margin of Error for 1036 = 2.576*sqrt(0.25/ 1036) = 0.0400162129 (too high)

Margin of Error for 1037 = 2.576*sqrt(0.25/1037) = 0.0399969141 (below 0.04- winner)