A survey of 50 U.S. adults and it was found that 5 planned to
file their taxes today (April 15).
(a) Find a 95% CI for for p (the proportion of US adults filing
their their taxes today today.)
(b) How large should the sample be if we are going to have a CI of
radius less than 0.01
Solution :
Given that,
n = 50
x = 5
a) Point estimate = sample proportion = = x / n = 5 / 50 = 0.10
1 - = 1 - 0.10 = 0.90
At 95% confidence level
= 1 - 95%
=1 - 0.95 =0.05
/2
= 0.025
Z/2
= Z0.025 = 1.960
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.96 (((0.10 * 0.90) / 50)
= 0.083
A 95% confidence interval for population proportion p is ,
± E
= 0.10 ± 0.083
= ( 0.017, 0.183 )
b) margin of error = E = 0.01
sample size = n = (Z / 2 / E )2 * * (1 - )
= (1.96 / 0.01 )2 * 0.10 * 0.90
= 3457.44
sample size = n = 3458
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