Question

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

Answer #1

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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