Question

you are a travel agent and wish to estimate with 90% confidence in the proportion of vacationers who use an online service or the internet to make travel reservations.
your estimate must be accurate to within 3% of the population proportion

a. find the minimum sample size needed to perform this estimation if you do not use any prior estimates for the sample proportion.

b.find the minimum sample size needed to perform this estimation if you assume that a prior estimate for the sample population is 0.82.

Answer 1

Solution :

Given that,

= 0.5 ( assume 0.5)

1 - = 1 - 0.5 = 0.5

margin of error = E =3 % = 0.03

At 90% confidence level

= 1 - 0.90 = 0.10

/2 =0.05

Z/2 = 1.645 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.645 / 0.03)2 * 0.5 * 0.5

=751.67

Sample size =752

(B)

Solution :

Given that,

= 0.82

1 - = 1 - 0.82= 0.18

margin of error = E =3 % = 0.03

At 90% confidence level

= 1 - 0.90 = 0.10

/2 =0.05

Z/2 = 1.645 ( Using z table )

Sample size = n = (Z/2 / E)2 * * (1 - )

= (1.645 / 0.03)2 * 0.82 * 0.18

=443.78

Sample size =444