Question

About 90% of young adult Internet users (ages 18 to 29) use social-networking sites.

(a) Suppose a sample survey contacts an SRS of 1400 young adult Internet users and calculates the proportion pˆp^ in this sample who use social-networking sites.

What is the approximate distribution of pˆp^?

μpˆμp^ (±±0.00001) = , σpˆσp^ (±±0.00001) =

What is the probability that pˆp^ is between 87% and 93%? This is the probability that pˆp^ estimates pp within 3%.

(Use software.)

P(0.87<pˆ<0.93)P(0.87<p^<0.93) (±±0.0001) =

(b) If the sample size were 5600 rather than 1400, what would be the approximate distribution of pˆp^?

μpˆμp^ (±±0.00001) = , σpˆσp^ (±±0.00001) =

What is the probability that pˆp^ is between 87% and 93%? This is the probability that pˆp^ estimates pp within 3%.

(Use software.)

P(0.87<pˆ<0.93)P(0.87<p^<0.93) (±±0.0001) =

Answer 1

Solution

Given that,

p = 0.90

1 - p = 1 - 0.90 = 0.10

a) n = 1400

= p = 0.90

  [p ( 1 - p ) / n] = [(0.90 * 0.10) / 1400 ] = 0.00802

P( 0.87 < < 0.93 )

= P[(0.87 - 0.90) / 0.00802 < ( - ) / < (0.93 - 0.90) / 0.00802 ]

= P(-3.74 < z < 3.74)

= P(z < 3.74) - P(z < -3.74)

Using z table,   

= 0.9999 - 0.0001

= 0.9998

b) n = 5600

= p = 0.90

  [p ( 1 - p ) / n] = [(0.90 * 0.10) / 5600 ] = 0.00401

P( 0.87 < < 0.93 )

= P[(0.87 - 0.90) / 0.00401 < ( - ) / < (0.93 - 0.90) / 0.00401 ]

= P(-7.48 < z < 7.48)

= P(z < 7.48) - P(z < -7.48)

Using z table,   

= 1 - 0

= 1