Question

The report indicates that 18.7% of Boston residents live below the poverty threshold. In a sample of 50 residents, there is probability 0.90 that the sample proportion living below the poverty threshold will fall between what two values, symmetric around the mean?

Answer 1

Solution

Given that,

p = 18.7% = 0.187

1 - p = 1 - 0.187 = 0.813

n = 50

= p = 0.187

=  [p( 1 - p ) / n] = [(0.187 * 0.813) / 50] = 0.0551

Using standard normal table,

P( -z < Z < z) = 90%

= P(Z < z) - P(Z <-z ) = 0.90

= 2P(Z < z) - 1 = 0.90

= 2P(Z < z) = 1 + 0.90

= P(Z < z) = 1.90 / 2

= P(Z < z) = 0.95

= P(Z < 1.645) = 0.95

= z  ± 1.645

Using z-score formula,

  = z *   +  

  = -1.645 * 0.0551 + 0.187

  = 0.096

  = 9.6%

Using z-score formula,

  = z *   +  

  = 1.645 * 0.0551 + 0.187

  = 0.278

  = 27.8%

The probability is 90% that the sample percentage will be contained above 9.6% and below 27.8%