Question

A study reports that 36​% of companies in Country A have three or more female board directors. Suppose you select a random sample of 100 respondents.

c. The probability is 95​% that the sample percentage of Country A companies having three or more female board directors will be contained within what symmetrical limits of the population​ percentage? The probability is 95​% that the sample percentage will be contained above nothing​% and below nothing​%. ​(Round to one decimal place as​ needed.)

Answer 1

Solution

Given that,

p = 36% = 0.36

1 - p = 1 - 0.36 = 0.64

n = 100

= p = 0.36

=  [p( 1 - p ) / n] = [(0.36 * 0.64) / 100] = 0.048

Using standard normal table,

P( -z < Z < z) = 95%

= P(Z < z) - P(Z <-z ) = 0.95

= 2P(Z < z) - 1 = 0.95

= 2P(Z < z) = 1 + 0.95

= P(Z < z) = 1.95 / 2

= P(Z < z) = 0.975

= P(Z < 1.96) = 0.975

= z  ± 1.96

Using z-score formula,

  = z *   +  

  = -1.96 * 0.048 + 0.36

  = 0.266

  = 26.6%

Using z-score formula,

  = z *   +  

  = 1.96 * 0.048 + 0.36

  = 0.454

  = 45.4%

The probability is 60% that the sample percentage will be contained above 26.6% and below 45.4%