Question

A health journal found that

56.2​%

of college students who lived in coed dormitories consumed alcohol weekly​ (compared with​ 26.5% who lived in​ single-sex dormitories). A random sample of 175

students who live in coed dormitories was selected. Complete part d below.

d. Identify the symmetric interval that includes 95​% of the sample proportions with a sample size of 175 if the true population proportion of students in coed dormitories who consume alcohol weekly is 56.2​%.

(%), (%)round to one decimal

Answer 1

Solution

Given that,

p = 56.2% = 0.562

1 - p = 1 - 0.562 = 0.438

n = 175

= p = 0.562

=  [p( 1 - p ) / n] = [(0.562 * 0.438) / 175] = 0.0375

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 * 0.0375 + 0.562

  = 0.489

  = 48.9%

Using z-score formula,

  = z *   +  

  = 1.96 * 0.0375 + 0.562

  = 0.636

  = 63.6%

The probability is 95% that the sample percentage will be contained above 48.9% and below 63.6%