Question

Ask Your Teacher Recent studies have shown that about 20% of American adults fit the medical definition of being obese. A large medical clinic would like to estimate what percentage of their patients are obese, so they take a random sample of 100 patients and find that 16 are obese. Suppose that in truth, the same percentage holds for the patients of the medical clinic as for the general population, 20%. If the clinic took repeated random samples of 100 observations and found the sample proportion who were obese, into what interval should those sample proportions fall about 95% of the time? (Round your answers to two decimal places. Between _____ and _____

Answer 1

Solution

Given that,

p = 0.16

1 - p = 1 - 0.16 = 0.84

n = 100

= p = 0.16

=  [p( 1 - p ) / n] = [(0.16 * 0.84) / 100] = 0.0367

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 * 0.0367 + 0.16

  = 0.09

Using z-score formula,

  = z *   +  

  = 1.96 * 0.0367 + 0.16

  = 0.23

95% of the time Between 0.09  and 0.23