Question

Recent studies have shown that about 16% 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 200 patients and find that 28 are obese. Suppose that in truth, the same percentage holds for the patients of the medical clinic as for the general population, 16%. Give a numerical value for each of the following concepts.

(a) The population proportion, p, of obese patients in the medical clinic.
p =  

(b) The proportion of obese patients, p̂, for the sample of 200 patients.
p̂ =  

(c) The standard error, s.e., of p̂. (Round your answer to four decimal places.)
s.e.(p̂) =  

(d) The mean of the sampling distribution of p̂.
mean =  

(e) The standard deviation, s.d., of the sampling distribution of p̂. (Round your answer to four decimal places.)
s.d.(p̂) =

Answer 1

Solution:

Given that,

a )The population proportion, p

p = 16% = 0.16

1 - p = 1 - 0.16 = 0.84

b ) The proportion of obese patients, p̂,

n = 200

x = 28

= x / n = 28 / 200 = 0.14

1 - = 1 - 0.14 = 0.86

c )  The standard error, s.e. is s.e( ₎

s.e ( )=  P ( 1 - P ) / n

=   0.16 * 0.84 / 200

= 0.0259

The standard error, s.e. = 0.0259

d) The mean of the sampling distribution of p̂

=   = 0.55

e ) The standard deviation, s.d., of the sampling distribution of p̂

=   ( 1 - ) / n

=   0.14 * 0.86 / 200

= 0.0245

= 0.0245