Question

In a simple random sample of 1600 people age 20 and over in a certain​ country, the proportion with a certain disease was found to be 0.110 ​(or 11.0​%).

Complete parts​ (a) through​ (d) below.

a. What is the standard error of the estimate of the proportion of all people in the country age 20 and over with the​ disease? SE est=Answer______ (Round to four decimal places as​ needed.)

b. Find the margin of​ error, using a​ 95% confidence​ level, for estimating this proportion. m= Answer______ ​(Round to three decimal places as​ needed.)

c. Report the​ 95% confidence interval for the proportion of all people in the country age 20 and over with the disease.

The​ 95% confidence interval for the proportion is Answer (______, ______) ​(Round to three decimal places as​ needed.)

d. According to a government​ agency, nationally, 15.4% of all people in the country age 20 or over have the disease. Does the confidence interval you found in part​ (c) support or refute this​ claim? Explain.

The confidence interval supports this​ claim, since the value Answer ( ______ ) is contained within the interval for the proportion. ​(Type an integer or a decimal. Do not​ round.)

Answer 1

Given,

Sample size : Number of people age 20 and over in the simple random sample : n=1600

Sample Proportion of people age 20 and over with certain disease : = 0.110

a.

standard error of the estimate of the proportion of all people in the country age 20 and over with the​ disease

Answer :

SEest = 0.0078

b. margin of​ error, using a​ 95% confidence​ level, for estimating this proportion.

Margin of error : m = Z/2 SEest

for 95% confidence level = (100-95)/100 =0.05

/2 = 0.05/2=0.025

Z/2 = Z0.025 = 1.96

Margin of error : m = Z/2 SEest = 1.96 x 0.0078 =0.015

Answer

m = 0.015

c. Report the​ 95% confidence interval for the proportion of all people in the country age 20 and over with the disease.

95% confidence interval for the proportion of all people in the country age 20 and over with the disease

= Sample proportion Margin of error = = 0.110.015 = (0.095,0.125)

The​ 95% confidence interval for the proportion is (0.095, 0.125) ​

(0.095, 0.125) ​

d. According to a government​ agency, nationally, 15.4% of all people in the country age 20 or over have the disease. Does the confidence interval you found in part​ (c) support or refute this​ claim? Explain.

15.4% of all people in the country age 20 or over have the disease i.e

Proportion of all people in the country age 20 or over have the disease: p = 15.4/100 =0.154

As 0.154 is outside the 95% confidence interval for the proportion ( 0.154 > 0.125) ​

The confidence interval does not support (refute)  this​ claim, since the value ( 0.154 ) is not contained within the interval for the proportion. ​