Question

In a simple random sample of 1000 people age 20 and over in a certain​ country, the proportion with a certain disease was found to be 0.160 ​(or 16.0​%). 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?

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

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

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

D. According to a government​ agency, nationally, 17.1​% 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 (refutes OR supports) this​ claim, since the value _ (is OR is not) contained within the interval for the proportion.

Answer 1

Solution:

We are given that: a simple random sample of 1000 people age 20 and over in a certain​ country, the proportion with a certain disease was found to be 0.160 ​(or 16.0​%).

That is: n = 1000 and

Part 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?

the standard error of the estimate of the proportion is given by:

Part B. Find the margin of​ error, using a​ 95% confidence​ level, for estimating this proportion.

We need to find z_c value for c=95% confidence level.

Find Area = ( 1 + c ) / 2 = ( 1 + 0.95) /2 = 1.95 / 2 = 0.9750

Look in z table for Area = 0.9750 or its closest area and find z value.

Area = 0.9750 corresponds to 1.9 and 0.06 , thus z critical value = 1.96

That is : Z_c = 1.96

Thus

Part 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:

Part D. According to a government​ agency, nationally, 17.1​% 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?

the confidence interval you found in part​ (c) support the claim, since the value 17.1% is contained within the interval for the proportion.