In a simple random sample of 800 people age 20 and over in a certain country, the proportion with a certain disease was found to be 0.150 (or 15.0%). 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? SEest = (Round to four decimal places as needed.) b. Find the margin of error, using a 95% confidence level, for estimating this proportion. m = (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 (Round to three decimal places as needed.) d. According to a government agency, nationally, 16.2% 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 this claim, since the value contained within the interval for the proportion. (Type an integer or a decimal. Do not round.)