1. Medicaid health-assistance programs are administered by the individual states

Question

1. Medicaid health-assistance programs are administered by the individual states, even though part of the funding is federal. The federal government requires that the states perform regular audits in order to assure that payments are accurate. One Florida hospital was recently audited by the Florida Department of Health and Rehabilitative Services (HRS). The HRS wants to estimate the mean size of the hospital's claims to within plus or minus $1.24 using a 99% confidence interval. No information was available for the standard deviation so a pilot sample of 18 Medicaid claims was done to calculate a sample standard deviation of S11.34. How many more claims must be sampled to meet HRS goals in estimating the true mean claim for this hospital? HRS has a budget for sampling 500 claims. If they want the estimate to remain within plus or minus S1.24, how confident would they then be in their interval estimate? If HRS decides they would rather sample 500 claims and remain 99% confident, how wide does the confidence interval now need to be?

Explanation / Answer

Question 1

sample standard deviation s = $11.34

Here margin of error = critical test statistic * standard error of proportion

critical test statistic for 99% CI for degree of freedom (18 -1= 17) = t17,0.01 = 2.8982

standard error of proportion = s/sqrt(n) = 11.34/sqrt(n)

Margin of error = 1.24

1.24 = 2.8982 * 11.34/sqrt(n)

sqrt(n) = 2.8982 * 11.34/1.24 = 26.5045

n = 702.4888

n = 703

so number of more sampled to be taken = 703 - 18 = 685

(b) Here n = 500

so standard error of sample means = s/sqrt(n) = 11.34/sqrt(500) = 0.50714

margin of error = $ 1.24

1.24 = 0.50714 * Critical test statistic

Critical test statistic = 1.24/0.50714 = 2.4451

Here confidence inteval = TDIST (t > 2.4451 ; 499) = 0.015

so the confidence interval is 0.015

(c) Here n = 500

Critical test statistic = t499,0.01 = 2.5857

Standard error = s/sqrt(n) = 11.34/sqrt(500) = 0.50714

Width of confidence interval = 2 * margin of error = 2 * 2.5857 * 0.50714 = $ 2.6226

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