Question

The article "Hospitals Dispute Medtronic Data on Wires" (The Wall Street Journal, February 4, 2010) describes several studies of the failure rate of defibrillators used in the treatment of heart problems. In one study conducted by the Mayo Clinic, it was reported that failures within the first 2 years were experienced by 18 of 89 patients under 50 years old and 13 of 362 patients age 50 and older. Assume that these two samples are representative of patients who receive this type of defibrillator in the two age groups. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of patients under 50 years old who experience a failure within the first 2 years. b. Construct and interpret a \(99 \%\) confidence interval for the proportion of patients age 50 and older who experience a failure within the first 2 years. c. Suppose that the researchers wanted to estimate the proportion of patients under 50 years old who experience this type of failure with a margin of error of \(0.03 .\) How large a sample should be used? Use the given study results to obtain a preliminary estimate of the population proportion.

Short answer

a) The 95% confidence interval for patients under 50 who experience a failure within the first 2 years is (0.124, 0.280). b) The 99% confidence interval for patients over 50 who experience a failure within the first 2 years is (0.014, 0.058). c) To estimate the proportion of patients under 50 who experience a failure within the first 2 years with a margin of error of 0.03, a sample size of 267 should be used.

Step-by-step solution

Calculate the sample proportions

For each age group, the sample proportion of failures (p) is calculated as \n\n For patients under 50: \(p_1 = 18 / 89 = 0.202\)\n For patients 50 and over: \(p_2 = 13 / 362 = 0.036\)

Construct 95% confidence interval for proportion of patients under 50

Using the formula for confidence interval construction, which is \(p ± (Z * sqrt(p(1-p)/n))\), the confidence interval calculation for group under 50 would be as follows: \n\n \(0.202 ± (1.96 * sqrt((0.202*0.798)/89)) = 0.202 ± 0.078\)\n\n Thus, the 95% confidence interval for the proportion of patients under 50 who experience a failure within the first 2 years is \((0.124, 0.280)\). This means, we're 95% confident that the true proportion of patients under 50 who experience a failure within the first 2 years lies between 12.4% and 28%.

Construct 99% confidence interval for proportion of patients 50 and over

Applying the same formula (replacing Z with 2.575 for a 99% confidence interval), the confidence interval calculation for group 50 and over would be as follows: \n\n \(0.036 ± (2.575 * sqrt((0.036 * 0.964) / 362)) = 0.036 ± 0.022 \)\n\n Thus, the 99% confidence interval for the proportion of patients 50 and older who experience a failure within the first 2 years is \((0.014, 0.058)\). This means, we're 99% confident that the true proportion of patients 50 and older who experience a failure within the first 2 years lies between 1.4% and 5.8%.

Estimate the required sample size

Given a margin of error \(E = 0.03\) and using the formula to estimate the necessary sample size which is \(n = (Z² * p * (1 - p)) / E²\). Here we will use our obtained preliminary estimate of the population proportion (0.202) and a Z-value for a 95% confidence level (1.96). \n\n \(n = (1.96² * 0.202 * (1 - 0.202)) / 0.03² = 266.97\)\n\n The sample size must be a whole number and it's always rounded up, so the required sample size is 267.