Question

A corrections researcher is examining prison health care coasts and wants to determine how long inmates survive once diagnosed with a particular form of cancer. Using data collected from a group of 45 patients with the disease, she observes an average survival time (time until death) of 32 months with a standard deviation of 9 months. Using a 95 percent level of confidence, estimate the population mean survival time. What would the 99% confidence interval be?

(6 pts total; 2pt each CI, 1pt each interpretation)

1. 95% Confidence Interval:

2. Interpretation:

3. 99% Confidence Interval:

4. Interpretation:

Answer 1

a)

sample mean, xbar = 32
sample standard deviation, s = 9
sample size, n = 45
degrees of freedom, df = n - 1 = 44

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.015

ME = tc * s/sqrt(n)
ME = 2.015 * 9/sqrt(45)
ME = 2.7

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (32 - 2.015 * 9/sqrt(45) , 32 + 2.015 * 9/sqrt(45))
CI = (29.2966 , 34.7034)

Therefore, based on the data provided, the 95% confidence interval for the population mean is 29.2966 < μ < 34.7034 which indicates that we are 95% confident that the true population mean μ is contained by the interval (29.2966 , 34.7034)

b)
sample mean, xbar = 32
sample standard deviation, s = 9
sample size, n = 45
degrees of freedom, df = n - 1 = 44

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.692

ME = tc * s/sqrt(n)
ME = 2.692 * 9/sqrt(45)
ME = 3.61

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (32 - 2.692 * 9/sqrt(45) , 32 + 2.692 * 9/sqrt(45))
CI = (28.3883 , 35.6117)

Therefore, based on the data provided, the 99% confidence interval for the population mean is 28.3883 < μ < 35.6117 which indicates that we are 99% confident that the true population mean μ is contained by the interval (28.3883 , 35.6117)