Confidence Interval for Population Mean Estimation

Question

The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systolic blood pressure is 34.3 mmHg (millimeters of mercury) for a sample of size 661 and a sample standard deviation 17.5 mmHg. How much of mmHg will lower for a typical patient's systolic blood pressure after taking the drug? Estimate with a 90% confidence.

Answer 1

In this problem, we are building a t-interval for a mean.

First, we should check that the conditions for inference have been met. Let's check them now. 1) The data comes from a simple random sample or well-designed experiment. Since the problem says "experimenter," we must assume that the experiment is appropriately randomized. 2) We need to know that the sample size is not more than 10% of the population size (N). Since n=661, as long as there are at least N=6610 people with high blood pressure in the population, we're good to go. 3) Finally, we need to know that the underlying population is normally distributed or that n>=30. Since n=661 >= 30, we meet this condition. Since all of the conditions for inference have been met, it is safe to proceed.

The formula for a single t-interval for a mean is xbar +/- t* s_x/sqrt(n), where xbar is the sample mean, t* is the critical value for a 90% CI, and s_x/sqrt(n) is the standard error. From the problem, we know that:

xbar = 34.3 mmHg

s_x/sqrt(n) = 17.5/sqrt(661) = 0.68067

The critical value t* for the problem is obtained from the inverse Student's t-distribution. Since we want a 90% CI, the two-tail probability is 10%, the one-tail probability is 5%, and the degrees of freedom are d.f.=n-1 = 660. Using a table, calculator, or Excel, we obtain t* = 1.64717

Putting everything together, we obtain 34.3 +/- 1.64717 (0.68067) = (33.18,35.42). This means we are 90% confident that the true average systolic blood pressure reduction lies in the given interval.

Let me know if you have any other questions.