Confidence Intervals

Question

1.    The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systolic blood pressure is 39.4 for a sample of size 1031 and standard deviation 6.7.

Estimate how much the drug will lower a typical patient's systolic blood pressure (using a 95% confidence level).

Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

__________ < μ < _________

Answer should be obtained without any preliminary rounding.

Answer 1

Hello Diana,

Let X represent the amount of reduction in systolic blood pressure for a randomly selected patient. We wish to produce a confidence interval with confidence level 95% to estimate μ (the population mean of X). The following information is given in the problem:

Sample Mean = X_m = 39.4

Sample Standard Deviation = s = 6.7

Sample Size = n = 1031

Since the sample size is large, we can base the confidence interval on the normal distribution. The confidence interval can be expressed as

X_m - z_(α/2)·s/√n < μ < X_m + z_(α/2)·s/√n

In the above expression,

α = 1 - .95 = .05

(The .95 comes from expressing the confidence level 95% in decimal form.) We then need to find the critical value z_(α/2) = z_(.05/2) = z_(.025). By definition, z_(.025) is the value of a standard normal random variable Z such the area to the right of z_(.025) is .025 Using a table of the standard normal distribution (or appropriate computer software), we find that

z_(.025) = 1.96

Finally, substituting the necessary values into the formula, we obtain

39.4 - 1.96·6.7/√1031 < μ < 39.4 + 1.96·6.7/√1031

39.4 - .40898 < μ < 39.4 + .40898

38.99102 < μ < 39.80898

39.0 < μ < 39.8 (rounded to the nearest tenth.)

Hope that helps. Let me know if you need any clarification.

William