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

William P.

08/20/19

Diana D.

thank you! I will be sending a message very soon!08/13/19