Question

The effectiveness of a blood-pressure drug is being investigated. An experimenter finds that, on average, the reduction in systolic blood pressure is 78.7 for a sample of size 529 and standard deviation 12.3 Estimate how much the drug will lower a typical patient's systolic blood pressure (using a 80% 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

Using a sample mean of 78.7, a standard deviation of 12.3, and a sample size of 529, the 80% confidence interval for the typical reduction in systolic blood pressure is estimated to be 78.0 < μ < 79.4.

Step-by-step explanation:

To estimate how much the drug will lower a typical patient's systolic blood pressure using an 80% confidence level, we use the sample mean (μ), the standard deviation (σ), and the sample size (n). The sample mean is given as 78.7, the standard deviation is 12.3, and the sample size is 529.

We need to find the Z-score that corresponds to an 80% confidence level. This Z-score typically represents the cutoff point on a standard normal distribution for the middle 80%, with 10% in each tail. To find the Z-score that corresponds to the confidence level, we refer to a standard normal distribution table or use statistical software. In this case, for an 80% confidence level, the Z-score is approximately 1.282.

To create the confidence interval, we use the formula:

Confidence Interval = Sample Mean ± (Z-score * (Standard Deviation / sqrt(n)))

Plugging in the numbers:

Confidence Interval = 78.7 ± (1.282 * (12.3 / sqrt(529)))

Simplifying further:

Confidence Interval = 78.7 ± (1.282 * (12.3 / 23))

Confidence Interval = 78.7 ± (1.282 * 0.535)

Confidence Interval = 78.7 ± 0.7

Therefore, the 80% confidence interval estimate for the typical reduction in systolic blood pressure is:

78.0 < μ < 79.4