Question

The effectiveness of a blood-pressure drug is being investigated. an experimenter finds that, on average, the reduction in systolic blood pressure is 47.3 for a sample of size 878 and standard deviation 15.9. 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).

Answer 1

Using the z-distribution, the estimate for how much the drug will lower a typical patient's systolic blood pressure is:

$46.6 \leq \mu \leq 48$

What is a z-distribution confidence interval?

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

In which:

• $\overline{x}$ is the sample mean.

• z is the critical value.

• n is the sample size.

• $\sigma$ is the standard deviation for the population.

In this problem, we have a 80% confidence level, hence$\alpha = 0.8$, z is the value of Z that has a p-value of $\frac{1+0.8}{2} = 0.9$, so the critical value is z = 1.28.

The other parameters are given by:

$\overline{x} = 47.3, \sigma = 15.9, n = 878$

Then the bounds of the interval are:

$\overline{x} - z\frac{\sigma}{\sqrt{n}} = 47.3 - 1.28\frac{15.9}{\sqrt{878}} = 46.6$

$\overline{x} + z\frac{\sigma}{\sqrt{n}} = 47.3 + 1.28\frac{15.9}{\sqrt{878}} = 48$

Hence the interval is:

$46.6 \leq \mu \leq 48$

More can be learned about the z-distribution at brainly.com/question/25890103

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