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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).

Sagot :

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

[tex]46.6 \leq \mu \leq 48[/tex]

What is a z-distribution confidence interval?

The confidence interval is:

[tex]\overline{x} \pm z\frac{\sigma}{\sqrt{n}}[/tex]

In which:

  • [tex]\overline{x}[/tex] is the sample mean.
  • z is the critical value.
  • n is the sample size.
  • [tex]\sigma[/tex] is the standard deviation for the population.

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

The other parameters are given by:

[tex]\overline{x} = 47.3, \sigma = 15.9, n = 878[/tex]

Then the bounds of the interval are:

[tex]\overline{x} - z\frac{\sigma}{\sqrt{n}} = 47.3 - 1.28\frac{15.9}{\sqrt{878}} = 46.6[/tex]

[tex]\overline{x} + z\frac{\sigma}{\sqrt{n}} = 47.3 + 1.28\frac{15.9}{\sqrt{878}} = 48[/tex]

Hence the interval is:

[tex]46.6 \leq \mu \leq 48[/tex]

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

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