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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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