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Sagot :
Using the z-distribution, it is found that the 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment) is (-0.234, -0.024).
What is a t-distribution confidence interval?
The confidence interval is:
[tex]\overline{x} \pm zs[/tex]
In which:
- [tex]\overline{x}[/tex] is the sample mean.
- z is the critical value.
- s is the standard error.
In this problem, we are given that z = 2.576, s = 0.0407. The sample mean is the difference of the proportions, hence:
[tex]\overline{x} = \frac{28}{162} - \frac{82}{272} = -0.129[/tex]
Then, the bounds of the interval are given by:
[tex]\overline{x} - zs = -0.129 - 2.576(0.0407) = -0.234[/tex]
[tex]\overline{x} + zs = -0.129 + 2.576(0.0407) = -0.024[/tex]
The 99% confidence interval to estimate the treatment effect of buproprion (placebo-treatment) is (-0.234, -0.024).
More can be learned about the z-distribution at https://brainly.com/question/25890103
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