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Sagot :
Using the z-distribution, as we are working with a proportion, it is found that the 95% confidence interval for the difference of the proportions is (-0.052, -0.0038).
What is the mean and the standard error of the distribution of differences?
For each sample, the mean and the standard error are given by:
[tex]p_C = 0.0256, s_C = \sqrt{\frac{0.0256(0.9744)}{586}} = 0.0065[/tex]
[tex]p_H = 0.0535, s_H = \sqrt{\frac{0.0535(0.9465)}{467}} = 0.0104[/tex]
Hence, for the distribution of differences, the mean and the standard error are given by:
[tex]p = p_C - p_H = 0.0256 - 0.0535 = -0.0279[/tex]
[tex]s = \sqrt{s_C^2 + s_H^2} = \sqrt{0.0065^2 + 0.0104^2} = 0.0123[/tex]
What is the confidence interval?
It is given by:
[tex]p \pm zs[/tex]
In which z is the critical value.
95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].
Then, the bounds of the interval are given as follows:
[tex]p - zs = -0.0279 - 1.96(0.0123) = -0.052[/tex]
[tex]p + zs = -0.0279 + 1.96(0.0123) = -0.0038[/tex]
The 95% confidence interval for the difference of the proportions is (-0.052, -0.0038).
More can be learned about the z-distribution at https://brainly.com/question/25890103
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