Using the z-distribution and the information given, it is found that the 89% confidence interval for the difference in the proportions that get sick with Selttiks vs the placebo is (-0.07, -0.002).
The proportions, and it's respective standard errors, are given by:
[tex]p_s = 0.13, s_s = \sqrt{\frac{0.13(0.87)}{560}} = 0.0142[/tex]
[tex]p_p = 0.166, s_p = \sqrt{\frac{0.166(0.834)}{560}} = 0.0157[/tex]
The distribution of the difference has mean and standard error given by:
[tex]\pi = p_s - p_p = 0.13 - 0.166 = -0.036[/tex]
[tex]s = \sqrt{s_s^2 + s_p^2} = \sqrt{0.0142^2 + 0.0157^2} = 0.021[/tex]
The interval is:
[tex]\pi \pm zs[/tex]
The critical value using the z-distribution, for a 89% confidence interval, is z = 1.598, hence:
[tex]\pi - zs = -0.036 - 1.598(0.021) = -0.07[/tex]
[tex]\pi + zs = -0.036 + 1.598(0.021) = -0.002[/tex]
The 89% confidence interval for the difference in the proportions that get sick with Selttiks vs the placebo is (-0.07, -0.002).
A similar problem is given at https://brainly.com/question/16807970
