x. Solution p_actual = 17 / 50 = .34 SE = sqrt(p * (1-p) / n) = sqrt(.3 * .7 / 5

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x.

Explanation / Answer

p_actual = 17 / 50 = .34 SE = sqrt(p * (1-p) / n) = sqrt(.3 * .7 / 50) = .0648 z = (.34 - .3) / .0648 = .617 --> p = .7324 Using =.05, we have tails of .05/2 = .025. Since .7324< 1 - .025 = .9725, we cannot reject the null hypothesis. 95% confidence interval --> z=1.96. Thus, the confidenceinterval = mean ± z*SE = .34 ± 1.96 * .0648 = .34± .127 = [.213, .467] Since .34 is contained in the interval, we cannot reject H0. To find the interval with an error of 2% or less, we need to find nsuch that 1.96 * sqrt(p * (1-p) / n) = .02 for n. I'll leavethis bit of algebra to you. If we don't know p, we assume the worst case scenario. p = .5gives the largest standard error (this can be easily proven usingelementary calculus). Thus, do the same thing as in d, butuse p = .5.

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