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To obtain a 90% confidenceintervalwith an approximatemargin of errorof 0.04, the student organization would need tosurveyapproximately 6,808students.In order to determine thesamplesize required to obtain a 90% confidence interval with an approximate margin of error of 0.04, we can use the formula:where n is the sample size, Z is the Z-score for the desired confidence level (in this case, Z = 1.65 for a 90% confidence level), p is the estimated proportion of thepopulationthat would support the proposal (assuming p = 0.5 for simplicity), and E is the desiredmarginof error (E = 0.04 in this case).Plugging in the values, we have:n = (2.7225 * 0.25) / 0.0016n = 10.890625 / 0.0016n = 6,807.890625For more such questions onErrorbrainly.com/question/29397166#SPJ4...