statistics
A researcher wishes to estimate the proportion of college students who cheat on exams. A poll of 490 college students showed that 33% of them had, or intended to, cheat on examinations. Find the margin of error for the 95% confidence interval.
solution I have but what do I need to omit?
Poll = 490 college students
33% revealed they had, intended to, and cheat on exams
Find the ME for 95% interval.
Solution: The critical value = 1.96 but it has to been rounded to 2 to calculate the 95% confidence interval. However, the 95% CI formula for proportion is p̂ ± 2 X √p̂(1p)/n.
The standard deviation = √ (0.33 * 0.67* 490)
= 10.41
E = (zs)/ √ (n) = 196 * 10.41)/ √ (490)
= 0.92
ME = (1.96) [√pq/n)]
= (1.96) [√ (0.33 * 0.67)/ 490]
= (1.96) [√0.00045]
= 1.96 * 0.021
= 1.981 ≈ 1.96
The sample proportion, p̂ = 33% or 0.33 is the best estimate of the population proportion. The formula used to determine the margin of error
is: E ≈ 2 √ [p̂ (1 p̂) / n]
E ≈ 2 √ [.33(1 .33) / 490]
E ≈ 2 √ (.2211 / 490)
E ≈ 2 √ 4.5122
E ≈ 2 * 0.02124
E ≈ 0.4248 (Answer)

95% confidence interval is equivalent to z = 1.96, so if you round to 2, then your calculations for E are almost correct. I think you missed a 0; I get E = 0.04248. The formula used to determine the margin of error is all you should need to answer the problem.
posted by MathGuru

0.0427
posted by Ashworth
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