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.
Margin of error = (1.96)[√(pq/n)]
...where p = .33, q = 1 - p, and n = 490
Note: 1.96 represents 95% confidence interval.
Plug the values into the formula and calculate.
I let you take it from here.
4.16%
To find the margin of error for the 95% confidence interval, we need the following information:
1. Sample size (n): The number of college students surveyed, which is given as 490.
2. Proportion (p): The proportion of college students who cheat on exams, which is given as 33% or 0.33.
The margin of error (E) can be calculated using the formula:
E = Z * sqrt((p * (1 - p)) / n)
where:
- Z is the Z-score corresponding to the desired confidence level. For a 95% confidence interval, the Z-score is approximately 1.96.
- sqrt is the square root function.
Let's plug in the values and calculate the margin of error:
E = 1.96 * sqrt((0.33 * (1 - 0.33)) / 490)
First, we calculate the inside part of the square root:
0.33 * (1 - 0.33) = 0.33 * 0.67 = 0.2211
Then, we calculate the square root:
sqrt(0.2211 / 490) ≈ 0.0174
Finally, we multiply by the Z-score:
E = 1.96 * 0.0174 ≈ 0.034
Therefore, the margin of error for the 95% confidence interval is approximately 0.034 or 3.4%.