# 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(1-p)/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)

1. 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
2. 0.0427

posted by Ashworth

First Name

## Similar Questions

1. ### MATH##\$\$&&^^**

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%
2. ### math

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%
3. ### math

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%
4. ### math

A researcher wishes to estimate the proportion of college students who cheat on exams. A poll of 400 college students showed that 33% of them had or intended to, cheat on examinations. Find the margin of error for the 95%
5. ### math

A researcher wishes to estimate the proportion of college students who cheat on exams. A poll of 560 college students showed that 27% of them had, or intended to, cheat on examinations. Find the 95% confidence interval. A. 0.2323
6. ### Math-Statistics

A marketing research company needs to estimate which of two soft drinks college students prefer. A random sample of n college students produced the following 98% confidence interval for the proportion of college students who
7. ### Statistics

In a random sample of 50 undergraduate students at a college, it was found that 44 students regularly access social networking websites from their college library. What is the margin of error for the true proportion of all
8. ### Statistics

What percentage of college students are attending a college in the state where they grew up? Let p be the proportion of college students from the same state as that in which the college resides. If no preliminary study is made to
9. ### Statistics

A random sample of 14 college female students revealed their average height was 66.1 inches. a) How many students should be in a samle if a 95% confidence interval is to have a margin of error of only plus/minus 1 inch? Is that
10. ### statistics

A researcher wishes to estimate the proportion of adults living in rural areas who own a gun. He wishes to achieve a margin of error of 2.8%. What is the minimum sample size needed?

More Similar Questions