A survey was conducted on a random sample of students at a college campus to study student opinions on the new honors program. Out of the students surveyed, 78% rated the new honors program as "very rigorous" with a margin of error of ±3% and a confidence interval of 90%.

What does the margin of error imply?

1. It can be concluded, with 90% confidence, that at least 75% of all students will rate the new honors program as "very rigorous."

2. It can be concluded, with 78% confidence, that at least 87% of all students will rate the new honors program as "very rigorous."

3. It can be concluded, with 78% confidence, that between 87% and 93% of all students will rate the new honors program as "very rigorous."

4. It can be concluded, with 90% confidence, that between 75% and 81% of all students will rate the new honors program as "very rigorous."

Statement (4) "It can be concluded, with 90% confidence, that between 75% and 81% of all students will rate the new honors program as "very rigorous." is correct.

What is the margin of error(MOE)?

It is defined as an error that provides an estimate of the percentage of errors in real statistical data.

The formula for finding the MOE:

[tex]\rm MOE = Z\times \dfrac{s}{\sqrt{n}}[/tex]

Where Z is the z-score at the confidence interval

s is the standard deviation

n is the number of samples.

We have:

Of the students surveyed, 78% rated the new honors program as "very rigorous" with a margin of error of ±3% and a confidence interval of 90%.

Lower confidence interval = 78-3 = 75%

Upper confidence interval = 78+3 = 81%

We can say it can be concluded, with 90% confidence, that between 75% and 81% of all students will rate the new honors program as "very rigorous."

Thus, statement (4) "It can be concluded, with 90% confidence, that between 75% and 81% of all students will rate the new honors program as "very rigorous." is correct.

Learn more about the Margin of error here:

brainly.com/question/13990500

#SPJ2