Question

3 select the correct answer. 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? a. it can be concluded, with 90% confidence, that at least 81% of all students will rate the new honors program as \very rigorous.\ b. it can be concluded, with 78% confidence, that between 87% and 93% of all students will rate the new honors program as \very rigorous.\ c. it can be concluded, with 78% confidence, that at least 87% of all students will rate the new honors program as \very rigorous.\ d. it can be concluded, with 90% confidence, that between 75% and 81% of all students will rate the new honors program as \very rigorous.\

Explanation:

Step1: Recall margin - of - error concept

The margin of error is used to create a confidence interval around the sample proportion. The confidence interval formula for a proportion is $\hat{p}\pm E$, where $\hat{p}$ is the sample proportion and $E$ is the margin of error.

Step2: Identify given values

We are given that $\hat{p} = 78\%=0.78$ and $E = 3\% = 0.03$, and the confidence level is $90\%$.

Step3: Calculate confidence interval

The lower limit of the confidence interval is $\hat{p}-E=0.78 - 0.03=0.75$ or $75\%$, and the upper limit is $\hat{p}+E=0.78 + 0.03=0.81$ or $81\%$. This means that with $90\%$ confidence, the true proportion of all students who rate the new honors program as "very rigorous" lies between $75\%$ and $81\%$.

Answer:

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