Question

(By-Hand) VCU recently conducted a survey of 800 randomly selected undergraduate students, and 286 indicated that they plan to travel outside of Virginia for Spring Break. Using this information, please construct and interpret a 90% confidence interval for the proportion of all VCU undergraduate students who plan to travel outside of Virginia for Spring Break. Be sure to check any assumptions before starting your analysis.

Answer 1

The 90% confidence interval for the proportion of all VCU undergraduate students planning to travel outside of Virginia for Spring Break is approximately 30.3% to 36.4%.

Step-by-step explanation:

To construct the confidence interval, we can use the formula for the confidence interval for a proportion :
`\(\hat{p} \pm z * \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)` , where
`\(\hat{p}\)` is the sample proportion, \(n\) is the sample size, and \(z\) is the critical value from the standard normal distribution corresponding to the desired confidence level.

Given that 286 out of 800 students plan to travel, the sample proportion \(\hat{p}\) is 286/800 = 0.3575. The critical value for a 90% confidence level from the standard normal distribution is approximately 1.645.

Substituting the values into the formula, we get:
`\(0.3575 \pm 1.645 * \sqrt{(0.3575 * (1-0.3575))/(800)}\).`

After calculations, the confidence interval is approximately 30.3% to 36.4%.

This means that we are 90% confident that the true proportion of all VCU undergraduate students planning to travel outside of Virginia for Spring Break lies between 30.3% and 36.4%. The interval gives a range of likely values for the population proportion based on the sample data, considering the sampling variability. Assumptions such as random sampling and independence of responses are assumed in this calculation to ensure the validity of the confidence interval.