Question

A week before final examinations, a random sample of 813 students from a large college was asked if they planned to study at least 10 hours for their final examinations. Based on this random sample, a 90% confidence interval for the proportion of all students at this college who plan to study at least 10 hours for their final examinations is 0.88 to 0.99. Which of the following provides the correct point estimate and margin of error for this interval?

a. point sstimate=0.935 margin of error=0.055
b. point estimate =0.11; margin of crror=0.77
c. None of these is correct
d. point estimate =0.88 : margin of cmor=0.99

Answer 1

The correct point estimate is 0.935 and the margin of error is 0.055 for the given confidence interval of 0.88 to 0.99, which corresponds to option (a).

Step-by-step explanation:

The question involves finding the correct point estimate and margin of error for a confidence interval regarding the proportion of students at a college who plan to study at least 10 hours for their final examinations. Given the 90% confidence interval of 0.88 to 0.99, the point estimate is calculated by averaging the upper and lower bounds of the confidence interval. The margin of error is determined by taking the difference between the upper bound and the point estimate (or the point estimate and the lower bound).

To calculate these values, we add the lower and upper bounds: 0.88 + 0.99 = 1.87. Then divide by 2 to find the point estimate: 1.87 / 2 = 0.935. The margin of error is the distance from this point estimate to one of the bounds; so, 0.99 - 0.935 = 0.055 or 0.935 - 0.88 = 0.055. Therefore, the correct point estimate is 0.935, and the margin of error is 0.055, making option (a) the correct choice.