Instruction 8-10
The Registrar of a university would like to estimate the proportion of the student population who owns a personal computer.In a sample of 500 students,417 own a personal computer.
-Referring to Instruction 8-10,we are 99% confident that between 79.11% and 87.69% of the student population own a personal computer.

Question

Quizplus · Accepted Answer

Since the sample proportion is 417/500 = 0.834, we can use a z-test for proportions to construct a confidence interval. Assuming a normal distribution for the sample proportion, with mean p and standard deviation sqrt[p(1-p)/n], we can use the standard normal distribution to find the z-scores corresponding to the upper and lower bounds of the interval.

Using a 99% confidence level, we find that the z-score for the lower bound is -2.576 and the z-score for the upper bound is 2.576. Substituting these values into the formula for the confidence interval gives:

p̂ ± z_{α/2} * sqrt(p̂(1-p̂)/n)

where p̂ = 0.834, z_{α/2} = 2.576, and n = 500.

Calculating the confidence interval, we get:

0.834 ± 2.576 * sqrt(0.834 * 0.166 / 500)

which simplifies to:

0.834 ± 0.037

Therefore, the 99% confidence interval for the proportion of the student population who owns a personal computer is (0.797, 0.872), or between 79.11% and 87.69%. Since the statement in the question matches this interval, the answer is A: true.

Instruction 8-10 the Registrar of a University Would Like to Estimate the Estimate