Question

in a college student poll it is of interest to estimate the proportion p of students in favor of changing from quarter system to a semester system. how many students should be polled so that we can estimate p to within 0.09 using a 9(% confidence interval

Answer 1

We need to poll at least 646 students to estimate the proportion (p) of students in favor of changing from a quarter system to a semester system with a 9% margin of error and a 99% confidence interval.

To estimate the proportion (p) of students in favor of changing from a quarter system to a semester system with a 9% margin of error and a 99% confidence interval, we can use the formula:

`[n = (z^2 p(1-p))/(E^2)]`

where (n) is the sample size, (z) is the z-score corresponding to the desired confidence level (99% in this case), (p) is the estimated proportion of students in favor of changing to a semester system (unknown), and (E) is the margin of error (0.09 in this case).

To find the value of (z), we can use a standard normal distribution table or a calculator to find the z-score corresponding to a 99% confidence level, which is approximately 2.576.

Substituting the given values into the formula, we get:

`[n = ((2.576)^2 p(1-p))/((0.09)^2)]`

We do not know the value of (p), so we can use a conservative estimate of (p = 0.5) to get the maximum sample size. Substituting this value, we get:

`[n = ((2.576)^2 (0.5)(1-0.5))/((0.09)^2)]`

Simplifying, we get:

`[n \approx 645.9]`

Rounding up to the nearest whole number, we get:

[n = 646]

Therefore, we need to poll at least 646 students to estimate the proportion (p) of students in favor of changing from a quarter system to a semester system with a 9% margin of error and a 99% confidence interval.