A confidence interval was used to estimate the proportion of statistics students that are female. A random sample of 72 statistics students generated the following 90% confidence interval: (0.438, 0.642). Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence? Explain/show how you obtain your answer

We need a sample of size at least 13.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

The margin of error is:

90% confidence interval: (0.438, 0.642).

The proportion estimate is the halfway point of these two bounds. So

95% confidence level

So , z is the value of Z that has a pvalue of , so .

Using the information above, what size sample would be necessary if we wanted to estimate the true proportion to within ±0.08 using 95% confidence?

We need a sample of size at least n.

n is found when M = 0.08. So

Rounding up

We need a sample of size at least 13.

hey there!

if she's saving for six months, let's divide the total number we need for the trip by six, dividing it into the amount that needs to be saved for each month.

500 ÷ 6 =

she needs to save at least this amount each month, so we should use this symbol: ≥

let x be the amount she saves each month.

your final answer is x ≥

hope this !

ok so the answer to your question is simple as long as you look at the different dimensions that contain the exemplary letters that combined in the right order give you words that eventually within the 5th dimension will conform and give you the correct answer to your question. hope i , stop looking at this easily, look at it with a spectral view.

step-by-step explanation:

told you