A study suggests that about 80% of undergraduate students are working at a job. Researchers plan to do a new study and use the new data to make a z interval on of students who are working at a job. They plan on using a confidence level of 90% and they want the margin of error to be no more than 1%. Let p represent the proportion of students who are working at a job and assume p 0.80, what is the smallest sample size required to obtain the desired margin of error?

The smallest sample size required to obtain the desired margin of error is 4330

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:

In this problem, we have that:

90% confidence level

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

What is the smallest sample size required to obtain the desired margin of error?

This is n when

So

Rounding up

The smallest sample size required to obtain the desired margin of error is 4330

i would you but i don't under stand the qustion

step-by-step explanation:

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: