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

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

a

step-by-step explanation: