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 = -4

step-by-step explanation:

- 20 - 4(x + 3) = 4x

=> -20 - 4x - 12 = 4x

=> - 32 -4x = 4x

=> -32 = 4x + 4x

=> - 32 = 8x

=> -32/8 = x

=> -4 = x

144.50

step-by-step explanation:

21.5 x 13.4=288.10 divide by 2