Question

A recent poll of 500 employees from a company of 1,300 employees was conducted to see how many of them believe the minimum wage should be raised. Of those polled, 435 feel that the minimum wage should be raised.


With a desired confidence level of 95%, and a corresponding z*-score of 1.96, what is the margin of error for this sample survey?


Complete the statements to find the margin of error.


The sample size in this problem is students.


The population proportion is estimated as .


When the margin of error is calculated using the


formula E = z* , to the nearest tenth of a percent, the result is %.

Answer 1

500

0.87

2.9

It’s right I got it wrong two times and the program told me those were the right answers

Answer 2

Answer:

The sample size in this problem is students is 500.

The population proportion is estimated as 0.87.

The margin of error for this sample survey is 2.9.

Step-by-step explanation:

It is given that a recent poll of 500 employees from a company of 1,300 employees was conducted.

It means the sample size in this problem is students is 500.

Of those polled, 435 feel that the minimum wage should be raised.

$p=\frac{435}{500}=0.87$

The population proportion is estimated as 0.87.

The formula for margin of error is

$M.E.=z^*\times \sqrt{\frac{p(1-p)}{n}}$

The desired confidence level of 95%, and a corresponding z*-score of 1.96. So, the margin of error for this sample survey is

$M.E.=1.96\times \sqrt{\frac{0.87(1-0.87)}{500}}$

$M.E.\approx 0.029$

$M.E.\approx 2.9\%$

Therefore the margin of error for this sample survey is 2.9.