Question

**Will fan & Medal**
A survey of factories in five northeastern states found that 10% of the 300 workers surveyed were satisfied with the benefits offered by their employers. The margin of error for the population proportion is ± BLANK %.
What is the blank?

Answer 1

http://www.dummies.com/how-to/content/how-to-calculate-the-margin-of-error-for-a-sample0.html

Answer 2

please help me set up the problem

Answer 3

what's the sample size?

Answer 4

I don't know

Answer 5

i need help finding it

Answer 6

try reading the first sentence and see what you come up with

Answer 7

10/300=.03?

Answer 8

https://onlinecourses.science.psu.edu/stat100/node/56

Answer 9

what does sample size mean to you? remember, a survey has been conducted and there were 300 respondents. 10 of the 300 respondents were satisfied with the benefits offered by their employers.

Answer 10

the 300 people are the sample size?

Answer 11

yes, very good. in the formula\[MoE = z^*\sqrt{\frac{\hat{p} \left( 1-\hat{p} \right)}{n}}\]n is the sample size. MoE is the Margin of Error.

Answer 12

what is the sample proportion? (this is \(\hat{p}\) in the formula)

Answer 13

is it 10% or 10/300=0.3?

Answer 14

\(\hat{p}\) is proportion... it can be given directly as a proportion or percentage, or calculated as the number (out of the total satisfying a given requirement) divided by the total.

hint: 10/300 is not appropriate and is it the percentage divided by the total.

.10*300 = 30 (which is 10% of 300) is the number of respondents who were satisfied by the benefits offered by their employers.

Answer 15

so p is 30?

Answer 16

proportion means ratio... percentage is also a ratio but with 100 as the denominator (always!)

to calculate the proportion of respondents who were satisfied with the benefits offered by their employers, we need the number of respondents who were satisfied with the benefits offered by their employers (call this x) AND the total number of respondents (the sample size).

\(\hat{p}=\Large{\frac{x}{n}}\)

in your problem, they've given it to you... it's the only percentage mentioned.

Answer 17

so 10%?

Answer 18

yep. next decide what value for \(z^*\) you'd leik to use. please check the second link I provided to see what value you think is appropriate

Answer 19

btw: you must use the decimal form of the percentage in the formula

Answer 20

so .10 for 10%?

Answer 21

Now I'm trying to find z I need help

Answer 22

yes, .10 for 10%

from the link:
Note: when you see a margin of error in a news report, it it almost always referring to a 95% confidence interval. But other levels of confidence are possible.

since your problem doesn't give a level of confidence, follow the suggestion above and use the appropriate \(z^*\) value.

Answer 23

the link has a table, relating confidence levels and \(z^*\) values

Answer 24

you could also use a standard normal table or your calculator

Answer 25

so does z have no value? do i just use z?

Answer 26

no, did you see table 10.1 in the link?

Answer 27

yeah i don't understand what z stands for in the table

Answer 28

well, forget that for just a second and let's make sure you get the correct value. what level of confidence should you use (according to the site and common practice)?

Answer 29

from the link:
Note: when you see a margin of error in a news report, it it almost always referring to a 95% confidence interval. But other levels of confidence are possible.

since your problem doesn't give a level of confidence, follow the suggestion above.

Answer 30

2.0 or more precisely 1.96?

Answer 31

i wasnt paying attention and completely missed what you said the confidence level was

Answer 32

wow im dumb sorry

Answer 33

you're an angel for putting up with my slow self

Answer 34

yes, you should use a 95% level of confidence, corresponding to a \(z^*\) value of 1.96


this 1.96 is how many standard deviations from the mean, in a normal distribution.