Question

Which sample size will produce a margin of error of ±6.3%?

152

114

252

550

Answer 1

@phi

Answer 2

@robtobey

Answer 3

@Directrix

Answer 4

@Loser66

Answer 5

@Mehek14
@MissSmartiez

Answer 6

They'll help @18jonea. I'm not good at mathematics.

Answer 7

The following may help:
http://www.dummies.com/how-to/content/how-sample-size-affects-the-margin-of-error.html

Answer 8

Can you help me work thrugh it @robtobey

Answer 9

@Astrophysics

Answer 10

@MissSmartiez

Answer 11

@phi

Answer 12

@Mehek14

Answer 13

@Kainui

Answer 14

we need more info to do this

Answer 15

like what this is all i got

Answer 16

@phi

Answer 17

I half remember that you need a level of confidence.

Answer 18

ok so how do i solve this @phi

Answer 19

We need more info or some assumptions. otherwise, I'm totally in the dark.

Answer 20

same

Answer 21

i can show you my notes

Answer 22

@phi

Answer 23

that might help

Answer 24

does that help

Answer 25

Yes, that helps a lot.

They are saying (I guess it's a "rule of thumb" ) that the margin of error
is about 1 / sqr(n)

Answer 26

to do the problem, first change 6.3% to a decimal . what is that ?

Answer 27

.063

Answer 28

now set that equal to 1/ sqr(n)
\[ 0.063= \frac{1}{\sqrt{n}}\]

and "solve for n"
do you know how to do that ?

Answer 29

no

Answer 30

can you solve for x if you had
\[ \frac{1}{x} = 2 \]
?

Answer 31

.5

Answer 32

i got 252 would that be right

Answer 33

you could multiply both sides by x to get
\[ \frac{x}{x} = 2 x \]
x/x is 1, so that is
\[ 1 = 2x\]
then divide both sides by 2
\[ \frac{1}{2} = x\]

Answer 34

for your problem, rename sqr(n) as x (temporarily)
\[ \frac{1}{x} = 0.063\]
can you solve for x ?

Answer 35

what

Answer 36

first multiply both sides by x

Answer 37

it will make sense in a minute

Answer 38

I got +- square root 15.87
n= 251.85
then round to 252

Answer 39

yes, that looks good

Answer 40

\[ \frac{1}{\sqrt{n}}= 0.063\]
multiply both sides by sqr(n)
\[ \frac{\sqrt{n}}{\sqrt{n}}= 0.063{\sqrt{n}} \\1=0.063{\sqrt{n}}\]

now divide both sides by 0.063
\[ \frac{1}{0.063}= \sqrt{n} \]
finally square both sides to get rid of the square root
\[ \left( \frac{1}{0.063}\right)^2= n \]

Answer 41

typing into google:
( 1/0.063)^2 =

we get 251.952633
or 252 rounded to the nearest whole number

Answer 42

When 876 voters were polled, 67% said they were voting yes on an initiative measure. Find

the margin of error and an interval that is likely to contain the true population proportion.

±29.6%; between 37.4% and 96.6%

±3.4%; between 63.6% and 70.4%

±34%; between 33% and 100%

±3.0%; between 64.0% and 70.0%
Can you help me with this one

Answer 43

@phi

Answer 44

do you have any more notes? I'm not sure if we use 876 or 67% of 876.

Answer 45

although, looking at the answer choices, it looks like we use the entire number in the sample.

in other words, the measure of error is 1/sqr(876)
what do you get ?

Answer 46

around .04

Answer 47

you need more decimals

Answer 48

then change to a percent by multiplying by 100

Answer 49

0.03378686892

Answer 50

around 3.4 % which is choice b

Answer 51

@phi

Answer 52

yes. and the 3.4% is the error in their 67%
so the range is 67-3.4 to 67+3.4
or 63.6% to 70.4%

Answer 53

so b is correct @phi

Answer 54

yes