Ronald W. answered • 12/13/14
Tutor
4.8
(4)
Mathematics, Tests, Science, Business, Computer Skills, 20 yrs exp.
Without knowing the Current estimate of the survey, we will have to leave answer in a variable form.
Use.
ME= z√p(1-p)/n
ME is margin of error, given .05
z is z score of 95%=1.96
p is not given. This would be the current estimate from previous surveys or what would be expected.
n is survey size
So plug in and solve for n.
.05=1.96√p(1-p/n)
.05/1.962 = p(1-p)n
.0006507 = p(1-p)/n
p(1-p)/.0006507 = n
Thats as far as we can go until given a current estimate (p) from the polling organization.
When you use .95 as p, just plug into equation.
.95(1-.95/0006507=72.99, or 73 person sample size
Typically confidence level has to be calculated as:
(1-.95)/2=.250
(1-.250)=.9750
Then look up on Standard Distribution Table .9750. You will see it is 1.9 down the left column and .06 over to the right. This is 1.96 ALWAYS at 95%.
If you had 99% confidence level.
(1-.99)/2=.005
(1-.005)=.9950
When you look this up it is 2.48 ALWAYS at 99% conf. Lvl.
That is why p is usually different from confidence level, but apparently you will use confidence level as p too.
ME=z(sqrt.p(1-p)/n)
.05=1.96(sqrt .95(1-.95)/n)=72.99 or 73
Ronald W.
Yes, if you had the value of p, the current estimate of what the polling company expects the answer to be, you will get a whole number for n.
Does the problem, or anywhere in your notes/textbook that would indicate what the polling company's current estimate of what it is their polling? That's what you need to solve for n.
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12/14/14
Ronald W.
Now if we assume that they mean the 95% is what they are calling their current estimate, plug in .95
.95(1-.95)/.0006507=n
.95(.05)/.006507=n
n=72.99 so 73 is sample size, assuming 95% is p
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12/14/14
Ivy C.
No, it doesn't tell the current estimate anywhere. But it must be 95% because 73 is right. So may I ask how you got the 73? Thank you for all your help. I have a final on all of this Thursday. I want to make sure I'm clear on this. If you're too busy, I completely understand. Thank you for taking all this time with me.
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12/14/14
Ronald W.
Using the formula
ME= z√p(1-p)/n
Look back at my first posting. The ME (margin of error is given, 5%, or 0.05
The confidence level of 95 has to be converted to Z score to get z. You either memorize that 95% =1.96, or calculate za/2 ?
The a= the area to the right of .95 (confidence level), so you must subtract it from 1 (1-.95=.05).
Then z=a/2 = .05/2=.025
Now subtract this from 1 to get area to the left of the bell curve.
1-.025=.9750
Then look up .9750 on the Normal Distribution Table. The value is 1.9 down and .06 over = 1.96.
It never changes so that's why I just memorize a confidence level of 95% =1.96. But if you have a 99% confidence level, you need to know how.
(1-.99)=.01/2=.005
(1-.005)=.995 then lookup.995 on table
It is 2.4 down and .08 over=2.48
So a 99% confidence level has za/2 of 2.48 ALWAYS
Just know how to calculate then memorize the confidence levels
But confidence level, as you can see, requires calculation so is usually not expressed as p in our equation. But the problem doesn't give you p, so they want you assume confidence level is the same as current estimate. Typically, that's not the case.
So given ME=.05, z=1.96, p=.95 we can find n (number of samples)
ME=z √p(1-p)/n
Then plug in.
.05=(1.96)√(.95)(1-.95)/n
Then solve for n as shown above, gives you 72.99. Which must be 73 since you cannot survey 72.99 people.
Hope that helps.
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12/14/14
Ronald W.
Plug in the formula
ME=z(sqrt p(1-p)/n)
ME = .05, z=1.96, p=.95, solve for n as shown above
But normally z has to be calculated. It is done this way:
(1-.95)/2=.250
(1-.250)=.9750
Then you have to look up .9750 on the Table of Standard Distribution.
You will see it is 1.9 down the left column and .06 over to the right.
So it is ALWAYS 1.96
But if you have a 99% confidence level:
(1-.99)/2= .005
(1-.005)=.995
Then look up .9950 on the Table
It is 2.4 down the left column and .08over to the right, 2.48
So 99% confidence level is 2.48 ALWAYS.
That is why typically confidence level is not the same as p, but in this problem, they want you to assume it is.
After setting up
.05=1.96(sqrt .95(1-.95)/n)
Solve for n. you get 72.99. Since you can't survey 72.99 people, the answer is 73.
Hope that helps.
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12/14/14
Ivy C.
THANK YOU SO MUCH!! Thank you, thank you. Merry Christmas.
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12/14/14
Ronald W.
Your welcome! And Merry Christmas to you as well.
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12/14/14
Ivy C.
12/14/14