Let A be the width of a 95% Confidence interval for population mean based on a sample

jules1234

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Let A be the width of a 95% Confidence interval for population mean based on a sample of size n=20 taken from a normal population with sigma=0.5
Let B b the width of a 95 confidence interval for population mean based on a sample size 3 times n taken from the same normal population with sigma=0.5
Calculate how much bigger A is then B, that is calculate the ratio A/B.

What I have done:

Width=Z(sima/SQRT(n))

For A- n is <30 soused Z=1.645 in the equation and subbed A for the width solve for A
For B- n=(3)(20)=60 which is>30 so used t=2.093 and got the equation B= 2.093*(.5/SQRT(60)) the soled for B
Using my values I found A/B but that wasn't right.

Thought
I think may have gone wrong in the equation because I am thinking that the equation is only for half the Confidence interval s wen set up he equation I should b using A/2=equation ad/2= equation
OR I am messing up the Z an t usage
OR both ha
 
Let A be the width of a 95% Confidence interval for population mean based on a sample of size n=20 taken from a normal population with sigma=0.5
Let B b the width of a 95 confidence interval for population mean based on a sample size 3 times n taken from the same normal population with sigma=0.5
Calculate how much bigger A is then B, that is calculate the ratio A/B.

What I have done:

Width=Z(sima/SQRT(n))

For A- n is <30 soused Z=1.645 in the equation and subbed A for the width solve for A
For B- n=(3)(20)=60 which is>30 so used t=2.093 and got the equation B= 2.093*(.5/SQRT(60)) the soled for B
Using my values I found A/B but that wasn't right.

Thought
I think may have gone wrong in the equation because I am thinking that the equation is only for half the Confidence interval s wen set up he equation I should b using A/2=equation ad/2= equation
OR I am messing up the Z an t usage
OR both ha

It's hard to follow what you say with so many typos, but you used the wrong z for a [two-tailed] 95% confidence interval, and I think you've also mixed up z and t as you suggest. What are the rules you were taught for deciding between z and t?
 
I was taught if the sample size is larger then 30 to use Z and if the sample size is less then 30 use t, so yes it appears that I have them mixed up. I didn't know there were different charts for two tailed tests! it there a two tailed chart for t to?
 
Also once finding the correct Z and t and putting them in the correct equations, which I think iv done now..... Is it still right in saying the equation (width/2)= z *(sigma/(SQRT(N)) or by using the two tailed test for z and t am I changing that to just (width)= z* (Sigma/SQRT(N))?
 
I was taught if the sample size is larger then 30 to use Z and if the sample size is less then 30 use t, so yes it appears that I have them mixed up. I didn't know there were different charts for two tailed tests! it there a two tailed chart for t to?

It's not typically a separate chart; rather, you have one chart in which for 95% with two tails, you know there is 2.5% in each tail, so you look up 2.5% or 97.5%, depending on how the table is organized. Many tables have column headings for both one- and two-tailed cases, or notes about it. This is true for both z and t.
 
Also once finding the correct Z and t and putting them in the correct equations, which I think iv done now..... Is it still right in saying the equation (width/2)= z *(sigma/(SQRT(N)) or by using the two tailed test for z and t am I changing that to just (width)= z* (Sigma/SQRT(N))?

No, the first version is right. Just use the right z: in your case, not 1.645 but 1.96, if I recall correctly.
 
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