Calculate required n for the confidence interval width to be doubled

[ausmathgenius420]

Question: A population proportion based on a sample size of 100 people has a width of w. Assuming other parameters remain equal ([imath]\hat{p}[/imath],[imath]z[/imath]), what is the sample size required for the population proportion to have a width of 2w?

*Multiple choice answers: 25,50,200,400
*Calculator free question

Confidence interval formula:

Decreasing the sample size will increase the width, hence n=25 or 50.

If you visualise the interval [imath](a,b)[/imath] on a number line, to double the width, each end point must be 50% further apart I.e. [imath]a\rightarrow a-0.5a[/imath] and [imath]b\rightarrow b+0.5b[/imath]

I don't know where to go from here?

 

[BigBeachBanana]

Question: A population proportion based on a sample size of 100 people has a width of w. Assuming other parameters remain equal ([imath]\hat{p}[/imath],[imath]z[/imath]), what is the sample size required for the population proportion to have a width of 2w?

*Multiple choice answers: 25,50,200,400
*Calculator free question

Confidence interval formula:
View attachment 33921
Decreasing the sample size will increase the width, hence n=25 or 50.

If you visualise the interval [imath](a,b)[/imath] on a number line, to double the width, each end point must be 50% further apart I.e. [imath]a\rightarrow a-0.5a[/imath] and [imath]b\rightarrow b+0.5b[/imath]

I don't know where to go from here?

It's an algebra problem. Hint:
[math]p+z\sqrt{\frac{p(1-p)}{n}}-\left(p-z\sqrt{\frac{p(1-p)}{n}}\right)=w[/math]

 

[ausmathgenius420]

It's an algebra problem. Hint:
[math]p+z\sqrt{\frac{p(1-p)}{n}}-\left(p-z\sqrt{\frac{p(1-p)}{n}}\right)=w[/math]

Can you explain further? I rearranged for n and it didn’t help.

 

[Harry_the_cat]

Have you simplified the LHS? Remember the original n is 100.

Don't rearrange for n.

 

[ausmathgenius420]

Have you simplified the LHS? Remember the original n is 100.

Don't rearrange for n.

---

[imath] w=z\frac{p(1-p)}{5} [/imath]

 

[BigBeachBanana]

Can you explain further? I rearranged for n and it didn’t help.

Since you call yourself a genius, I trust that you can simplify the LHS.

[math]p+z\sqrt{\frac{p(1-p)}{n}}-\left(p-z\sqrt{\frac{p(1-p)}{n}}\right)=w\\ 2z\sqrt{\frac{p(1-p)}{n}}=w[/math]
Now we seek [imath]n^*[/imath] such that we have an extra 2 on the LHS.
[math]\red{2}\cdot 2z\sqrt{\frac{p(1-p)}{n}}=2z \sqrt{\frac{p(1-p)}{n^*}}=\red{2}\cdot w[/math]
Can you finish from here?

 

[ausmathgenius420]

Since you call yourself a genius, I trust that you can simplify the LHS.

[math]p+z\sqrt{\frac{p(1-p)}{n}}-\left(p-z\sqrt{\frac{p(1-p)}{n}}\right)=w\\ 2z\sqrt{\frac{p(1-p)}{n}}=w[/math]
Now we seek [imath]n^*[/imath] such that we have an extra 2 on the LHS.
[math]\red{2}\cdot 2z\sqrt{\frac{p(1-p)}{n}}=2z \sqrt{\frac{p(1-p)}{n^*}}=\red{2}\cdot w[/math]
Can you finish from here?

[math]\frac{2z\sqrt{p(1-p)}}{10}=\frac{4z\sqrt{p(1-p)}}{n}[/math][math]n=20?[/math]
20 is not one of the solutions.

P.S. The username is a joke

 

[BigBeachBanana]

[math]\frac{2z\sqrt{p(1-p)}}{10}=\frac{4z\sqrt{p(1-p)}}{n}[/math][math]n=20?[/math]
20 is not one of the solutions.

P.S. The username is a joke

Ignore the 100. It's an algebra problem no need to plug in 100. What do I need to do to n to get an extra 2?
[math]2z\sqrt{\frac{p(1-p)}{n}} \implies \red{2}\cdot 2z\sqrt{\frac{p(1-p)}{n}}?[/math]

 

[ausmathgenius420]

Ignore the 100. It's an algebra problem no need to plug in 100. What do I need to do to n to get an extra 2?
[math]2z\sqrt{\frac{p(1-p)}{n}} \implies \red{2}\cdot 2z\sqrt{\frac{p(1-p)}{n}}?[/math]

What do I need to do to n to get an extra 2?

Do you mean w? Multiply by 2
I'm not following along

 

[BigBeachBanana]

Do you mean w? Multiply by 2
I'm not following along

Let me rephrase it.

[math]\red{2}\cdot 2z\sqrt{\frac{p(1-p)}{n}} = \red{2}w\\ 2z\sqrt{\frac{p(1-p)}{n^*}}=\red{2}w\\[/math]What's [imath]n^*[/imath] in terms of [imath]n?[/imath]

 

[ausmathgenius420]

Let me rephrase it.

[math]\red{2}\cdot 2z\sqrt{\frac{p(1-p)}{n}} = \red{2}w\\ 2z\sqrt{\frac{p(1-p)}{n^*}}=\red{2}w\\[/math]What's [imath]n^*[/imath] in terms of [imath]n?[/imath]

I don't know + I'm not sure where the two went in the second line
Sorry

 

[BigBeachBanana]

I'm not sure where the two went in the second line.

That's for you to figure out... Here's an extra step.
[math]\red{2}\cdot 2z\sqrt{\frac{p(1-p)}{n}}=2z\sqrt{\frac{\red{4}p(1-p)}{n}}=2z\sqrt{\frac{p(1-p)}{n^*}}[/math]I'll ask you again. What's [imath]n^*[/imath] in terms of [imath]n?[/imath]

 

[ausmathgenius420]

That's for you to figure out... Here's an extra step.
[math]\red{2}\cdot 2z\sqrt{\frac{p(1-p)}{n}}=2z\sqrt{\frac{\red{4}p(1-p)}{n}}=2\sqrt{\frac{p(1-p)}{n^*}}[/math]I'll ask you again. What's [imath]n^*[/imath] in terms of [imath]n?[/imath]

So: [math]n^*=\frac{n}{4}[/math][math]n=25[/math]Was the z meant to disappear?

 

[BigBeachBanana]

So: [math]n^*=\frac{n}{4}[/math][math]n=25[/math]Was the z meant to disappear?

Correct and no it was a typo (edited above).
To summarize, you've shown that for a sample size [imath]n[/imath] with a confidence interval [imath]w[/imath], the required sample size to keep the same population proportion and a width [imath]2w[/imath] is [imath]\frac{n}{4}[/imath].

 

[ausmathgenius420]

Correct and no it was a typo (edited above).
To summarize, you've shown that for a sample size [imath]n[/imath] with a confidence interval [imath]w[/imath], the required sample size to keep the same population proportion and a width [imath]2w[/imath] is [imath]\frac{n}{4}[/imath].

Could you suggest an alternate way to solve the problem? I'm struggling to grasp that way.

[math]\frac{2z\sqrt{p(1-p)}}{10}=\frac{4z\sqrt{p(1-p)}}{n}[/math][math]n=20?[/math]
20 is not one of the solutions.

Why didn't this work?

 

[BigBeachBanana]

Could you suggest an alternate way to solve the problem? I'm struggling to grasp that way.

Maybe this will help.
\(\displaystyle \text{If } 2\sqrt{\frac{p(1-p)}{n}}=w,\, \text{then } 2\sqrt{\frac{p(1-p)}{n/4}}=2\cdot w\)

What you should have is [imath]n^*[/imath], instead of [imath]n[/imath] in your equation.

[math]\frac{2z\sqrt{p(1-p)}}{10}=\frac{4z\sqrt{p(1-p)}}{\sqrt{n^*}}\implies n^*=400\\ nx=n^* \implies 100x=400 \implies x=4 \implies n^*=\frac{n}{4}[/math]

 

[BigBeachBanana]

Maybe this will help.
\(\displaystyle \text{If } 2\sqrt{\frac{p(1-p)}{n}}=w,\, \text{then } 2\sqrt{\frac{p(1-p)}{n/4}}=2\cdot w\)

What you should have is [imath]n^*[/imath], instead of [imath]n[/imath] in your equation.

[math]\frac{2z\sqrt{p(1-p)}}{10}=\frac{4z\sqrt{p(1-p)}}{\sqrt{n^*}}\implies n^*=400\\ nx=n^* \implies 100x=400 \implies x=4 \implies n^*=\frac{n}{4}[/math]

Addendum: The reason you can't have just have [imath]n[/imath] on the RHS is that you multiplied it by 2, but did not do the same on the LHS so the equality no longer holds. Therefore, you need to call it [imath]n^*[/imath], and need to figure out what's [imath]n^*[/imath] in terms of [imath]n.[/imath]

 

[ausmathgenius420]

nx=n∗⟹100x=400⟹x=4⟹n∗=4n

That would make [imath]n^*=4n[/imath]

 

[BigBeachBanana]

That would make [imath]n^*=4n[/imath]

I made a whoopsie, but what you're trying to do is similar to what we did except you're not balancing the equation correctly somewhere.

[math]2z\sqrt{\frac{p(1-p)}{n}}=w[/math]Now, if I want [imath]2w[/imath], then [math]2\cdot 2z\sqrt{\frac{p(1-p)}{n}}=2z\sqrt{\frac{4p(1-p)}{n}}=2z\sqrt{\frac{p(1-p)}{n/4}}=2w[/math]
Or "your way"
If [math]\frac{2z\sqrt{p(1-p)}}{10}=w[/math], and I want [math]2\cdot \frac{2z\sqrt{p(1-p)}}{10} =\frac{2z\sqrt{p(1-p)}}{10/2}.[/math] where [imath]10/2=\sqrt{n}/2 = \sqrt{n/4}[/imath]

 

[Harry_the_cat]

As BBB said in Post #2:

Simplifying the LHS gives:
\(\displaystyle w = 2z\sqrt{\frac{p(1-p)}{n}}\)