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 ($\hat{p}$,$z$), 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 $(a,b)$ on a number line, to double the width, each end point must be 50% further apart I.e. $a\rightarrow a-0.5a$ and $b\rightarrow b+0.5b$

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 ($\hat{p}$,$z$), 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 $(a,b)$ on a number line, to double the width, each end point must be 50% further apart I.e. $a\rightarrow a-0.5a$ and $b\rightarrow b+0.5b$

I don't know where to go from here?

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

 

[ausmathgenius420]

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

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.

---

$w=z\frac{p(1-p)}{5}$

 

[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.

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

 

[ausmathgenius420]

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

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

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

P.S. The username is a joke

 

[BigBeachBanana]

$$\frac{2z\sqrt{p(1-p)}}{10}=\frac{4z\sqrt{p(1-p)}}{n}$$$$n=20?$$
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?
$$2z\sqrt{\frac{p(1-p)}{n}} \implies \red{2}\cdot 2z\sqrt{\frac{p(1-p)}{n}}?$$

 

[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?
$$2z\sqrt{\frac{p(1-p)}{n}} \implies \red{2}\cdot 2z\sqrt{\frac{p(1-p)}{n}}?$$

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.

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

 

[ausmathgenius420]

Let me rephrase it.

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

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.
$$\red{2}\cdot 2z\sqrt{\frac{p(1-p)}{n}}=2z\sqrt{\frac{\red{4}p(1-p)}{n}}=2z\sqrt{\frac{p(1-p)}{n^*}}$$I'll ask you again. What's $n^*$ in terms of $n?$

 

[ausmathgenius420]

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

So: $$n^*=\frac{n}{4}$$$$n=25$$Was the z meant to disappear?

 

[BigBeachBanana]

So: $$n^*=\frac{n}{4}$$$$n=25$$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 $n$ with a confidence interval $w$, the required sample size to keep the same population proportion and a width $2w$ is $\frac{n}{4}$.

 

[ausmathgenius420]

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

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

$$\frac{2z\sqrt{p(1-p)}}{10}=\frac{4z\sqrt{p(1-p)}}{n}$$$$n=20?$$
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 $n^*$, instead of $n$ in your equation.

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

 

[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 $n^*$, instead of $n$ in your equation.

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

Addendum: The reason you can't have just have $n$ 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 $n^*$, and need to figure out what's $n^*$ in terms of $n.$

 

[ausmathgenius420]

nx=n∗⟹100x=400⟹x=4⟹n∗=4n

That would make $n^*=4n$

 

[BigBeachBanana]

That would make $n^*=4n$

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.

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

 

[Harry_the_cat]

As BBB said in Post #2:

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