# Conceptual Question Dealing with Confidence Intervals

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1. Apr 8, 2015

### antonisz

1. The problem statement, all variables and given/known data
1.) When we collect more data for our confidence interval
(A) Confidence Increases
(B) Confidence Decreases
(C) n decreases
(D) The length of the interval decreases
(E) The length of the interval increases

2.) We wish to determine a confidence interval for the population mean. If we could lower the population standard deviation prior to collecting our data
(A) Confidence Increases
(B) Confidence Decreases
(C) n decreases
(D) The length of the interval decreases
(E) The length of the interval increases

2. Relevant equations
Z∝/2σ / √n

3. The attempt at a solution

I attempted to solve to test with fake numbers.

n = 10, σ = .5, Z = .15811
n = 10, σ = .25, Z = .07906

So, the lower the population, the length of the interval decreases.

n = 10, σ = .5, Z = .15811
n = 15, σ = .5, Z = .12909

So, as the data increases, the confidence increases.

I'm not sure if this is correct, or if is even correct at all.

Thank you!

2. Apr 9, 2015

### Simon Bridge

You did not write any equations under "relevant equations". Equations have an "=" sign in them.

The problem does not ask you about what happens when you change the population.
The value of "n" would be the sample size, which we expect to be much smaller than the population size.

What does it mean to say that "confidence increases" or "confidence decreases" in this context?
When you collect more data, what happens to "n"? What happens to the confidence value? What happens to the confidence interval "length"?
(I would usually think of confidence intervals having a width rather than a length.)

3. Apr 9, 2015

### RUber

If you collect more data, doesn't that mean that you have collected more samples--i.e. n has increased?
When you talk about confidence, you are normally saying that you are $1-\alpha$% confident that the true parameter (pop. mean) is in the confidence interval created by $\overline{x} \pm \frac{\sigma Z_{\alpha/2}}{\sqrt{n}}$. You choose the confidence based on what value for $\alpha$ you use.
You could make a backward argument that if you kept the same interval, you might be more confident that the mean would be in it...but that's not guaranteed, since additional samples may not imply the same mean at all.

To make this as plain as possible:
Using the equation for the confidence interval of a population mean:
$\overline{x} \pm \frac{\sigma Z_{\alpha/2}}{\sqrt{n}}$.
You have no control over $\overline{x}$, so let's just focus on the components of the interval range.
$\frac{\sigma Z_{\alpha/2}}{\sqrt{n}}$.
Changing anything on the top of the fraction will change the size of the interval proportionally. I.e. increase sigma, you increase the size of the interval. Use a smaller alpha, |Z| increases and the interval increases.
Changing anything on the bottom of the fraction will change the size of the interval with inverse proportionality. So, if sqrt(n) goes down, the size of the interval increases.

4. Apr 19, 2015

### Chuckstabler

First of all, you have to determine what the equation is for a confidence interval. There are a couple ways of doing this, but the method that is the easiest to understand conceptually is as follows:

The confidence interval is given via the following equation: xbar]+- Zcrit * (SD/sqrtN). The critical Z value can be found in any Z table. If the test is one tailed, the critical z value can be found without dividing the alpha level by 2, as all of that alpha is going to be concentrated on one side of the distribution, whereas a two tailed test will have it evenly distributed on each side of the distribution. We will assume the test is two tailed. In this hypothetical example, we will use an alpha level of 0.05, which corresponds to a critical Z value of 1.96. The sample mean in our example will be 10, SD will = 2.5, and N =10 Let's first solve the second part of the equation, SD/sqrtN, which ~ 0.79. We multiply this by the critical Z value of 1.96 to get 1.55. Therefore the confidence interval = 10+/- 1.55, or 8.45 - 11.55.

In the next example I will use N = 20, with the exact same numbers for every other variable of interest. SD/sqrtN in this case ~ 0.56. 0.56 * 1.96 ~1.1. Therefore the confidence interval is 10 +/- 1.1, or 8.9 - 11.1. As you can see, increasing the sample size allows for a more precise estimate of the true population mean.

The precise relationship between sample size and the difference between two samples that differ only in N is given by the equation R2 =R1 * (sqrtN1/sqrtN2). To cut the confidence interval in half, you have to increase sample size by a factor of 4. To reduce the confidence interval by a factor of 10, you have to increase sample size by a factor of 100! As you can see, there are diminishing returns to increasing sample size. This is why you rarely see sample sizes beyond a certain point, there is little to be gained (in most cases) by increasing sample size past a certain point.