# Large samples confidence interval for difference in means

• kingwinner
In summary, the two cases for large samples confidence interval for difference in means are: - If the population variances are known to be equal, then the confidence interval for \mu_1 - \mu_2 simplifies to - If the population variances are unknown but known to be equal, then the confidence interval for \mu_1 - \mu_2 is (\overline X_1 - \overline X_2) \pm z_{\frac{\alpha}2} \sqrt{\, \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} }
kingwinner
The following distinguishes TWO cases for large samples confidence interval for difference in means:
http://www.geocities.com/asdfasdf23135/stat11.JPG

where Sp^2 is the pooled estimate of the common variance, n1 is the sample size from the first population, n2 is the sample size from the second population, and z_alpha/2 is 100(1-alpha/2) th percentile of the standard normal.
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It seems to me that case 1 is a special case of case 2 with the population variances being equal. If this is the case, the formula for case 2 should reduce to the formula for case 1 when the population variances are equal. However, I have no way of seeing it being the case.
[aside: I am trying to cut down on the number of formulas that I have to memorize. Instead of two different formulas, if case 2 contains case 1, then I only have to memorize the general case 2 formula which is nice.]

Could somebody please show me how I can reduce case 2 to case 1?
Any help would be appreciated!

Last edited:
Suppose you knew the two population variances. The confidence interval for $$\mu_1 - \mu_2$$ would look like this.

$$(\overline X_1 - \overline X_2) \pm z_{\frac{\alpha}2} \sqrt{\, \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} }$$

This is true for any $$\sigma_1^2$$ and $$\sigma_2^2$$. If the two variances are the same, call the common value $$\sigma^2$$. The confidence interval simplifies to

$$(\overline X_1 - \overline X_2) \pm z_{\frac{\alpha}2} \sqrt{\,\sigma^2 \left(\frac 1 {n_1} + \frac 1 {n_2} \right) }$$

so if you actually know the real variances, the two formulae are equivalent.

However, in practice you don't know the real variances, and you have to make due with the sample variances. Since, even if you are willing to assume the population variances are equal, there is no reason to expect the sample variances will be equal,
instead of using them individually they are pooled to obtain $$s^2_p$$. Then the interval is

\begin{align*} (\overline X_1 - \overline X_2) &\pm z_{\frac{\alpha}2} \sqrt{\, \frac{s_p^2}{n_1} + \frac{s_p^2}{n_2} } \Rightarrow \\ (\overline X_1 - \overline X_2) &\pm z_{\frac{\alpha}2} \sqrt{\, s_p^2 \left(\frac 1 {n_1} + \frac 1 {n_2} \right)} \end{align*}

So we have,

Case 1:
The variances are unknown but known to be equal.

Case 2:
The variance are unknown and not-known to be equal (they could be equal, but we just don't have this additional information).

So they are really two separate cases. Am I right?

But what is the real point of using the "pooled estimate"? Is it going to give a better estimate? (i.e. when the population variances are unkown but known to be equal, will the case 1 formula give a narrower (better) interval than case 2?)

Thanks!

## 1. What is a large sample confidence interval for difference in means?

A large sample confidence interval for difference in means is a statistical tool used to estimate the difference between two population means based on a sample of data. It provides a range of values within which we can be confident that the true difference between the two population means lies.

## 2. Why is a large sample size important for calculating a confidence interval for difference in means?

A large sample size is important because it allows for more precise estimates and reduces the margin of error in the confidence interval. This means that we can have a higher level of confidence in the accuracy of the estimated difference between the two population means.

## 3. How is a large sample confidence interval for difference in means calculated?

A large sample confidence interval for difference in means is calculated by using the formula: CI = (x̄1 - x̄2) ± Zα/2 * √[(s12 / n1) + (s22 / n2)], where x̄1 and x̄2 are the sample means, s1 and s2 are the sample standard deviations, n1 and n2 are the sample sizes, and Zα/2 is the critical value from the standard normal distribution for the desired level of confidence.

## 4. What is the significance of the confidence level in a large sample confidence interval for difference in means?

The confidence level represents the probability that the true difference between the two population means falls within the calculated confidence interval. For example, a 95% confidence level means that there is a 95% chance that the true difference between the two population means lies within the calculated interval.

## 5. How can a large sample confidence interval for difference in means be used in research or data analysis?

A large sample confidence interval for difference in means can be used to compare the means of two populations and determine if there is a statistically significant difference between them. This information can be useful in making decisions, identifying trends, and drawing conclusions in various fields such as psychology, economics, and healthcare.

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