# 6. Confidence Intervals for the Population Mean

## i. describe properties of Student's t-distribution and calculate and interpret its degrees of freedom;

## j. calculate and interpret a confidence interval for a population mean, given a normal distribution with 1) a known population variance, 2) an unknown population variance, or 3) an unknown variance and a large sample size;

How is a confidence interval constructed and what does each parameter mean?

`Confidence interval = Point Estimate +- Reliability Factor * Standard error`

- The point estimate is the value of a sample statistic of the population parameter.

- The reliability factor is a number based on the sampling distribution of the point estimate and the degree of confidence (1 - α).

- Standard error refers to the standard error of the sample statistic that is used to produce the point estimate.

Where does the point estimate lie on the confidence intervals range?

It always lies exactly in the middle because it is the best estimate for `\mu`

.

What is a z-statistic?

It is a z-score, which is the number of standard deviations traveled. It is the value obtained from a z-table. If a population is normally distributed with a known variance, a z-statistic is used as the reliability factor to construct confidence intervals for the population mean.

What three values are used to construct a confidence interval for `\mu`

?

1. The sample mean (`m`

)

2. The value of `z`

(which depends on the level of confidence)

3. The standard error of the mean (`\sigma_m`

)

What is the formula for a confidence interval and how is it described?

$$m - z \sigma_m <= \mu <= m + z \sigma_m$$

- The confidence interval has `\mu`

for it's center.

- It extends a distance equal to the product of `z`

in both directions

Given the following information about a normal distribution, how would you calculate a 95% confidence interval? You are given 10 scores out of a population. `\sigma = 100`

, `N = 10`

, `\bar{m} = 530`

, `\sigma_m = 100 / sqrt(10) = 31.62`

- Step 1: Find the z-score for a 95% confidence interval (look up a z-table, find the z-score which has 2.5% on each side) = 1.96

- Apply the formula for confidence interval:

- $$m - z \sigma_m <= \mu <= m + z \sigma_m$$

- Lower limit = `530 - (1.96)(31.62)`

= 468.02

- Upper limit = `530 + (1.96)(31.62)`

= 591.98

- This means you can be 95% certain that the population mean is between `468 <= \mu <= 592`

When it comes to confidence intervals, do they get smaller or bigger given a larger sample size?

They get smaller - the closer the sample size approaches in size to the population, the reduced range of the confidence intervals. E.g., you can estimate the mean of the population with more precision.

When should a T-distribution be used rather than the z-distribution to calculate confidence intervals?

If you know the population mean `\mu`

and standard deviation `\sigma`

then use the z-distribution (z-scores). If the sample size >30 then use the z-distribution (because the central limit theorem says you'll get a pretty good approximation of `\mu`

). If you don't know `\mu`

or `\sigma`

AND you have a sample size <30, then use the t-distribution.

If you have a non-normal distribution, can you use z-scores or t-scores to estimate confidence interval?

- Yes, it works for z-score (but you need a sample size >30 or to know the `\mu`

/`\sigma`

)

- No, it does not work for t-score.

When using a t-distribution are the confidence intervals larger or smaller? Why?

T-distribution confidence intervals are wider than z-distribution because the `\mu`

and `\sigma`

are estimated.

What is the formula to calculate the confidence interval when population mean `\mu`

and standard deviation `\sigma`

are not known?

- Same as when the variables *are* known, but substitute the t-score for the z-score

- $$m - (t)(\s_m) <= \bar{m} <= m + (t)(\s_m)$$

##### Source:

- CFA