# Confidence Intervals

Intro to Confidence Intervals for One Mean (Sigma Known)
10:36 — by jbstatistics

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## Key Questions

See explanation.

#### Explanation:

An estimate of a population parameter may be expressed in two ways:

POINT ESTIMATION

A point estimate of a population parameter is a single value of a statistic.

Example,

The sample mean $\overline{x}$ is a point estimate of the population mean Î¼. Similarly, the sample proportion p is a point estimate of the population proportion P.

INTERVAL ESTIMATION

An interval estimate is defined by two numbers, between which a population parameter is said to lie.

Example

$a < x < b$ is an interval estimate of the population mean $\mu$. It indicates that the population mean is greater than a but less than $b$.

In any estimation problem, we need to obtain both a point estimate and an interval estimate. The point estimate is our best guess of the true value of the parameter, while the interval estimate gives a measure of the accuracy of that point estimate by providing an interval that contains plausible values.

• There are total 3 cases
if sigma is known(doesn't matter n is large or small) then use this formula
$\overline{x} - {z}_{\text{(alpha/2)}} \cdot \frac{\sigma}{\sqrt{n}}$ for lower confidence interval limit
$\overline{x} + {z}_{\text{(alpha/2)}} \cdot \frac{\sigma}{\sqrt{n}}$ for upper confidence interval limit

if n is >30 and sigma is unknown then use this formula
$\overline{x} - {z}_{\text{(alpha/2)}} \cdot \frac{S}{\sqrt{n}}$ for lower confidence interval limit
$\overline{x} + {z}_{\text{(alpha/2)}} \cdot \frac{S}{\sqrt{n}}$for upper confidence interval limit

if n is <=30 sigma is unknown then use this formula
$\overline{x} - {t}_{\text{(alpha/2,df=n-1)}} \cdot \frac{s}{\sqrt{n}}$for lower confidence interval limit
$\overline{x} + {t}_{\text{(alpha/2,df=n-1)}} \cdot \frac{s}{\sqrt{n}}$ for upper confidence interval limit

now consider the example for large sample size
n=3,500
xbar=174.4
S=38.7
The Z value for 95% confidence is Z=1.96
$174.4 - 1.96 \cdot \frac{38.7}{\sqrt{3500}}$
$174.4 + 1.96 \cdot \frac{38.7}{\sqrt{3500}}$
hence
$173.11 \le \mu \le 175.68$

now consider the example for small sample size
n=10
xbar=176
s=33
the degrees of freedom (df) = n-1 = 9. The t value for 95% confidence with df = 9 is t = 2.262.
$176 - 2.262 \cdot \frac{33}{\sqrt{10}}$
$176 + 2.262 \cdot \frac{33}{\sqrt{10}}$
hence
$152.39 \le \mu \le 190.60$

• Suppose the confidence interval is 95%.

This interval then represents the range within which you would expect a sample mean to fall in 95% of random samples.

It does not mean that the probability is 95%, or that a specific value will fall within this interval in 95% of cases.

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