Statistics - Interval Estimation

Interval estimation is the use of sample data to calculate an interval of possible (or probable) values of an unknown population parameter, in contrast to point estimation, which is a single number.

Formula

${\mu = \bar x \pm Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt n}}$

Where −

${\bar x}$ = mean
${Z_{\frac{\alpha}{2}}}$ = the confidence coefficient
${\alpha}$ = confidence level
${\sigma}$ = standard deviation
${n}$ = sample size

Example

Problem Statement:

Suppose a student measuring the boiling temperature of a certain liquid observes the readings (in degrees Celsius) 102.5, 101.7, 103.1, 100.9, 100.5, and 102.2 on 6 different samples of the liquid. He calculates the sample mean to be 101.82. If he knows that the standard deviation for this procedure is 1.2 degrees, what is the interval estimation for the population mean at a 95% confidence level?

Solution:

The student calculated the sample mean of the boiling temperatures to be 101.82, with standard deviation ${\sigma = 0.49}$. The critical value for a 95% confidence interval is 1.96, where ${\frac{1-0.95}{2} = 0.025}$. A 95% confidence interval for the unknown mean.

${ = ((101.82 - (1.96 \times 0.49)), (101.82 + (1.96 \times 0.49))) \\[7pt] \ = (101.82 - 0.96, 101.82 + 0.96) \\[7pt] \ = (100.86, 102.78) }$

As the level of confidence decreases, the size of the corresponding interval will decrease. Suppose the student was interested in a 90% confidence interval for the boiling temperature. In this case, ${\sigma = 0.90}$, and ${\frac{1-0.90}{2} = 0.05}$. The critical value for this level is equal to 1.645, so the 90% confidence interval is

${ = ((101.82 - (1.645 \times 0.49)), (101.82 + (1.645 \times 0.49))) \\[7pt] \ = (101.82 - 0.81, 101.82 + 0.81) \\[7pt] \ = (101.01, 102.63)}$

An increase in sample size will decrease the length of the confidence interval without reducing the level of confidence. This is because the standard deviation decreases as n increases.

Margin of Error

The margin of error ${m}$ of interval estimation is defined to be the value added or subtracted from the sample mean which determines the length of the interval:

${Z_{\frac{\alpha}{2}}\frac{\sigma}{\sqrt n}}$

Suppose in the example above, the student wishes to have a margin of error equal to 0.5 with 95% confidence. Substituting the appropriate values into the expression for ${m}$ and solving for n gives the calculation.

${ n = {(1.96 \times \frac{1.2}{0.5})}^2 \\[7pt] \ = {\frac{2.35}{0.5}^2} \\[7pt] \ = {(4.7)}^2 \ = 22.09 }$

To achieve 95% interval estimation for the mean boiling point with total length less than 1 degree, the student will have to take 23 measurements.

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