A point estimate of the mean of a population is determined by calculating the mean of a sample drawn from the population. The calculation of the mean is the sum of all sample values divided by the number of values.
$$ \large\displaystyle \bar{X}=\frac{\sum\nolimits_{i=1}^{n}{{{x}_{i}}}}{n}$$
Where $- \bar{X} -$ is the mean of the n individual xi values.
The larger the sample the more accurate the estimate. If n equals the number of items in the population the same formula calculates the population mean, $- \mu -$.
Estimate of location of $- \mu -$
The same of n items also can provide an interval that indicates where the population mean, $- \mu -$, resides with a stated probability or confidence. If we say the unknown population mean has a 90% chance of residing within an interval, we are using a 90% confidence interval. This also means there is a 10% chance $- \mu -$ falls outside the interval.
The formula for the confidence interval about an estimated mean is dependent on whether or not we know the population standard deviation, $- \sigma -$. Generally we do not know $- \sigma -$ and use an estimated standard deviation, s, using the sample values. The sample’s mean confidence interval is
$$ \large\displaystyle \bar{X}\pm {{t}_{\alpha /2,\text{ }n-1}}\frac{s}{\sqrt{n}}$$
Where $- {{t}_{\alpha /2,\text{ }n-1}} -$ is the t-distribution with 1-$- \alpha -$ confidence and n-1 degrees of freedom value. s is the sample standard deviation and n is the number of values in the sample.
If we know the population standard deviation, $- \sigma -$, then we can use the normal distribution instead of the t-distribution and the known $- \sigma -$ value instead of the estimate s.
I have long wondered that if we know the population standard deviation we probably also know the population mean and why would we be calculating a confidence interval. Not sure of a situation where I would use the known standard deviation, if you know of a situation, do let me know.
The formula for the confidence interval about a sample mean with a known population standard deviation is
$$ \large\displaystyle \bar{X}\pm {{Z}_{\alpha /2}}\frac{\sigma }{\sqrt{n}}$$
Where $- {{Z}_{\alpha /2}} -$ is the number of standard deviations from the center to the point when the area under the standard normal equals the $- 1-{\alpha }/{2}\; -$ percentile.
Example
Let say we have four readings of tensile strength of new polymer blend based on polypropylene, 32.3, 34.7, 32,6, and 33.5 MPa. Determine the estimate of the population mean and a 95% confidence interval about the point estimate.
The mean of the four values is 33.275 Mpa. This is the point estimate of the mean.
The sample standard deviation, s, is 1.08 MPa, which we will need for the confidence interval calculation.
$$ \large\displaystyle \bar{X}\pm {{t}_{\alpha /2,\text{ }n-1}}\frac{s}{\sqrt{n}}=33.275\pm 3.182\left( \frac{1.08}{2} \right)=31.56\text{ and }34.99$$
The critical value is from the t-distribution table with n-1 = 3 degrees of freedom, and $- \alpha -$ = 0.05 / 2 = 0.025 Practice problems like this using the same calculator and references you expect to use during the exam.
Related:
Statistical Confidence (article)
Reliability with confidence (article)
Tolerance Intervals for Normal Distribution Based Set of Data (article)
Why did you not show your working step after the other
Hi Sunday, not sure I understand the comment. cheers, Fred