Estimation of population mean
If the sample size is small (n<30) the arithmetic mean of small samples are not normally distributed. In such conditions, student's t distribution must be utilized to estimate the population mean.
In this case
Population mean µ = x¯ ± tS_{x¯}
x¯ = Sample mean
S_{x¯} = s/√n
S = standard deviation of samples =
for small samples.
n = sample size
v = n - 1 degrees of freedom.
The value of t is acquired from student's t distribution tables for the essential confidence level
Illustration
A random sample of 12 items is taken and is found to have a mean weight of 50 gram and a standard deviation of 9 gram
What is the mean weight of population
a) Along with 95 percent confidence
b) Along with 99 percent confidence
Solution
S = 9; v = n - 1 = 12 - 1 = 11;
S_{x¯ }= s/√n = 9/√12
µ = x¯ ± t S_{x¯}
At 95 percent confidence level
µ = 50 ± 2.262
= 50 ± 5.72 grams
Hence we can state with 95 percent confidence that the population mean is among 44.28 and 55.72 gram
At 99 percent confidence level
µ = 50 ± 3.25 (9/√12)
= 50 ± 8.07 gram
Therefore we can state with 99 percent confidence that the population mean is between 41.93 and 58.07 grams
Note: To employ the t distribution tables it is significant to find the degrees of freedom (v = n - 1). In the illustration above v = 12 - 1 = 11
From the tables we find that at 95 percent confidence level against 11 and under 0.05, the value of t = 2.201