Unit 2: statistical estimation

Example 1. From a random sample of 36 Addis Ababa civil service personnel, the mean age and the sample standard deviation were found to be 40 years and 4.5 years respectively. Construct a 95 percent confidence interval for the mean age of civil servants in Addis.


Solution: The given information can be written as under:
n = 36, = 40 years, = 4.5 years.
95% confidence interval means  = 0.05 and
/2= 0.025 and Z/2 = Z0.025 = 1.96 (as per the normal curve area table).
Thus, 95% confidence interval for the mean age of population is:
Z_/2
or 40 1.96
or 40 (1.96) (0.75)
or 40 1.47 years

i.e we are 95% confident that the population mean is expected in the range (38.53, 41.47)

Example 2: The foreman if ABC Mining Company has estimated the average quantity of iron ore extracted to be 36.8 tons per shift and the sample standard deviation to be 2.8 tons per shift, based upon a random selection of 4 shifts. Construct a 90% confidence interval around this estimate.


Solution: As the standard deviation of population is not known and the sample size is small, we shall use the t-distribution for finding the required confidence interval about the population mean. The given information can be written as:
= 36.8 tons per shift, S or = 2.8 tons per shift n = 4 degrees of freedom = n-1 = 4-1 = 3 and the critical value of ‘t’ for 90 percent confidence interval or at 10% level of significance is  = 10% = 0.1 and /2 = 0.05 and t/2 = t0.05 = 2.353 (as per the table of t-distribution)
Thus, 90% confidence interval for population mean is
t_/2^(n-1) = 36.8 2.353 x
= 36.8 3.294 tons per shift
= (33.506, 40.094)


Example 3: In a random selection of 64 of the 2400 intersections in a city, the mean number of car accidents per year was 3.2 and the sample standard deviation was 0.8. Obtain the 90% confidence interval for the mean number of accidents per intersection per year.
Solution: Given
N = 2400 (this means the population is finite)
n = 64 (large)
= 3.2
= 0.8,  = 0.1  /2 = 0.05 and Z0.05 = 1.64
using the confidence interval formula for large sample and finite population given by:
Z/2{ }
Substituting the given values, one can obtain the interval as 3.2 0.16 accidents per intersection.