LARGE SAMPLES
These are samples that have a sample size greater than 30(that is n>30)
(a) Estimation of population mean
Here we suppose that if we take a large sample from a population then the mean of the population is extremely close to the mean of the sample
Steps to follow to estimate the population mean having:
i. Take a random sample of n items where (n>30)
ii. Calculate sample mean (x¯) and standard deviation (S)
iii. Calculate the standard error of the mean by using the following formular
S_{x¯} = s/√n
Whereas S_{x¯}= Standard error of mean
S = standard deviation of the sample
n = sample size
iv. Choose a confidence level for illustration: 95 percent or 99 percent
v. Estimate the population mean as under
Population mean µ = x¯ ± (Appropriate number) × S_{x¯}
'Appropriate number' means confidence level for illustration, at 95 percent confidence level is 1.96 this number is generally denoted by Z and is acquired from the normal tables.
Illustration
The quality department of a wire manufacturing company periodically chooses a sample of wire specimens in order to test for breaking strength. Past experience has displayed that the breaking strengths of a specific type of wire are normally distributed along with standard deviation of 200 kilogram (kg). A random sample of 64 specimens gave a mean of 6200 kilogram (kg). Find out the population mean at 95 percent level of confidence
Solution
Population mean = x¯ ± 1.96 S_{x¯}
Note that sample size is already n > 30 whereas s and are described hence step i), ii) and iv) are provided.
Now: x¯ = 6200 kilogram (kg)
S_{x¯}= s/√n = 200/√64 = 25
Population mean = 6200 ± 1.96(25)
= 6200 ± 49
= 6151 to 6249
At 95 percent level of confidence, population mean will be in among 6151 and 6249