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