Preliminary sample of 100 labourers was selected from a population of 5000 labourers by simple random sampling. it was found that 40 of the selected labourers opt for a new incentive scheme. how large a sample must be selected to have a precision of ± 5% with 95% confidence ?

n=369

Step-by-step explanation:

1) Notation and definitions

number of the selected labourers opt for a new incentive scheme.

random sample taken

estimated proportion of the selected labourers opt for a new incentive scheme.

true population proportion of the selected labourers opt for a new incentive scheme.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

2) Solution tot he problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by and . And the critical value would be given by:

The margin of error for the proportion interval is given by this formula:  

   (a)  

And on this case we have that and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

  (b)  

And replacing into equation (b) the values from part a we got:

And rounded up we have that n=369

60%