Important formula for numerical problems

Determination of sample size in research

1. For quantitative variables

$$n_0=\frac{t^2_a/_2σ^2}{d^2}$$

Then to estimate ‘n’ by using the relation.

$$n=\frac{n_0}{1+\frac{n_0}{N}}$$

Where (σ)=population standard deviation.

ta/2=critical value for two tail test at α level of significance.

D=described error in estimation of the population means =\(\overline{X}\)-μ

But for, large population, N→∞

$$\frac{n_0}{N}→0$$

Then it becomes,

$$n=n_0$$

2. For attributes with two possible outcomes.

Then the relation takes the form

$$n_0=\frac{t^2_a/_2PQ}{d^2}$$

Where P= the proportion of an outcome of a dichotomous attribute in the population.

$$Q=1-P.$$

3. For sample size based on C.V

$$n_0=\frac{t^2_a/_2(CV^2)}{P^2}$$

Where,C.V=coefficient of variation

P=% of error in estimation of the sample mean.

Confidence limit.

1.Sometimes on the basis of sample mean observed, the limit of the value of unknown population mean are to be predicted. The limit is called confidence limits.

2. The range of the limit is called confidence interval and the estimation done is called interval estimation.

$$C.I=\overline{X}±t_a/_2S.t$$

$$Where\,\overline{X}=mean$$

$$C.I=Condidence\,Interval$$

$$S.E=\frac{σ}{\sqrt{n}}=standard\,error$$

Numerical

1. It is intended to conduct a survey of disabled persons in a certain locality of Nepal. The prior known disability rate was 0%. What should be the sample size for the survey if the desired error of estimation is 5%?

$$solution$$

$$Given$$

$$P=10%$$

$$P=\frac{10}{100}=0.1$$

$$Q=1-P$$

$$Desired\,error(d)=5\,percent \,%$$

$$(d)=\frac{5}{100}=0.05$$

When the level of significance is not given, generally it assumed to be 5%.

$$Then,α=5()$$

We have,

$$n_0=\frac{t^2_a/_2PQ}{d^2}$$

$$n_0=\frac{1.96^2×0.1×0.9}{(0.05)^2}$$

$$n_0=144$$

Again,we know that,

$$N=n=\frac{n_0}{1+\frac{n_0}{N}}$$

$$For\,population\,size\,N→∞\,,\frac{n_0}{N}→0$$

$$Hence\,n=n_0$$

$$N=144$$

3. A quality control engineer wants to estimate the fraction of defectives in a large lot of film cartridge from previous experience she feels that the actual fraction of defective should be somewhere around 0.05. How large a sample she should take if she wants to estimate the true fraction within 0.01, using 95% probability?

$$P=0.05$$

$$q=1-0.05=0.95$$

$$Desired\,error\,of\,estimation\,(d)=0.01$$

$$F0r\,α\,5\,percent\,,t_1/_2=1.96$$

$$SAmple\,size\,(n)=?$$

We have,

$$n_0=\frac{t^2_a/_2PQ}{d^2}$$

$$n_0=\frac{1.96^2×0.05×0.95}{(0.01)^2}$$

$$=1824.76$$

Again we have,

$$n=\frac{n_0}{1+\frac{n_0}{N}}$$

$$For\,large\,,population\,\,size\,N→∞\,,\frac{n_0}{N}→0$$

$$n=n_0$$

$$N=1824.76$$

Reference.

Kerlinger, F.N. Foundation of Behavioural Research. New Delhi: Surjeet Publication, 2000.

Kothari, C.R. Research Methodology. India: Vishwa Prakashan, 1990.

Singh, M.L. and J.M Singh. Understanding Research Methodology. 1998.

Singh, Mrigendra Lal. Understanding Research Methodology. Nepal: National Book centre, 2013.