Question

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 ?

Answer 1

Answer:

$n=\frac{0.4(1-0.4)}{(\frac{0.05}{1.96})^2}=368.79$  

n=369

Step-by-step explanation:

1) Notation and definitions

$X=40$ number of the selected labourers opt for a new incentive scheme.

$n=100$ random sample taken

$\hat p=\frac{40}{100}=0.4$ estimated proportion of the selected labourers opt for a new incentive scheme.

$p$ 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

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

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 $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

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

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$    (a)  

And on this case we have that $ME =\pm 0.05$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$   (b)  

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

$n=\frac{0.4(1-0.4)}{(\frac{0.05}{1.96})^2}=368.79$  

And rounded up we have that n=369

Answer 2

The size of the sample that must be selected to have a precision of ± 5% with 95% confidence is 369

How to find the margin of error of sample proportion?

For large enough sample, let the population proportion of a quantity be denoted by random variable $p$

Then, we get:

$p \sim N(\hat{p}, \sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}})$

where

• $\hat{p}$ = estimated (mean value) proportion of that quantity, and
• n = size of sample drawn.

It is visible that as we increase the value of n, the standard deviation decreases, therefore, forcing the values of population proportion to be closer to the estimated proportion.

Margin of error is the distance between the mean and one of the end point of the confidence interval(assuming its equal on  both the sides of the mean). The margin of error with level of significance $\alpha$ is calculated as:

$MOE = Z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

where

$Z_{\alpha/2}$

is the critical value of the test statistic for level of significance $\alpha$

For the considered case, we have following facts:

• Size of the preliminary sample = 100
• The precision needed = Margin of error = 5% =0.05
• Confidence level = 95%
• Count of labors in sample who opt for a new incentive scheme = 40

Thus, if we denote p = proportion of labors opting for a new incentive scheme in the considered population, then,

$\hat{p} = 40/100 = 0.4$ (estimate from the sample about the proportion of such labors who opt for new incentive scheme to the total count of labors of the sample).

For 95% confidence interval, level of significance is 100% - 95% = 5% = 0.05

At this level of significance, the critical value of Z is ±1.96

Let the needed sample size be 'n', then:

$MOE = Z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\0.05= \pm 1.96 \sqrt{\dfrac{0.4(1-04)}{n}}\\\\n = \dfrac{0.4 \times 0.6}{(0.05/1.96)^2} \approx 369$

Thus, the size of the sample that must be selected to have a precision of ± 5% with 95% confidence is 369

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