Question

Searchers are planning a study to estimate the impact on crop yield when no-till is used in combination with residue retention and crop rotation. Based on data from a meta-analysis of 610 small farms, the researchers estimate there is a 2.5 % decline in crop yield when no-till is used with residue retention and crop rotation. Suppose the researchers want to produce a 95 % confidence interval with a margin of error of no more than 3.5 % .
1. Determine the minimum sample size required for this study?

Answer 1

Answer:

n=77

Step-by-step explanation:

Notation and definitions

$\hat p=0.025$ or 2.5%, estimated proportion for decline in crop yield when no-till is used with residue retention and crop rotation

$p$ true population proportion for decline in crop yield when no-till is used with residue retention and crop rotation

n (variable of interest) represent the sample size required

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

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

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 $ME =\pm 0.035$ or 3.5% 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)  

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:

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

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

$n=\frac{0.025(1-0.025)}{(\frac{0.035}{1.96})^2}=76.44$  

And rounded up we have that n=77