Coffee markets that conform to organic standards focus on the environmental aspects of coffee growing, such as the use of shade trees and a reduced reliance on chemical pesticides. A study of organic coffee growers was published in Food Policy (Vol. 36, 2010). In a representative sample of 845 coffee growers from southern Mexico, 417 growers were certified to sell to organic coffee markets while 77 growers were transitioning to become organic certified. In the United States, 60% of coffee growers are organic certified. Is there evidence to indicate that fewer than 60% of the coffee growers in southern Mexico are either organic certified or transitioning to become organic certified? State your conclusion so that there is only a 5% chance of making a Type I error.

In the United States, the proportion of coffee growers who got an organic certification is 60%.

As per an article published in Food Policy (Vol. 36, 2010), out of 845 coffee growers from southern Mexico, 494 (=417+77) coffee growers either got an organic certification or transitioning to become organically certified.

That is

The size of the sample is$n=845$

The sample proportion is

$$\begin{gathered} \hat{p}=\frac{494}{845} \\ =0.585 \end{gathered}$$

Where is the sample proportion of coffee growers who are either got an organic certification or transitioning to become organically certified.

We have to test whether coffee growers have either got organic certification or are transitioning to become certified organic is lesser than 60%.

The null and alternative hypotheses are given as

$H_0:p=0.60$

That is, in southern Mexico, the true proportion of coffee growers who are either got an organic certification or transitioning to become organically certified is not less than 60%.

And

$H_a:p<0.60$

That is, in southern Mexico, the true proportion of coffee growers who are either got an organic certification or transitioning to become organically certified is fewer than 60%.

The test statistic for testing these hypotheses is

$$\begin{gathered} Z=\frac{\hat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}} \\ =\frac{0.585-0.60}{\sqrt{\frac{0.60\left(1-0.60\right)}{845}}} \\ =\frac{-0.015}{\sqrt{0.000284024}} \\ =-0.89 \end{gathered}$$

Here

$\alpha :$The level of significance (chance of making a type I error)

$\alpha =.05$

Using the standard normal table, the critical value at the 5% significance level is -1.645.

We can see that

$Z=-0.89>-1.645$

Hence, we failed to reject the null hypothesis.

At a 5% significance level, we do not have sufficient evidence to conclude that fewer than 60% of the coffee growers in southern Mexico are either got an organic certification or transitioning to become organically certified.