Question

Pesticide residue in corn products. The U.S. Environmental Protection Agency (EPA) bans using the cancer-causing pesticide ethylene dibromide (EDB) as a fumigant for grain- and flour-milling equipment. EDB was once used to protect against infestation by microscopic roundworms called nematodes. The EPA sets maximum safe levels for DB presence in raw grain, flour, cake mixes, cereals, bread, and other grain products on supermarket shelves and in warehouses. Tests of the 3,000 corn-related products sold in one state indicated that 15 of a random sample of 175 had EDB residues above the safe level. Will more than 7% of the corn-related products in this state have to be removed from shelves and warehouses? Explain.

Short answer

It is observed that 95% confidence intervals do not include zero, so more than 7% of the corn-related products in this state have.

Step-by-step solution

Given information

Let N denotes the corn-related product sold in one state that is N = 3000

The sample size is n = 175

Let x denote the EDB residues at the above safe level, which is x = 15

Calculating the confidence interval

The true proportion is given by,

$\hat{p}=\frac{x}{n}$

Therefore,

$$\begin{gathered} \hat{p}=\frac{15}{175} \\ =0.086 \end{gathered}$$

The 95% confidence interval for the true proportion is calculated using the following formula,

$\hat{\mathbf{p}}\mathbf{\pm }\mathbf{2}\mathbf{\times }\sqrt{\frac{\hat{\mathbf{p}}\left(1-\hat{p}\right)}{\mathbf{n}}}\mathbf{\times }\sqrt{\frac{\mathbf{N}\mathbf{-}\mathbf{n}}{\mathbf{N}}}$

Therefore,

$$\begin{gathered} \hat{p}\pm 2\times \sqrt{\frac{\hat{p}\left(1-\hat{p}\right)}{n}}\times \sqrt{\frac{N-n}{N}}=0.086\pm 2\times \sqrt{\frac{0.086\left(1-0.086\right)}{175}}\times \sqrt{\frac{3000-175}{3000}} \\ =0.086\pm 0.0411 \\ =\left(0.0449,0.1271\right) \end{gathered}$$

From the above result, it is observed that 95% confidence intervals do not include zero, so more than 7% of the corn-related products in this state have