Salmonella in yield. Salmonella infection is the most common bacterial foodborne illness in the United States. How current is Salmonella in yield grown in the major agricultural region of Monterey, California? Experimenters from the U.S. Department of Agriculture (USDA) conducted tests for Salmonella in yield grown in the region and published their results in Applied and Environmental Microbiology (April 2011). In a sample of 252 societies attained from water used to wash the region, 18 tested positive for Salmonella. In an independent sample of 476 societies attained from the region's wildlife (e.g., catcalls), 20 tested positive for Salmonella. Is this sufficient substantiation for the USDA to state that the frequency of Salmonella in the region's water differs from the frequency of Salmonella in the region's wildlife? Use a = .01 to make your decision

Let p₁ be the proportion of Salmonella in the region's water and p₂ be the proportion of Salmonella in the region's wildlife. The hypotheses are given below:

Null hypothesis:

H₁: P₁ – P₂ = 0

There is no evidence that the prevalence of Salmonella in the region's water differs from the prevalence of Salmonella in the region's wildlife.

Alternative hypothesis:

$H_e$: P₁ – P₂ ≠ 0

There is evidence that the prevalence of Salmonella in the region's water differs from the prevalence of Salmonella in the wildlife.

calculate the critical value.

Let the confidence position be0.99.

1-α = 0.99

α = 1-0.99

= 0.01

$\frac{\alpha }{2}=0.005$

From Excursus Table II, the value of $z_{\frac{a}{2}}$is given below.

$\begin{array}{l}{z}_{\frac{a}{2}}=Z_{0.005}\\ =2.58\end{array}$

$So,thevalueof{z}_{\frac{{}_a}{2}}is2.58.$

Rejection region:

If Z >$z_{\frac{a}{2}}$(= 2.58), then reject the null hypothesis.

its Z > $z_{\frac{a}{2}}$(= -2.58), then reject the null hypothesis.

Calculate the value of ${\overline{P}}_1$and ${\overline{P}}_2$

Consider α = 0.10, n₁= 476,x₁ = 18, and x₂ = 20.

The value of ${\overline{P}}_1$is obtained as shown below.

$\begin{array}{l}{\overline{P}}_1=\frac{x_1}{n_1}\\ =\frac{18}{252}\\ =0.0714\end{array}$

The value of ${\overline{P}}_2$is obtained as shown below.

$\begin{array}{l}{\overline{P}}_2=\frac{x_2}{n_2}\\ =\frac{20}{252}\\ =0.0420\end{array}$

The value of $\overline{P}$is obtained below.

$\begin{array}{l}\overline{P}=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{18+20}{252+476}\\ =\frac{38}{728}\\ =0.0522\end{array}$

Calculate the test statistic (z).

The formula for z is given below.

$Z=\frac{{\overline{p}}_1-{\overline{p}}_2}{\sqrt{\overline{p}\overline{q}(\frac{1}{n}+\frac{1}{n})}}$

There $\overline{P}{}_1$= 0.0714, ${\overline{P}}_2$=0.0420, $\overline{P}$ = 0.0522, n₁=252 and n₂ =476.

$\begin{array}{l}\mathrm{Z}=\frac{0.0714-0.0420}{0.0522(1-0.0522)(\frac{1}{252}+\frac{1}{476})}\\ =\frac{0.0294}{0.0174}\\ =1.69\end{array}$

The critical value is 2.58, and the value of z is 1.69.

Here, the value of z is lesser than the value of$z_{\frac{a}{2}}$.

That is, z (= 1.69) <$z_{\frac{a}{2}}$(=2.58).

So, by the rejection rule, don't reject the null hypothesis ($H_0$)

Interpretation

Thus, it can be concluded that there is no evidence to reject the null hypothesis$H_0$ at α = 0.01.

Hence, there is no evidence that the prevalence of Salmonella in the region's water differs from the prevalence of Salmonella in the region's wildlife.

From the above steps, there is no evidence to reject the null hypothesis at α = 0.01.