For a particular variety of wheat, it has been estimated that 5 percent of plants suffered an illness. A sample of 600 plants of the same variety of wheat was observed and found that 50 plants were infected with disease. Test if the sample the results were consistent with the population.

Let's denote:
- p as the proportion of plants suffering from illness in the population
- \hat{p} as the proportion of plants suffering from illness in the sample

Given that in the population it has been estimated that 5% of plants suffered an illness, so p = 0.05 .

In the sample of 600 plants, there were 50 plants infected with the disease, so the proportion in the sample is:
\hat{p} = \frac{50}{600} = \frac{1}{12} \approx 0.0833

To test if the sample results were consistent with the population, we will perform a hypothesis test:
- Null Hypothesis ( H_0 ): The proportion of plants suffering from illness in the sample is the same as in the population ( \hat{p} = p )
- Alternative Hypothesis ( H_1 ): The proportion in the sample is significantly different from the population ( \hat{p} \neq p )

We will use the Z-test for proportions to test this hypothesis. The formula for the Z-score is:
Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}
where n is the sample size, p is the population proportion, and \hat{p} is the sample proportion.

Plugging in the values:
Z = \frac{0.0833 - 0.05}{\sqrt{\frac{0.05(1-0.05)}{600}}} \approx \frac{0.0333}{0.0064} \approx 5.20

The critical Z-score for a 2-tailed test at a 5% significance level is approximately \pm 1.96 .

Since |5.20| > 1.96 , we reject the null hypothesis.

Therefore, the sample results are not consistent with the population.

\boxed{\text{Answer}: \text{The sample results are not consistent with the population.}}