Question

Tumbleweed, commonly found in the western United States, is the dried structure of certain plants that are blown by the wind. Kochia, a type of plant that turns into tumbleweed at the end of the summer, is a problem for farmers because it takes nutrients away from soil that would otherwise go to more beneficial plants. Scientists are concerned that kochia plants are becoming resistant to the most commonly used herbicide, glyphosate. In 2014, 19. 7 percent of 61 randomly selected kochia plants were resistant to glyphosate. In 2017, 38. 5 percent of 52 randomly selected kochia plants were resistant to glyphosate. You will conduct a significance test to determine if the data provides convincing statistical evidence, at the level of a = 0. 05, that there has been an increase in the proportion of all kochia plants that are resistant to glyphosate?

Answer 1

Using the z-distribution, as we are working with a proportion, it is found that since the test statistic is less than the critical value for the right-tailed test, there is not enough evidence to conclude that there has been an increase in the proportion of all kochia plants that are resistant to glyphosate.

In this problem, we consider that:

• $p_1$ is the proportion in 2014.

• $p_2$ is the proportion is 2017.

What are the hypothesis tested?

At the null hypothesis, we test if there has been no increase, that is, the subtraction is of 0.

$H_0: p_2 - p_1 = 0$

At the alternative hypothesis, we test if there has been an increase, hence:

$H_1: p_2 - p_1 \neq 0$.

What is the distribution of the difference of sample proportions?

The proportions and standard errors are given by:

$p_1 = 0.197, s_1 = \sqrt{\frac{0.197(0.803)}{61}} = 0.0509$

$p_2 = 0.385, s_2 = \sqrt{\frac{0.385(0.615)}{52}} = 0.0675$

Hence, the distribution has mean and standard error given by:

$\overline{p} = p_2 - p_1 = 0.0675 - 0.0509 = 0.0166$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.0509^2 + 0.0675^2} = 0.0845$

What is the test statistic?

It is given by:

$z = \frac{\overline{p} - p}{s}$

In which p = 0 is the value tested at the null hypothesis, hence:

$z = \frac{0.0166}{0.0845}$

$z = 0.2$

What is the decision?

Considering a right-tailed test, as we are testing if the proportion is greater than a value, with a significance level of 0.05, the critical value is of $z^{\ast} = 1.645$

Since the test statistic is less than the critical value for the right-tailed test, there is not enough evidence to conclude that there has been an increase in the proportion of all kochia plants that are resistant to glyphosate.

More can be learned about the z-distribution at https://brainly.com/question/26454209