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 ko:hia 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 selecteikochia plants were resistant to glyphosate. do the data provide convincing statistical evidence, at the leve of 0.05, that there has been an increase in the proportion of all kochia plants that are resistant to glyphosale?

Answer 1

Since we are dealing with a proportion, it is determined by using the z-distribution that there is insufficient evidence to draw the conclusion that the proportion of all kochia plants that are resistant to glyphosate has increased because the test statistic is less than the critical value for the right-tailed test.

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:

$\bar{p}=p_{2}-p_{1}=0.0675-0.0509=0.0166\\s=\sqrt{s_{1}^{2}+s_{2}^{2}}=\sqrt{0.0509^{2}+0.0675^{2}}=0.0845$

What is the test statistic?

It is given by:

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

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 $z^{*}=1.645$

• There is insufficient information to establish that the fraction of all kochia plants that are resistant to glyphosate has increased because the test statistic is smaller than the crucial value for the right-tailed test.

Learn More about the z-distribution here: brainly.com/question/26454209

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