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% of 61 randomly selected kochia plants were resistant to glyphosate. In 2017 , 38. 5% of 52 randomly selected kochia plants were resistant to glyphosate. Do the data provide convincing statistical evidence, at the level of α=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, it is found that the data provides convincing evidence that the proportion has increased.

What are the hypotheses tested?

At the null hypotheses, it is tested if the proportion has not increased, that is, the difference of the 2017 proportion and the 2014 proportions is not positive, hence:

$H_0: p_2 - p_1 \leq 0$

At the alternative hypothesis, it is tested if the proportion has increased, hence:

$H_1: p_2 - p_1 > 0$

What are the mean and the standard error for the distribution of differences?

For each sample, they are given by:

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

$p_2 = 0.383, s_2 = \sqrt{\frac{0.385(0.615)}{52}} = 0.0675$

Hence, for the distribution of differences, they are given by:

$\overline{p} = p_2 - p_1 = 0.383 - 0.197 = 0.186$

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

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:

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

$z = \frac{0.186 - 0}{0.0845}$

z = 2.2

What is the decision?

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

Since the test statistic is greater than the critical value for the left tailed test, the data provides convincing evidence that the proportion has increased.

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