Question

in june of 2020, a study was performed to determine the extent to which trees in south carolina had been infested by asian longhorn beetles. researchers believed that more than 12% of trees had been infested. after randomly selecting 1408 south carolina trees, the researchers determined that 197 had been infested by the asian longhorn beetle. at the 0.05 significance level, test the researchers claim.
a. enter the null hypothesis for this test.
$h_0$:?
?

b. enter the alternative hypothesis for this test.
$h_1$:?
?

c. is the original claim located in the null or alternative hypothesis?
select an answer

d. what is the test statistic for the given statistics?

e. what is the p - value for this test?

f. what is the decision based on the given statistics?
select an answer

g. what is the correct interpretation of this decision?
using a 5% level of significance, there select an answer sufficient evidence to select an answer the claim that more than 12% of trees had been infested.

Explanation:

Step1: Define null and alternative hypotheses

The null hypothesis $H_0$ is a statement of no - effect or no difference. The alternative hypothesis $H_1$ is what we are trying to find evidence for. The claim is that more than 12% of trees are infested. Let $p$ be the proportion of infested trees.
$H_0:p\leq0.12$
$H_1:p > 0.12$

Step2: Calculate the sample proportion $\hat{p}$

The sample proportion $\hat{p}=\frac{x}{n}$, where $x = 197$ (number of infested trees) and $n=1408$ (sample size).
$\hat{p}=\frac{197}{1408}\approx0.14$

Step3: Calculate the test - statistic $z$

The formula for the test - statistic in a one - sample proportion test is $z=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}}$, where $p_0 = 0.12$ (the hypothesized proportion under the null hypothesis), $\hat{p}\approx0.14$, and $n = 1408$.
$z=\frac{0.14 - 0.12}{\sqrt{\frac{0.12\times(1 - 0.12)}{1408}}}=\frac{0.02}{\sqrt{\frac{0.12\times0.88}{1408}}}=\frac{0.02}{\sqrt{\frac{0.1056}{1408}}}=\frac{0.02}{\sqrt{0.0000750007}}\approx\frac{0.02}{0.00866}\approx2.31$

Step4: Calculate the p - value

Since this is a right - tailed test, the p - value is $P(Z>z)$. Using the standard normal distribution table, $P(Z > 2.31)=1 - P(Z\leq2.31)$. From the standard normal table, $P(Z\leq2.31)=0.9896$, so the p - value is $1 - 0.9896 = 0.0104$.

Step5: Make a decision

The significance level $\alpha=0.05$. Since the p - value ($0.0104$) is less than $\alpha = 0.05$, we reject the null hypothesis.

Step6: Interpret the decision

Using a 5% level of significance, there is sufficient evidence to support the claim that more than 12% of trees had been infested.

Answer:

a. $H_0:p\leq0.12$
b. $H_1:p > 0.12$
c. Alternative hypothesis
d. $z\approx2.31$
e. $p - value\approx0.0104$
f. Reject the null hypothesis
g. Using a 5% level of significance, there is sufficient evidence to support the claim that more than 12% of trees had been infested.