Question

Suppose that 10% of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in this area are randomly selected, and 30 are found to be infested with whitefly.
(a) Assuming that the experiment satisfies the conditions of the binomial experiment, do the data indicate that the proportion of infested fields is greater than expected? Use the p-value approach, and test using a 5% significance level. (Round your answer to four decimal places.)p-value = Conclusion:H0 is not rejected. There is sufficient evidence to indicate that the proportion of infested fields is larger than expected.H0 is rejected. There is sufficient evidence to indicate that the proportion of infested fields is larger than expected. H0 is rejected. There is insufficient evidence to indicate that the proportion of infested fields is larger than expected.H0 is not rejected. There is insufficient evidence to indicate that the proportion of infested fields is larger than expected.

Answer 1

Answer:

H0 is rejected. There is sufficient evidence to indicate that the proportion of infested fields is larger than expected.

Step-by-step explanation:

Let p be the proportion of fields infested with whitefly.

$H_{0}$: p=0.10

$H_{a}$: p>0.10

test statistic of the sample proportion ca be found as:

z=$\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

• p(s) is the sample proportion of fields infested with whitefly (0.30)

• p is the proportion assumed under null hypothesis. (0.10)

• N is the sample size (100)

Then z=$\frac{0.30-0.10}{\sqrt{\frac{0.10*0.90}{100} } }$ ≈ 6,66

Since p(z)<0.0001 <0.05 . We can reject the null hypothesis.