4) Consider a (ictitious) species of plant that can be categorized as Short (S)

Question

4) Consider a (ictitious) species of plant that can be categorized as Short (S) or Tall (T) and as resistant (R) sease. Use a-0.05 (0.8-pt) or nonresistant (NR) to a certain d Short (S) Tall D Total 40 18 42 Resistant (R) 18 40 60 Nonresistant (NR) Total 60 100 a. Please determine whether resistance to certain disease is associated with tree height. (0.8pt) b. Please determine whether short trees have a greater tendency to resist certain disease. (0.8pt) c. What ane the sample size conditions of a valid chi-squane fest? (0.6-pt)

Explanation / Answer

Answer

Question.No.4

Part (a)

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho?: The two variables are independent i.e Resistant to certain disease is not associated with tree height

Ha?: The two variables are dependent i.e.Resistant to certain disease is associated with tree height

This corresponds to a Chi-Square test of independence.

Rejection Region

Based on the information provided, the significance level is lpha = 0.05?=0.05 , the number of degrees of freedom isdf=(2?1)×(2?1)=1, so then the rejection region for this test is [R = {chi^2: chi^2 > 3.841}]

Test Statistics

The Chi-Squared statistic is computed as follows:

[chi^2 = sum_{i=1}^n rac{(O_{ij}-E_{ij})^2}{E_{ij}}= 2.25+1.5+1.5+1 = 6.25]

Decision about the null hypothesis

Since it is observed that [chi^2 = 6.25 > chi_c^2 = 3.841] , it is then concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables are dependent, at the 0.05 significance level.

Therefore

we could say that the;

Resistant to certain disease is associated with tree height.i.e. resistant and tree hight are dependent of each other.

Question.No.4 (C)

Determing whether short trees have a greater tendency to resist certain disease.

Solution -

For sample 1, we have that the sample size is N1?=40, the number of favorable cases is X1?=22, so then the sample proportion is [hat p_1 = rac{X_1}{N_1} = rac{ 22}{ 40} = 0.55]

For sample 2, we have that the sample size is N2?=60, the number of favorable cases is X2?=18, so then the sample proportion is [hat p_2 = rac{X_2}{N_2} = rac{ 18}{ 60} = 0.3]

The value of the pooled proportion is computed as [ar p = rac{ X_1+X_2}{N_1+N_2}= rac{ 22 + 18}{ 40+60} = 0.4]

Also, the given significance level is ?=0.05.

The following null and alternative hypotheses need to be tested:

Ho:p1?=p2?

Ha:p1?>p2?

This corresponds to a right-tailed test, for which a z-test for two population proportions needs to be conducted.

Based on the information provided, the significance level is?=0.05, and the critical value for a right-tailed test is zc?=1.64.

The rejection region for this right-tailed test is [R = {z: z > 1.64}]

The z-statistic is computed as follows:

[z = rac{hat p_1 - hat p_2}{sqrt{ ar p(1-ar p)(1/n_1 + 1/n_2)}} = rac{ 0.55 - 0.3}{sqrt{ 0.4cdot(1-0.4)(1/40 + 1/60)}} = 2.5]

Since it is observed that z=2.5>zc?=1.64, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0062, and since p=0.0062<0.05, it is concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis Ho is rejected.Therefore, there is enough evidence to claim that population proportion p1? is greater than p2?, at the 0.05 significance level.

i.e Short trees have a greater tendency to resist certain disease.

Question.No.4

Part (c)

The sample size conditions of a valid chi- square test

Note:

Please don't frame such a long question as one question because we are provided with a limited time frame.

plz do like it

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.