Question

PART 1

Two methods are compared to inoculate or infect a corn fungus strain known as huitlacoche. In a first stage of the study, the experimenter wants to determine which of the methods generates the highest percentage of infection. The method A consists in cutting the tip of the ear to apply the strain, and in the method B the strain is injected transversely. Of 41 ears inoculated with the method A, 20 were infected, that is, they generated huitlacoche; Meanwhile, 38 ears of corn inoculated with the method B they became infected 27.

Cob	% Coverage (Texcoco)	% Coverage (Celaya)	Weight in grams (Texcoco)	Weigth in grams (Celaya)	Difference
1	60	95	122.6	231.8
2	40	100	182.74	346.74
3	95	70	203.45	231.41
4	55	40	84.03	141.49
5	40	35	128.46	149.69
6	20	100	31.85	291.28
7	10	30	12.81	86.03
8	10	100	57.05	158.74
9	55	100	145.83	167.25
10	15	100	49.49	120.89
11	35	25	103.66	19.7
12	25	15	95.05	22.08
13	70	85	125.02	137.02
14	20	15	40.57	28.76
15	20	30	19.36	24.87

Part 2

With respect to the problem described in part 1, the best method of inoculation was applied to two varieties of maize in two locations. Once the ear is infected, it is important to measure the final percentage of the surface of the ear that was covered by the fungus, as well as the weight in grams of the huitlacoche. The results for variety 2 of maize, obtained in 15 ears of Texcoco and in 15 ears of Celaya are the same as in the table in part 1.

a) Can we say that the percentage of coverage of the fungus is higher in Celaya than in Texcoco? Test the appropriate hypothesis for the means

b) Use a scatter diagram (graph type X-Y) to verify if there is a linear relationship between the percentage of coverage of the ear with grams of huitlacoche.

c) Ignore the coverage and test the equality of the average production of huitlacoche in the two locations.

d) It is evident that the greater the coverage, the greater the production of huitlacoche, would there be a way of knowing with these data whether a similar production of huitlacoche in both locations corresponds to the same coverage? Argue your answer.

Note: The exercises must be solved in two ways. The first is without using the excel data analysis and the second using the excel data analysis. Remember that you have to show step by step explaining clearly.

Answer 1

PART 1

We have to compare the two sample proportions.

The two methods A and B have been applied.

The proportion infected by the method A

p_a = 20/41

The proportion infected by the method B

pb = 27/38

1) The null and the alternative hypothesis for testing if the method B is better than method A is given as follows.

H₀ = p_a- p_b = 0

H1 = pa - pb <0

p^* = x1+ x2/ n1+n2

The test statistic is given as

The critical value at the 0.05 level of significance is -1.605

Hence we reject the null hypothesis.

Hence we say that the method B is better than method A.

PART 2

(a) To test if the percentage coverage of fungus is higher in celaya than in texcoco.

x1 be the average coverage of fungus in texcoco = 38

x2 be the average coverage of fungus in celaya = 62.67

The null and the alternative hypothesis is given as

H0 : x1 = x2

H1 : x1 <x2

The test statistic for testing the following hypothesis is given as

Which falls in the critical region for 5 % level of significance for a left tailed test

Hence we reject the null hypothesis.

The percentage covergae of fungus in celaya is greater than in texcoco.

(b)

In the case of celaya the scatter plot between the percentage coverage and the weight of huitlacoche

(c) To test if the average production of huitlacoche is same in both the locations.

The null and alternative hypothesis is given as follow

H0 : x1 = x2

H1 :

x1 = 93.46 is the average production of texcoco

x2 = 143.85 is the average production of celaya

The test statistic is given as

Hence we reject the null hypothesis.