Question

The pathogen Phytophthora capsici causes bell peppers to wilt and die. Because bell peppers are an important commercial crop, this disease had undergone a great deal of agriculture research. It is thought that too much water aids the spread of the pathogen. Two fields are under study. The first step in the research project is to compare the mean soil water content for the two fields (Source: Journal of Agricultural, Biological, and Environmental Statistics, Vol. 2, Number 2).

Soil Water Content (% by volume)
Field A samples	10.2	10.7	15.5	10.4	9.9	10.0	16.6	15.1	15.2	13.8
	14.1	11.4	11.5	11.0
Field B samples	8.1	8.5	8.4	7.3	8.0	7.1	13.9	12.2	13.4	11.3
	12.6	12.6	12.7	12.4	11.3

Use a 5% level of significance to test the claim that Field A has a higher average soil water content than field B. (Note: on the TI calculator output make sure that X-bar1 = 12.529 and X-bar2 = 10.653).

1. State the null (H₀) and alternate (H_A) hypotheses.

2. Check the assumptions

3. Give the test statistic.

4. Give the P-value and the conclusion reached about the null hypothesis based on the P-value.

5. Summarize the final conclusion in the context of the claim.

6. Also construct a 95% confidence interval for the mean difference. Does this confidence interval include the value of 0? Should it? Explain

Answer 1

For Field A :

∑x = 175.4

∑x² = 2272.02

n1 = 14

Mean , x̅1 = Ʃx/n = 175.4/14 = 12.5286

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(2272.02-(175.4)²/14)/(14-1)] = 2.3940

For Field B :

∑x = 159.8

∑x² = 1785.68

n2 = 15

Mean , x̅2 = Ʃx/n = 159.8/15 = 10.6533

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(1785.68-(159.8)²/15)/(15-1)] = 2.4389

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a) Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 > µ2

b) Assumptions:

The data follow a normal distribution

The variance of the two population are equal

The sample is random and independent.

c) Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((14-1)*2.394² + (15-1)*2.4389²) / (14+15-2) = 5.8439

Test statistic:

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (12.5286 - 10.6533) / √(5.8439*(1/14 + 1/15)) = 2.0874

d) df = n1+n2-2 = 27

p-value = T.DIST.RT(2.0874, 27) = 0.0232

Decision:

p-value < α, Reject the null hypothesis

e) Conclusion:

There is enough evidence to conclude that Field A has a higher average soil water content than field B at 0.05 significance level

f) 95% Confidence interval for the mean difference:

At α = 0.05 and df = n1+n2-2 = 27, two tailed critical value, t-crit = T.INV.2T(0.05, 27) = 2.052

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((14-1)*2.394² + (15-1)*2.4389²) / (14+15-2) = 5.84392

Lower Bound = (x̅1 - x̅2) - t-crit*√(S²p*(1/n1 +1/n2)) = (12.5286 - 10.6533) - 2.052*√(5.8439*(1/14 + 1/15)) = 0.0320

Upper Bound = (x̅1 - x̅2) + t-crit*√(S²p*(1/n1 +1/n2)) = (12.5286 - 10.6533) + 2.052*√(5.8439*(1/14 + 1/15)) = 3.7185

0.032 < µ1 - µ2 < 3.7185

As the confidence interval do not contain 0. and both the value are positive we can say that Field A has a higher average soil water content than field B.