Some researchers have conjectured that stem-pitting disease in peach-tree seedli

Question

Some researchers have conjectured that stem-pitting disease in peach-tree seedlings might be controlled with weed and soil treatment. An experiment is conducted to compare peach-tree seedling growth when the soil and weeds are treated with one of two herbicides. In a field containing 20 seedlings, 10 are randomly selected throughout the field and assigned to receive herbicide A. The remainder of the seedlings is assigned to receive herbicide B. Soil and weeds for each seedling are treated with the appropriate herbicide, and at the end of the study period, the height in centimeters is recorded for each seedling. The following results are obtained: Herbicide A = 94.5 cm s1 = 10 cm Herbicide B = 109.1 cm s2 = 9 cm 1.

What is a 90% confidence interval for u2 –u 1? a. 14.6 ± 7.00 b. 14.6 ± 7.38 c. 14.6 ± 7.80 d. 14.6 ± 9.62

What can we say about the value of the P-value? a. P-value < 0.01 b. 0.01 < P-value < 0.05 c. 0.05 < P-value < 0.10 d. P-value > 0.10

Please answer both questions with some explanation,

Explanation / Answer

(a)

n1 = 10

n2 = 10

x1-bar = 109.1

x2-bar = 94.5

s1 = 9

s2 = 10

% = 90

Degrees of freedom = n1 + n2 - 2 = 10 + 10 -2 = 18

Pooled s = (((n1 - 1) * s1^2 + (n2 - 1) * s2^2)/DOF) = (((10 - 1) * 9^2 + ( 10 - 1) * 10^2)/(10 + 10 -2)) = 9.513148795

SE = Pooled s * ((1/n1) + (1/n2)) = 9.51314879522022 * ((1/10) + (1/10)) = 4.254409477

t- score = 1.734063592

Width of the confidence interval = t * SE = 1.73406359230939 * 4.25440947723653 = 7.377416581

The 90% confidence interval is 14.6 ± 7.38 (Option b)

(b)

Data:        

n1 = 10       

n2 = 10       

x1-bar = 109.1       

x2-bar = 94.5       

s1 = 9       

s2 = 10       

Hypotheses:        

Ho: 1 = 2        

Ha: 1 2        

Decision Rule:        

= 0.1       

Degrees of freedom = 10 + 10 - 2 = 18      

Lower Critical t- score = -1.734063592       

Upper Critical t- score = 1.734063592       

Reject Ho if |t| > 1.734063592       

Test Statistic:        

Pooled SD, s = [{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] =   (((10 - 1) * 9^2 + (10 - 1) * 10^2)/(10 + 10 -2)) =     9.513

SE = s * {(1 /n1) + (1 /n2)} = 9.51314879522022 * ((1/10) + (1/10)) = 4.254409477      

t = (x1-bar -x2-bar)/SE = 3.431733611       

p- value = 0.002974816 (Option a)

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