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. Use an alpha level of 0.05. The following results are obtained:

Herbicide A

Sample size: 94.5 cm s1: 10 cm

Herbicide B

Sample size: 109.1 s2: 9 cm

a) What df will be used?

b) Find the t* that would be used for a 90%, 95%, and 99% confidence level

c) Make a 95% confidence interval for the difference in plant height.

d) Interpret your confidence interval

e) Calculate the appropriate test statistic

f) what is the p-value?

Answer 1

a)

using conservative degree of freedom ,DF = min(n1-1 , n2-1 )=   9

b)

for 90% CI

t*=1.833

for 95% CI

t* = 2.262

for 99% CI

t* = 3.250

(excel formula =t.inv(α/2,df)

c)

mean of sample 1,    x̅1=   94.5
standard deviation of sample 1,   s1 =    10
size of sample 1,    n1=   10
      
mean of sample 2,    x̅2=   109.1
standard deviation of sample 2,   s2 =    9
size of sample 2,    n2=   10
      
difference in sample means =    x̅1-x̅2 =    -14.6000

α=0.05

t*=2.262

margin of error, E =    t*SE =    9.6241
              
difference of means =    x̅1-x̅2 =    -14.600      
confidence interval is               
Interval Lower Limit=   (x̅1-x̅2) - E = -24.2241
Interval Upper Limit=   (x̅1-x̅2) + E =     -4.9759

d)

we are 95% confident that true difference in plant height lies within confidence interval

e)

t-test for two means is used

difference in sample means =    x̅1-x̅2 =    -14.6000
      
std error , SE =    √(s1²/n1+s2²/n2) =    4.2544
      
t-statistic =    ((x̅1-x̅2)-µd)/SE =    -3.43

f)

p-value =        0.0075 [excel function: =t.dist.2t(3.43 ,9) ]