1.At the start of the growing season, two types of fertilizer were applied to la
ID: 3134970 • Letter: 1
Question
1.At the start of the growing season, two types of fertilizer were applied to large plots of land, growing ponderosa pine seedlings. Organic fertilizer was applied to seven plots and inorganic was applied to eight plots. After one year, the plots were harvested and the total biomass measurements (in grams) per plot were recorded:
Organic
Inorganic
570
634
482
482
592
493
593
445
630
558
513
460
512
583
560
a) Is the biomass after one year significantly different by fertilizer type? Use a 0.05 level of significance. Assume s21 = s22.
b)Construct a 95% confidence interval for the difference between the two fertilizer types. Interpret its meaning.
c) Re-do (a) and (b), assuming s21 s22.
Organic
Inorganic
570
634
482
482
592
493
593
445
630
558
513
460
512
583
560
Explanation / Answer
1.
a)
Formulating the null and alternative hypotheses,
Ho: u1 - u2 = 0
Ha: u1 - u2 =/ 0
At level of significance = 0.05
As we can see, this is a two tailed test.
Calculating the means of each group,
X1 = 569.8571429
X2 = 514.75
Calculating the standard deviations of each group,
s1 = 54.18618696
s2 = 57.12079931
Thus, the pooled standard deviation is given by
S = sqrt[((n1 - 1)s1^2 + (n2 - 1)(s2^2))/(n1 + n2 - 2)]
As n1 = 7 , n2 = 8
Then
S = 55.78554896
Thus, the standard error of the difference is
Sd = S sqrt (1/n1 + 1/n2) = 28.87176453
As ud = the hypothesized difference between means = 0 , then
t = [X1 - X2 - ud]/Sd = 1.90868635
Getting the critical value using table/technology,
df = n1 + n2 - 2 = 13
tcrit = +/- 2.160368656
Getting the p value using technology,
p = 0.078722145
As |t| < 2.160, and P > 0.05, we FAIL TO REJECT THE NULL HYPOTHESIS.
Hence, the biomass after one year not significantly different by fertilizer type at 0.05 level. [CONCLUSION]
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b)
For the 0.95 confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.025
t(alpha/2) = 2.160368656
lower bound = [X1 - X2] - t(alpha/2) * sD = -7.266512298
upper bound = [X1 - X2] + t(alpha/2) * sD = 117.480798
Thus, the confidence interval is
( -7.266512298 , 117.480798 )
Hence, we are 95% confident that the true mean difference in mean biomass after one year for organic and inorganic fertilizers is between 7.266512298 and 117.480798 grams. [ANSWER]
*****************************
c)
Formulating the null and alternative hypotheses,
Ho: u1 - u2 = 0
Ha: u1 - u2 =/ 0
At level of significance = 0.05
As we can see, this is a two tailed test.
Calculating the means of each group,
X1 = 569.8571429
X2 = 514.75
Calculating the standard deviations of each group,
s1 = 54.18618696
s2 = 57.12079931
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 7
n2 = sample size of group 2 = 8
Thus, df = n1 + n2 - 2 = 13
Also, sD = 28.76277445
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = 1.915918889
where uD = hypothesized difference = 0
Now, the critical value for t is
tcrit = +/- 2.160368656
Also, using p values,
p = 0.077628323
As |t| < 2.160, and P > 0.05, WE FAIL TO REJECT THE NULL HYPOTHESIS.
Hence, the biomass after one year not significantly different by fertilizer type at 0.05 level. [CONCLUSION]
******************************************
For the 0.95 confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.025
t(alpha/2) = 2.160368656
lower bound = [X1 - X2] - t(alpha/2) * sD = -7.031053531
upper bound = [X1 - X2] + t(alpha/2) * sD = 117.2453392
Thus, the confidence interval is
( -7.031053531 , 117.2453392 )
Hence, we are 95% confident that the true mean difference in mean biomass after one year for organic and inorganic fertilizers is between -7.031053531 and 117.2453392 grams. [ANSWER]
*****************************************************
Hi! If you use another method/formula in calculating the degrees of freedom in this t-test, please resubmit this question together with the formula/method you use in determining the degrees of freedom. That way we can continue helping you! Thanks!
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