n systematic and random error? 5) Using the description and data below, perform
ID: 3353068 • Letter: N
Question
n systematic and random error? 5) Using the description and data below, perform a statistical analysis on state whether the data below. Clearly er you are using either a student's t-test or a paired t-test and why you have chosen to use that test. Use Table 1 on page 11 (Show your work by han of the lab handout as needed. d. You do not need to show calculation of averages or basic ition/subtraction, but I should be able to see your usage of proper equations.) An Indiana engineering student has inherited dozens of farming plots from his grandfather, an illustrious corn and soy bean farmer. Since this student doesn't have time to spend growing corn himself, he has found a team to manage the land on his behalf. Looking through his grandfather's notes, he found some old data gathered from several growing seasons. The land management team has asked which of these two fertilizers should be used for the soil in Indiana. This engineering student has no idea how to do statistics and has come to you for help. Be sure to tell him which fertilizer is better, and why they should use that one for this land. Hint -a two-sided test only tells you if something is DIFFERENT. A one-sided test ill tll you if something is larger or smaller.Explanation / Answer
We will be conducting paired t-test to compare two means (of corn produced) because we have two samples (Fertilizer A and Fertilizer B) in which observations in one sample can be paired (by growing year) with observations in the other sample.
Null hypothesis: d = 0
Alternative hypothesis: d > 0
where d is the mean difference of corn produced from Fertilizer A and Fertilizer B.
Let di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.
Average of the difference, d = 1.243
s = sqrt [ ((di - d)2 / (n - 1) ]
s = sqrt[ 2.5971/(7-1) ] = sqrt(0.4329) = 0.6579
SE = s / sqrt(n) = 0.6579 / [ sqrt(7) ]
SE = 0.6579/2.6458 = 0.2487
DF = n - 1 = 7 -1 = 6
t = (d - D)/ SE = (1.243 - 0)/0.2487 = 4.998
The P-value is the probability that a t statistic having 6 degrees of freedom is more extreme than 4.998; that is, greater than 4.998.
We use the t Distribution Calculator to find P(t > 4.998) = 0.0012. Thus, the P-value = 0.0012
Growing Year Corn produced with Fertilizer A Corn produced with Fertilizer B Difference 1996 14.2 12.4 +1.8 1997 14.8 14.3 +0.5 1998 15.2 13.5 + 1.7 1999 16 14.2 +1.8 2000 14.5 12.8 +1.7 2001 13.9 13.6 +0.3 2002 15.4 14.5 +0.9Related Questions
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