Question

1. Marta has data from 12 streams, which range from glaciated to non-glaciated – from which she has estimated the spring contribution of both the water runoff and dissolved organic carbon yield (units are % for both).  She wants to make a point within her manuscript that the annual contribution of runoff in the spring not equal to that of the dissolved organic carbon yield for the same season.  Can you help her out?

Site	Glaciated?	Runoff	DOC yield
1	Y	46	52
2	Y	49	53
3	Y	59	63
4	Y	56	61
5	Y	57	63
6	Y	56	62
7	N	67	72
8	N	60	64
9	N	53	55
10	N	61	58
11	N	60	67
12	N	67	68

1. What is the null and alternative hypothesis?
2. Is the data normally distributed (and does it matter for your t-test?)
3. Which type of t-test and why and calculate the t-statistic (by hand, show your work)
4. Now in R – code & results
5. What can you conclude based on this analysis?

Answer 1

Sol:

Null hypothesis states that mean water runoff and dissolved organic carbon yield are equal

Ho:

Alternative hypothesis states that mean water runoff and dissolved organic carbon yield are different

Ha:

Solutionb:

yes for t test

sample sizes to be less than 30 and

sample should be random and indpedent.

Should follow normal ditribution

Run off follows normal distribution

DOC yield follow normal dsitribution

Solutionc:

Get the descriptive stats for both the run off and DOC yield

Runoff		DOC yield
Mean	57.58333333	Mean	61.5
Median	58	Median	62.5
Standard Deviation	6.302356462	Standard Deviation	6.112580172
Count	12	Count	12

t =x1bar-x2bar/sqrt(s1^2/n1+s2^2/n2)

t=(57.58333333-61.5)/sqrt(6.302356462^2/12+6.112580172^2/12)

t=-1.545

Solutiond:

Rcode:

Runoff <- c(46,49,59,56,57,56,67,60,53,61,60,67)
DOC_yield <- c(52,53,63,61,63,62,72,64,55,58,67,68)
shapiro.test(Runoff)
shapiro.test(DOC_yield)
hist(Runoff)
hist(DOC_yield)
t.test(Runoff,DOC_yield)

Output of t est:

data: Runoff and DOC_yield
t = -1.5454, df = 21.979, p-value = 0.1365
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-9.173150 1.339817
sample estimates:
mean of x mean of y
57.58333 61.50000

Solutione:

t statsitci form R=-1.5454

which is same as we calculated

p=0.1365

p>0.05

Conclusion:

There is no sufficient evidence at 5% level of significance to conclude that

annual contribution of runoff in the spring not equal to that of the dissolved organic carbon yield for the same season.