Little Bear
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Homework Statement
Dissolved O2 Content. Industrial wastes and sewage dumped into our rivers and streams absorb oxygen and thereby reduce the amount of dissolved oxygen available for fish and other forms of aquatic life. One state agency requires a minimum of 5 parts per million (ppm) of dissolved oxygen in order for the oxygen content to be sufficient to support aquatic life. Six water specimens taken from a river at a specific location during the low-water season (July) gave readings of 4.9, 5.1, 4.9, 5.0, 5.0 and 4.7 ppm of dissolved oxygen. Do the data provide sufficient evidence to indicate that the dissolved oxygen content is less than 5 ppm? Test using alpha = .05.
Homework Equations
n = 6
sum of x samples = x1+x2...+xn
ample mean = (x1+ x2...+xn)/n
Sum of squares = x1^2+x2^2+...+xn^2
s^2 = [sum of squares – (sum of x samples)^2/n]/[n-1]
s = sqrt(s^2)
popn mean = 5 (parts ppm)
t = [sample mean – popn mean]/[s/sqrt(n)]
The Attempt at a Solution
sum of x samples = [4.9 + 5.1 + 4.9 + 5.0 + 5.0 + 4.7] = 29.6
sample mean = sum of x samples/6 = 29.6/6 = 4.93
sum of squares(x^2) = [4.9^2+ 5.1^2+ 4.9^2+ 5.0^2+ 5.0^2+ 4.7^2]
= 24.01 + 26.01 + 24.01 + 25 + 25 + 22.09
= 146.12
s^2= [Sum of x^2– (Sum of x samples)^2/n]/[n-1]
= [146.12 – (29.6^2/6)]/[6-1]
= [146.12 – 146.03]/5
= 0.09/5
= 0.018
s = sqrt(0.018)
s = 0.134
t = [sample mean – popn mean]/[s/sqrt(n)]
= [4.93 – 5]/[0.134/sqrt(6)]
= -0.07/0.0547
= -1.28
The answer at the back of the book is:
No; t = -1.195.
So, what did I do wrong in my calculations?