Industrial wastes and sewage dumped into our rivers and streams absorb oxygen an
ID: 3312688 • Letter: I
Question
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, 5.2, 5.0, 5.1, 4.9, 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 = 0.05. (Round your answers to three decimal places.)
3. Test statistic: t =
4. Rejection region: If the test is one-tailed, enter NONE for the unused region.
t <
t >t <
Explanation / Answer
Given that,
population mean(u)=5
sample mean, x =4.9667
standard deviation, s =0.1751
number (n)=6
null, Ho: =5
alternate, H1: <5
level of significance, = 0.05
from standard normal table,left tailed t /2 =2.015
since our test is left-tailed
reject Ho, if to < -2.015
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =4.9667-5/(0.1751/sqrt(6))
to =-0.4658
| to | =0.4658
critical value
the value of |t | with n-1 = 5 d.f is 2.015
we got |to| =0.4658 & | t | =2.015
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :left tail - Ha : ( p < -0.4658 ) = 0.33046
hence value of p0.05 < 0.33046,here we do not reject Ho
ANSWERS
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null, Ho: =5
alternate, H1: <5
test statistic: -0.4658
critical value: -2.015
decision: do not reject Ho
p-value: 0.33046
reject Ho, if to < -2.015
we support the claim that is less than 5