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, 5.0, 5.0, 5.1, 4.9, and 4.8 ppm of dissolved oxygen. Do the data provide sufficient evidence to indicate that the dissolved oxygen content is less than 5 ppm? Test using α = 0.05.
State the null and alternative hypotheses.
H0: μ ≠ 5 versus Ha: μ = 5
H0: μ < 5 versus Ha: μ > 5
H0: μ = 5 versus Ha: μ ≠ 5
H0: μ < 5 versus Ha: μ = 5
H0: μ = 5 versus Ha: μ < 5
State the test statistic. (Round your answer to three decimal places.)
t =
State the rejection region. (If the test is one-tailed, enter NONE for the unused region. Round your answers to three decimal places.)
t >
t <
State the conclusion.
H0 is rejected. There is sufficient evidence to conclude that the dissolved oxygen content is less than 5 ppm.
H0 is rejected. There is insufficient evidence to conclude that the dissolved oxygen content is less than 5 ppm.
H0 is not rejected. There is insufficient evidence to conclude that the dissolved oxygen content is less than 5 ppm.
H0 is not rejected. There is sufficient evidence to conclude that the dissolved oxygen content is less than 5 ppm.
From the data,
sample mean = 4.96
Sample standard deviation s = 0.10488
H0: = 5 Vs. Ha: < 5
Test statistics
t = - / (S / sqrt(n) )
= 4.95 - 5 / (0.10488 / sqrt(6) )
= -1.168
t critical values at 0.05 level with 5 df = -2.571,2.571
The rejection region = Reject H0 if t < -2.571 or t > 2.571
Since test statistics t falls in non-rejection region, we do not have sufficient evidence to reject H0.
We conclude that
H0 is not rejected. There is insufficient evidence to conlcude that the dissolved oxygen content is less than 5 ppm
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