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 5.0, 5.0, 4.9, 4.9, 5.0, and 4.6 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 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. 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. You may need to use the appropriate appendix table or technology to answer this question.

Answer #1

Solution:

Here, we have to use a one-tailed t-test for a population mean. The null and alternative hypotheses for this test are given as below:

H_{0}: µ = 5 versus H_{a}: µ < 5

This is a one/lower/left tailed test.

From given sample we have

Sample mean = Xbar = 4.9

Sample standard deviation = S = 0.154919334

Sample size = n = 6

Degrees of freedom = n – 1 = 6 – 1 = 5

Level of significance = α = 0.05

Test statistic = t = (Xbar - µ)/[S/sqrt(n)]

t = (4.9 – 5)/[ 0.154919334/sqrt(6)]

t = (4.9 – 5) / 0.0632

t = -1.5811

Test statistic = t = -1.581

Critical t value = -2.0150 (by using t-table)

Rejection region = Reject H_{0} if t < -2.015

Here, t-test statistic is not less than critical t value, so we
do not reject the null hypothesis H_{0}.

Conclusion:

H_{0} is not rejected.

There is insufficient evidence to conclude that the dissolved oxygen content is less than 5 ppm.

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
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aquatic life. Six water specimens taken from a river at a specific
location during the low-water season (July) gave readings of 4.9,
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Industrial wastes and sewage dumped into our rivers and streams
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aquatic life. A pollution control inspector suspected that a river
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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. A pollution control inspector suspected that a river
community was releasing amounts of semitreated sewage into a river.
To check his theory, he...

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. A pollution control inspector suspected that a river
community was releasing amounts of semitreated sewage into a river.
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