In: Statistics and Probability
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 drew five randomly selected specimens of river water at a location above the town, and another five below. The dissolved oxygen readings (in parts per million) are as follows.
Above Town | 4.9 | 5.3 | 4.9 | 4.9 | 5.1 |
---|---|---|---|---|---|
Below Town | 5.1 | 4.6 | 4.8 | 4.9 | 4.9 |
(a) Do the data provide sufficient evidence to indicate that the
mean oxygen content below the town is less than the mean oxygen
content above? Test using α = 0.05. (Use
μ1 for the population mean for the above town
location and μ2 for the population mean for the
below town location.)
State the null and alternative hypotheses.
H0: (μ1 − μ2) = 0 versus Ha: (μ1 − μ2) > 0
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 insufficient evidence to indicate that the mean content of oxygen below town is less than the mean content above town.
H0 is not rejected. There is sufficient evidence to indicate that the mean content of oxygen below town is less than the mean content above town.
H0 is not rejected. There is insufficient evidence to indicate that the mean content of oxygen below town is less than the mean content above town.
H0 is rejected. There is sufficient evidence to indicate that the mean content of oxygen below town is less than the mean content above town.
(b) Suppose you prefer estimation as a method of inference.
Estimate the difference in the mean dissolved oxygen contents (in
ppm) for locations above and below the town. Use a 95% confidence
interval. (Use μ1 − μ2.
Round your answers to three decimal places.)
ppm to ppm
Above ( X ) | Below ( Y ) | |||
4.9 | 0.0144 | 5.1 | 0.0576 | |
5.3 | 0.0784 | 4.6 | 0.0676 | |
4.9 | 0.0144 | 4.8 | 0.0036 | |
4.9 | 0.0144 | 4.9 | 0.0016 | |
5.1 | 0.0064 | 4.9 | 0.0016 | |
Total | 25.1 | 0.128 | 24.3 | 0.132 |
To Test :-
H0 :-
H1 :-
Test Statistic :-
Test Criteria :-
Reject null hypothesis if
Critical value
Rejection region t > 1.895
Result :- Fail to Reject Null Hypothesis
Conclusion :- Accept Null Hypotesis
There is no sufficient evidence to suppport the claim that the mean oxygen content below the town is less than the mean oxygen content above.
H0 is not rejected. There is insufficient evidence to indicate that the mean content of oxygen below town is less than the mean content above town.
Part b)
Confidence interval :-
95% Confidence interval is ( -0.139 , 0.459 )