Industrial wastes and sewage dumped into our rivers and streams absorb oxygen an
ID: 3316279 • 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. 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.
(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 = 0.05. (Use 1 for the population mean for the above town location and 2 for the population mean for the below town location. Round your answers to three decimal places.)
1-2. Null and alternative hypotheses:
H0: (1 2) = 0 versus Ha: (1 2) > 0H0: (1 2) = 0 versus Ha: (1 2) 0 H0: (1 2) = 0 versus Ha: (1 2) < 0H0: (1 2) < 0 versus Ha: (1 2) > 0H0: (1 2) 0 versus Ha: (1 2) = 0
3. Test statistic: t =
4. Rejection region: If the test is one-tailed, enter NONE for the unused region.
5. Conclusion:
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.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 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.
(b) Suppose you prefer estimation as a method of inference. Estimate the difference in the mean dissolved oxygen contents for locations above and below the town. Use a 95% confidence interval. (Round your answers to three decimal places.)
to ppm
Explanation / Answer
PART A.
Given that,
mean(x)=4.94
standard deviation , s.d1=0.2302
number(n1)=5
y(mean)=4.88
standard deviation, s.d2 =0.1643
number(n2)=5
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, = 0.05
from standard normal table,left tailed t /2 =2.132
since our test is left-tailed
reject Ho, if to < -2.132
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =4.94-4.88/sqrt((0.05299/5)+(0.02699/5))
to =0.4744
| to | =0.4744
critical value
the value of |t | with min (n1-1, n2-1) i.e 4 d.f is 2.132
we got |to| = 0.47438 & | t | = 2.132
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:left tail - Ha : ( p < 0.4744 ) = 0.67002
hence value of p0.05 < 0.67002,here we do not reject Ho
ANSWERS
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1. null, Ho: u1 = u2
2. alternate, H1: u1 < u2
3. test statistic: 0.4744
critical value: -2.132
4. decision: do not reject Ho
p-value: 0.67002
5. we do not have evidence to indicate that the mean oxygen content below the town is less than the mean oxygen
PART B.
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 4.94-4.88) ± t a/2 * sqrt((0.053/5)+(0.027/5)]
= [ (0.06) ± t a/2 * 0.126]
= [-0.291 , 0.411]
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interpretations:
1. we are 95% sure that the interval [-0.291 , 0.411] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion