Industrial wastes and sewage dumped into our rivers and streams absorb oxygen an
ID: 3317413 • 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 Above Town 4.8 5.3 5.1 4.8 5.0 Below Town .0 4.6 5.0 4.7 5. (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 : 411-r2) = 0 versus Ha : (M1-2) > 0 O Hg: (1-r2) = 0 versus Hai (M1-12)Explanation / Answer
Given that,
mean(x)=5
standard deviation , s.d1=0.2121
number(n1)=5
y(mean)=4.86
standard deviation, s.d2 =0.1949
number(n2)=5
since our test is right-tailed
reject Ho, if to > 2.132
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =5-4.86/sqrt((0.04499/5)+(0.03799/5))
to =1.0868
test statistic: 1.0868
TRADITIONAL METHOD
given that,
mean(x)=296
standard deviation , s.d1=12
number(n1)=10
y(mean)=315
standard deviation, s.d2 =18
number(n2)=25
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((144/10)+(324/25))
= 5.231
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.05
from standard normal table, two tailed and
value of |t | with min (n1-1, n2-1) i.e 9 d.f is 2.262
margin of error = 2.262 * 5.231
= 11.832
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (296-315) ± 11.832 ]
= [-30.832 , -7.168]
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DIRECT METHOD
given that,
mean(x)=296
standard deviation , s.d1=12
sample size, n1=10
y(mean)=315
standard deviation, s.d2 =18
sample size,n2 =25
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 = [( 296-315) ± t a/2 * sqrt((144/10)+(324/25)]
= [ (-19) ± t a/2 * 5.231]
= [-30.832 , -7.168]
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interpretations:
1. we are 95% sure that the interval [-30.832 , -7.168] 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