1. The urinary fluoride concentration (parts per million) was measured both for
ID: 3317306 • Letter: 1
Question
1. The urinary fluoride concentration (parts per million) was measured both for a sample of livestock grazing in an area previously exposed to fluoride pollution and for a similar sample grazing in an unpolluted region. Assume unequal variances for the two populations. Unpolluted 14.2 18.3 17.2 18.4 25.0 (a) (10 pts) Does the data indicate strongly that the true average fluoride concentration for livestock grazing in the polluted region is larger than for the unpolluted region? Use the appropriate test at level 0.01 (b) (5 pts) Construct a 99% confidence interval with a lower bound for the difference in the two population means polluted unpollutedExplanation / Answer
ANSWER A.
Given that,
mean(x)=19.9714
standard deviation , s.d1=2.0097
number(n1)=7
y(mean)=18.62
standard deviation, s.d2 =3.9499
number(n2)=5
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.01
from standard normal table,right tailed t /2 =3.747
since our test is right-tailed
reject Ho, if to > 3.747
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =19.9714-18.62/sqrt((4.03889/7)+(15.60171/5))
to =0.7028
| to | =0.7028
critical value
the value of |t | with min (n1-1, n2-1) i.e 4 d.f is 3.747
we got |to| = 0.70281 & | t | = 3.747
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > 0.7028 ) = 0.26046
hence value of p0.01 < 0.26046,here we do not reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 0.7028
critical value: 3.747
decision: do not reject Ho
p-value: 0.26046
we don't have evidence that polluted region is larger than for the unpolluted region.
ANSWER 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 = [( 19.9714-18.62) ± t a/2 * sqrt((4.039/7)+(15.602/5)]
= [ (1.351) ± t a/2 * 1.923]
= [-7.501 , 10.204]
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interpretations:
1. we are 99% sure that the interval [-7.501 , 10.204] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population proportion
99% CI with a lower bound value is -7.50
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