2. (25 points) An article contains measurements on samples of coal from several

Question

2. (25 points) An article contains measurements on samples of coal from several counties from Kentucky. In units of percent ash, five samples from Knott County had an average aluminum dioxide (AO2) content of 32.17 and a standard deviation of 2.23. Six samples from Lesile County had an average Al02 content of 26.48 and a standard deviation 2.02. S, : 22.17 . S( : 2.23 /ny : b . y-26 2. (a) Can we conclude that the variance of AlO2 content of coal sample differs between two counties? i). State hypothesis. i). Compute P-value. ii). What is your conclusion? from the two counties. (b) Find a 98% confidence interval for the difference in A102 contend of coal sam ples

Explanation / Answer

PART A.
Given that,
mean(x)=32.17
standard deviation , s.d1=2.23
number(n1)=5
y(mean)=26.48
standard deviation, s.d2 =2.02
number(n2)=6
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.025
from standard normal table, two tailed t /2 =3.495
since our test is two-tailed
reject Ho, if to < -3.495 OR if to > 3.495
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =32.17-26.48/sqrt((4.9729/5)+(4.0804/6))
to =4.397
| to | =4.397
critical value
the value of |t | with min (n1-1, n2-1) i.e 4 d.f is 3.495
we got |to| = 4.39694 & | t | = 3.495
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 4.3969 ) = 0.012
hence value of p0.025 > 0.012,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 4.397
critical value: -3.495 , 3.495
decision: reject Ho
p-value: 0.012
sample differs between 2 countries

PART B.
given that,
mean(x)=32.17
standard deviation , s.d1=2.23
sample size, n1=5
y(mean)=26.48
standard deviation, s.d2 =2.02
sample size,n2 =6
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 = [( 32.17-26.48) ± t a/2 * sqrt((4.973/5)+(4.08/6)]
= [ (5.69) ± t a/2 * 1.294]
= [0.841 , 10.539]

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interpretations:
1. we are 98% sure that the interval [0.841 , 10.539] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population proportion

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