Question

A geologist collected 20 different ore samples, all of the same weight, and randomly divided them into two groups. The titanium contents of the samples, found using two different methods, are listed in the table below.
Method 1 Method 2
0.012   0.013   0.014   0.014    0.014 0.012   0.017   0.013   0.012   0.014
0.012   0.010   0.014   0.011   0.013 0.012   0.017   0.012   0.014   0.016
(a)
Use an appropriate method to test for a significant difference in the average titanium contents using the two different methods. (Use μ1 for Method 1 and μ2 for Method 2. Use α = 0.05.)

State the test statistic. (Round your answer to three decimal places.)

t = _________

(b)
Determine a 95% confidence interval estimate for (μ1 − μ2).  (Round your answers to four decimal places.)

___________ to ____________

Answer 1

a.
Given that,
mean(x)=0.0135
standard deviation , s.d1=0.0015
number(n1)=10
y(mean)=0.0131
standard deviation, s.d2 =0.0022
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =0.0135-0.0131/sqrt((0/10)+(0/10))
to =0.475
| to | =0.475
critical value
the value of |t α| with min (n1-1, n2-1) i.e 9 d.f is 2.262
we got |to| = 0.47505 & | t α | = 2.262
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.475 ) = 0.646
hence value of p0.05 < 0.646,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 0.475
critical value: -2.262 , 2.262
decision: do not reject Ho
p-value: 0.646
we do not have enough evidence to support the claim that significant difference in the average titanium contents using the two different methods.
b.
TRADITIONAL METHOD
given that,
mean(x)=0.0135
standard deviation , s.d1=0.0015
number(n1)=10
y(mean)=0.0131
standard deviation, s.d2 =0.0022
number(n2)=10
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((0/10)+(0/10))
= 0.0008
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 9 d.f is 2.2622
margin of error = 2.262 * 0.0008
= 0.0019
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (0.0135-0.0131) ± 0.0019 ]
= [-0.0015 , 0.0023]
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DIRECT METHOD
given that,
mean(x)=0.0135
standard deviation , s.d1=0.0015
sample size, n1=10
y(mean)=0.0131
standard deviation, s.d2 =0.0022
sample size,n2 =10
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 = [( 0.0135-0.0131) ± t a/2 * sqrt((0/10)+(0/10)]
= [ (0.0004) ± t a/2 * 0.0008]
= [-0.0015 , 0.0023]
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interpretations:
1. we are 95% sure that the interval [-0.0015 , 0.0023] 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