Two methods for measuring the molar heat of fusion of water are being compared.

Question

Two methods for measuring the molar heat of fusion of water are being compared. Ten measurements made by method A have a mean of 6.02 kilojoules per mole (kJ/mol) with a standard deviation of 0.02 (kJ/mol). Five measurements made by method B have a mean of 6.00 kJ/mol and a standard deviation of 0.01 kJ/mol.

a-Can you conclude that the mean measurements for method A is greater than that of method B?

b-Repeat this test based on the knowledge that the population variances were equal.

c-Find an 80% two sided confidence interval based on the knowledge of “equal variances

PLEASE EXPLAIN EACH STEP CLEARLY

Explanation / Answer

a.
WHEN POPULATION S.D ARE UNEQUAL
Given that,
mean(x)=6.02
standard deviation , s.d1=0.02
number(n1)=10
y(mean)=6
standard deviation, s.d2 =0.01
number(n2)=5
null, Ho: u1 = u2
alternate,the mean measurements for method A is greater than that of method B H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =2.132
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 =6.02-6/sqrt((0.0004/10)+(0.0001/5))
to =2.582
| to | =2.582
critical value
the value of |t | with min (n1-1, n2-1) i.e 4 d.f is 2.132
we got |to| = 2.58199 & | t | = 2.132
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 2.582 ) = 0.0306
hence value of p0.05 > 0.0306,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 2.582
critical value: 2.132
decision: reject Ho
p-value: 0.0306
we have evidence that the mean measurements for method A is greater than that of method B

b. WHEN POPULATION S.D ARE EQUAL
Given that,
mean(x)=6.02
standard deviation , s.d1=0.02
number(n1)=10
y(mean)=6
standard deviation, s.d2 =0.01
number(n2)=5
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.771
since our test is right-tailed
reject Ho, if to > 1.771
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (9*0.0004 + 4*0.0001) / (15- 2 )
s^2 = 0.0003
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=6.02-6/sqrt((0.0003( 1 /10+ 1/5 ))
to=0.02/0.0095
to=2.1082
| to | =2.1082
critical value
the value of |t | with (n1+n2-2) i.e 13 d.f is 1.771
we got |to| = 2.1082 & | t | = 1.771
make decision
hence value of | to | > | t | and here we reject Ho
p-value: right tail -ha : ( p > 2.1082 ) = 0.02749
hence value of p0.05 > 0.02749,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 2.1082
critical value: 1.771
decision: reject Ho
p-value: 0.02749

c.
CI = x1 - x2 ± t a/2 * Sqrt(S^2(1/n1+1/n2))
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
Value Pooled variance S^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=6.02
Standard deviation( sd1 )=0.02
Sample Size(n1)=10
Mean(x2)=6
Standard deviation( sd2 )=0.01
Sample Size(n2)=5
S^2 = (9*0 + 4*0) / (15- 2 )
S^2 = 0.000308
CI = [ ( 6.02-6) ± t a/2 * 0.018 Sqrt( 1/10+1/5)]
= [ (0.02) ± t a/2 * 0.018 * Sqrt( 0.3) ]
= [ (0.02) ± 2.000995 * 0.018 * Sqrt( 0.3) ]
= [ 0.001 , 0.039 ]

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