Question

The deflection temperature under load for two different types of plastic pipe is being investigated. Two random samples of 15 pipe specimens are tested, and the deflection temperatures observed are as follows (?? 0?).

Type_01: 206 188 187 194 193 207 185 189 213 192

210 194 178 205

Type_02: 177 197 206 201 180 176 185 200 197 192 198

188 189 203 192

a.Write down the null and alternative hypotheses to test if the deflection temperature under load for Type_02 pipe exceeds that of Type_01. 


b.What is the type of statistical test that is appropriate? Explain. 


c.Test the hypotheses at ? = 0.05, using the critical value approach. What is your decision on null hypothesis(?0)? Interpret the test results. 


d.What is the ? ?????for the test? Test the hypotheses using the ? ?????. Do you get the same answer as part (c)? 


e.Construct a 95% confidence interval for the mean difference of deflection temperatures. Test the hypotheses using the confidence interval approach. Do you get the same answer as in parts (c) and (d)? 


Answer 1

Given that,
mean(x)=192.0667
standard deviation , s.d1=9.4375
number(n1)=15
y(mean)=195.7857
standard deviation, s.d2 =10.5916
number(n2)=14
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.771
since our test is right-tailed
reject Ho, if to > 1.771
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =192.0667-195.7857/sqrt((89.06641/15)+(112.18199/14))
to =-0.9957
| to | =0.9957
critical value
the value of |t α| with min (n1-1, n2-1) i.e 13 d.f is 1.771
we got |to| = 0.9957 & | t α | = 1.771
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value:right tail - Ha : ( p > -0.9957 ) = 0.83121
hence value of p0.05 < 0.83121,here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 > u2
b.
test statistic: -0.9957
c.
critical value: 1.771
decision: do not reject Ho
d.
p-value: 0.83121
we do not have enough evidence to support the claim that if the deflection temperature under load for Type_02 pipe exceeds that of Type_01.
e.
TRADITIONAL METHOD
given that,
mean(x)=192.0667
standard deviation , s.d1=9.4375
number(n1)=15
y(mean)=195.7857
standard deviation, s.d2 =10.5916
number(n2)=14
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((89.066/15)+(112.182/14))
= 3.735
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 13 d.f is 2.16
margin of error = 2.16 * 3.735
= 8.068
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (192.0667-195.7857) ± 8.068 ]
= [-11.787 , 4.349]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=192.0667
standard deviation , s.d1=9.4375
sample size, n1=15
y(mean)=195.7857
standard deviation, s.d2 =10.5916
sample size,n2 =14
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 = [( 192.0667-195.7857) ± t a/2 * sqrt((89.066/15)+(112.182/14)]
= [ (-3.719) ± t a/2 * 3.735]
= [-11.787 , 4.349]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-11.787 , 4.349] 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

from part (c),
Given that,
mean(x)=192.0667
standard deviation , s.d1=9.4375
number(n1)=15
y(mean)=195.7857
standard deviation, s.d2 =10.5916
number(n2)=14
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.16
since our test is two-tailed
reject Ho, if to < -2.16 OR if to > 2.16
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =192.0667-195.7857/sqrt((89.06641/15)+(112.18199/14))
to =-0.9957
| to | =0.9957
critical value
the value of |t α| with min (n1-1, n2-1) i.e 13 d.f is 2.16
we got |to| = 0.9957 & | t α | = 2.16
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 != -0.9957 ) = 0.338
hence value of p0.05 < 0.338,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -0.9957
critical value: -2.16 , 2.16
decision: do not reject Ho
p-value: 0.338
we do not have enough evidence to support the claim that if the deflection temperature under load the mean difference of deflection temperatures.