Question

The detection temperature under load for two different types of plastic pipe is being investigated. Two random samples of 15 pipe specimens are tested, and the detection temperatures observed are as follows (in °F):

Type 1: 206, 188, 205, 187, 194, 193, 207, 185, 189, 213, 192, 210, 194, 178, 205

Type 2: 177, 197, 206, 201, 180, 176, 185, 200, 197, 192, 198, 188, 189, 203, 192

Do the data support the claim that the deflection temperature under load for type 1 pipe exceeds that of type 2? Use the 0.05 significance level.

a) In Minitab, calculate “Descriptive Statistics” for the two samples. Copy and paste the output here.

b) State the hypotheses.

c) List the assumptions. Check the assumptions are satisfied. In Minitab, make two normal probability plots, one for each sample. Copy and paste the graphs into your report. Do these plots support of the assumption of normality? Why?

d) Calculate the test statistic.

e) Find the p-value.

f) Make a decision.

g) State the conclusion in terms of the problem.

h) In Minitab, perform the appropriate test. Copy and paste the output into your report. In your output, circle the hypotheses, test statistic, and p-value. They should match your answers from above.

i) In Minitab, perform a “Power and Sample Size” calculation to answer this question. Copy and paste the output into your report. If the mean deflection temperature for type 1 pipe exceeds that of type 2 by as much as 5 degree F, it is important to detect this difference with probability 0.9. Is the choice of ??1 = ??2 = 15 adequate? If not, how many pipe specimens should have been used? Use ?? = 0.05.

Answer 1

a.

Descriptive Statistics: Type-1, Type-2

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
Type-1 15 0 196.40 2.71 10.48 178.00 188.00 194.00 206.00 213.00
Type-2 15 0 192.07 2.44 9.44 176.00 185.00 192.00 200.00 206.00

b.

Given that,
mean(x)=196.4
standard deviation , s.d1=10.4799
number(n1)=15
y(mean)=192.0667
standard deviation, s.d2 =9.4375
number(n2)=15
null, Ho: u1 < u2
alternate, H1: u1 > u2
level of significance, ? = 0.05
from standard normal table,right tailed t ?/2 =1.761
since our test is right-tailed
reject Ho, if to > 1.761
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =196.4-192.0667/sqrt((109.8283/15)+(89.06641/15))
to =1.19
| to | =1.19
critical value
the value of |t ?| with min (n1-1, n2-1) i.e 14 d.f is 1.761
we got |to| = 1.19002 & | t ? | = 1.761
make decision
hence value of |to | < | t ? | and here we do not reject Ho
p-value:right tail - Ha : ( p > 1.19 ) = 0.12691
hence value of p0.05 < 0.12691,here we do not reject Ho
ANSWERS
---------------
null, no diffrence between themHo: u1 = u2
alternate, H1: u1 > u2
test statistic: 1.19
critical value: 1.761
decision: do not reject Ho
p-value: 0.12691,
no evidence t0 support support the claim that the deflection temperature under
load for type 1 pipe exceeds that of type 2

h.

Two-Sample T-Test and CI: Type-1, Type-2

Two-sample T for Type-1 vs Type-2

N Mean StDev SE Mean
Type-1 15 196.4 10.5 2.7
Type-2 15 192.07 9.44 2.4

Difference = ? (Type-1) - ? (Type-2)
Estimate for difference: 4.33
95% lower bound for difference: -1.87
T-Test of difference = 0 (vs >): T-Value = 1.19 P-Value = 0.122 DF = 27