Question

2. (2 pts) For the following scenarios,

• describe what the mean parameter µ represents in each scenario, and

• set up H0 and Ha related to µ (i.e. What hypotheses would you test to assess the specification/claim/belief?) Do not perform the hypothesis tests. a.

A random sample of 30 pieces of acetate fiber has a sample mean absorbency of 15% with a sample standard deviation of 1.5%. Is there strong evidence that this fiber has a true mean absorbency of less than 18%?

b. A manufacturer of a synthetic fishing line claims that its product has an exact mean breaking strength of 9 kilograms (no more, no less). A fishing fanatic takes a random sample of n=20 fishing line specimens and computes a sample average strength of 8.5 kilograms with sample standard deviation of 0.65 kilogram. Is there strong evidence that the mean breaking strength is 9 kg?

For part b, notice the the evidence we're looking for is 9 kg; it's not bigger or less.

Answer 1

2.

a.
Given that,
population mean(u)=18
sample mean, x =15
standard deviation, s =1.5
number (n)=30
null, Ho: μ=18
alternate, H1: μ<18
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.699
since our test is left-tailed
reject Ho, if to < -1.699
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =15-18/(1.5/sqrt(30))
to =-10.9545
| to | =10.9545
critical value
the value of |t α| with n-1 = 29 d.f is 1.699
we got |to| =10.9545 & | t α | =1.699
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :left tail - Ha : ( p < -10.9545 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: μ=18
alternate, H1: μ<18
test statistic: -10.9545
critical value: -1.699
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that this fiber has a true mean absorbency of less than 18%
b.
Given that,
population mean(u)=9
sample mean, x =8.5
standard deviation, s =0.65
number (n)=20
null, Ho: μ=9
alternate, H1: μ!=9
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.093
since our test is two-tailed
reject Ho, if to < -2.093 OR if to > 2.093
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =8.5-9/(0.65/sqrt(20))
to =-3.4401
| to | =3.4401
critical value
the value of |t α| with n-1 = 19 d.f is 2.093
we got |to| =3.4401 & | t α | =2.093
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -3.4401 ) = 0.0027
hence value of p0.05 > 0.0027,here we reject Ho
ANSWERS
---------------
null, Ho: μ=9
alternate, H1: μ!=9
test statistic: -3.4401
critical value: -2.093 , 2.093
decision: reject Ho
p-value: 0.0027
we have enough evidence to support the claim that the mean breaking strength is 9 kg.
we're looking for is 9 kg; it's not bigger or less.