How heavy a load (pounds) is needed to pull apart pieces of Douglas fir 4 inches

Question

How heavy a load (pounds) is needed to pull apart pieces of Douglas fir 4 inches long and 1.5 inches squar? Here are the data from students doing a laboratory exercise: 33190 31860 32590 26520 33280 32320 33020 32030 30460 32700 23040 30930 32720 33650 32340 24050 30170 31300 28730 31920 We are willing to regard the wood pieces prepared for the lab session as an SRS of all similar pieces of Douglas fir. Engineers also commonly assume that characteristics of materials vary Normally. Suppose that the strength of pieces of wood like these follows a Normal distribution with standard deviation 3000 pounds.

a) Is there statistically significant evidence at the alpha = 0.10 level against the hypothesis that the mean is 32500 pounds for the two-sided alternative?

b) Is there statistically significant evidence at the alpha = 0.10 level against the hypothesis that the mean is 31500 pounds for the two-sided alternative?

Explanation / Answer

Q1.
Given that,
population mean(u)=32500
standard deviation, =3000
sample mean, x =30841
number (n)=20
null, Ho: =32500
alternate, H1: !=32500
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 30841-32500/(3000/sqrt(20)
zo = -2.47309
| zo | = 2.47309
critical value
the value of |z | at los 10% is 1.645
we got |zo| =2.47309 & | z | = 1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -2.47309 ) = 0.01339
hence value of p0.1 > 0.01339, here we reject Ho
ANSWERS
---------------
null, Ho: =32500
alternate, H1: !=32500
test statistic: -2.47309
critical value: -1.645 , 1.645
decision: reject Ho
p-value: 0.01339
Q2.
Given that,
population mean(u)=32500
standard deviation, =3000
sample mean, x =31500
number (n)=20
null, Ho: =32500
alternate, H1: !=32500
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 31500-32500/(3000/sqrt(20)
zo = -1.49071
| zo | = 1.49071
critical value
the value of |z | at los 10% is 1.645
we got |zo| =1.49071 & | z | = 1.645
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -1.49071 ) = 0.13604
hence value of p0.1 < 0.13604, here we do not reject Ho
ANSWERS
---------------
null, Ho: =32500
alternate, H1: !=32500
test statistic: -1.49071
critical value: -1.645 , 1.645
decision: do not reject Ho
p-value: 0.13604

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