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 square? Here are data from students doing a laboratory exercise:

33,190, 31,860, 32,590, 26,520, 33,280, 32,320, 33,202, 32,030, 30,460, 32,700, 23,040, 30,930, 32,720, 33,650, 32,340, 24,050, 30,170, 31,300, 28,730, 31,920

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.

Explanation / Answer

PART A.
Given that,
population mean(u)=32500
standard deviation, =3000
sample mean, x =30850.1
number (n)=20
null, Ho: =32500
alternate, H1: (not equal)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 = 30850.1-32500/(3000/sqrt(20)
zo = -2.45953
| zo | = 2.45953
critical value
the value of |z | at los 10% is 1.645
we got |zo| =2.45953 & | 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 != -2.45953 ) = 0.01391
hence value of p0.1 > 0.01391, here we reject Ho
ANSWERS
---------------
null, Ho: =32500
alternate, H1: (not equal)32500
test statistic: -2.45953
critical value: -1.645 , 1.645
decision: reject Ho
p-value: 0.01391

there is statistically significant evidence that the mean is 32,500 pounds

PART B.
null, Ho: =31500
alternate, (not equal)31500
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 = 30850.1-31500/(3000/sqrt(20)
zo = -0.96881
| zo | = 0.96881
critical value
the value of |z | at los 10% is 1.645
we got |zo| =0.96881 & | 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 != -0.96881 ) = 0.33264
hence value of p0.1 < 0.33264, here we do not reject Ho
ANSWERS
---------------
null, Ho: =31500
alternate, H1: (not equal)31500
test statistic: -0.96881
critical value: -1.645 , 1.645
decision: do not reject Ho
p-value: 0.33264

there is evidence that the mean is 31,500 pounds

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