An article in the ASCE Journal of Energy Engineering (1999, Vol. 125, pp. 59-75)

Question

An article in the ASCE Journal of Energy Engineering (1999, Vol. 125, pp. 59-75) describes a study of the thermal inertia properties of autoclaved aerated concrete used as a building material. Five samples of the material were tested in a structure, and the sample mean and the sample standard deviation of interior temperatures (in C) were 23.0, s 0.40 Test the hypotheses H : 22.5 versus H #22.5, using = 0.05. Also compute the p-value 6. for this test. Answer all parts. Critical (rejection) region Observed value of the test statistic: Conclusion: P-value (show your computation or say what calculator procedure you used):

Explanation / Answer

Given that,
population mean(u)=22.5
sample mean, x =23
standard deviation, s =0.4
number (n)=5
null, Ho: =22.5
alternate, H1: !=22.5
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.776
since our test is two-tailed
reject Ho, if to < -2.776 OR if to > 2.776
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =23-22.5/(0.4/sqrt(5))
to =2.7951
| to | =2.7951
critical value
the value of |t | with n-1 = 4 d.f is 2.776
we got |to| =2.7951 & | t | =2.776
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.7951 ) = 0.0491
hence value of p0.05 > 0.0491,here we reject Ho
ANSWERS
---------------
null, Ho: =22.5
alternate, H1: !=22.5

reject Ho, if to < -2.776 OR if to > 2.776

test statistic: 2.7951
critical value: -2.776 , 2.776
decision: reject Ho
p-value: 0.0491

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