10.1.27-T Assume that a simple random sample has been selected and test the give
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Question
10.1.27-T Assume that a simple random sample has been selected and test the given claim. Identify the null and alternative hypotheses, test statistic, P-value, and state the final conclusion that addresses the original claim. Listed below are brain volumes in cm3 of unrelated subjects used in a study, Use a 0.05 significance level to test the claim that the population of brain volumes has a mean equal to 1100.6 cm3 962 1028 1271 1078 1071 1174 1067 1346 1101 1203 What are the hypotheses? A. Ho : -1100.6 crm3 B. Ho' = 1100.6 cm 3 Ha: #1100.6 cm 3 OD. Ho:-1100.6 cm 3 Ha:
Explanation / Answer
Given that,
population mean(u)=1100.6
sample mean, x =1130.1
standard deviation, s =117.1063
number (n)=10
null, Ho: =1100.6
alternate, H1: !=1100.6
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =1130.1-1100.6/(117.1063/sqrt(10))
to =0.7966
| to | =0.7966
critical value
the value of |t | with n-1 = 9 d.f is 2.262
we got |to| =0.7966 & | t | =2.262
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 0.7966 ) = 0.4462
hence value of p0.05 < 0.4462,here we do not reject Ho
ANSWERS
---------------
null, Ho: =1100.6
alternate, H1: !=1100.6
test statistic: 0.797
critical value: -2.262 , 2.262
decision: do not reject Ho
p-value: 0.4462
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