12 13 14 14 15 15 15 16 1616 16 16 17 1717 18 18 18 18 19 19 19 20 21 A large so
ID: 3360154 • Letter: 1
Question
12 13 14 14 15 15 15 16 1616 16 16 17 1717 18 18 18 18 19 19 19 20 21 A large software development firm recently relocated its facilities. Top management is interested in fostering good relations with their new local community and has encouraged their professional employees to engage in local service activities. They want to test if the firm's professionals volunteer an average of more than 15 hours per month. If this is not the case, they will institute an incentive program to increase community involvement. A random sample of 24 professionals reported are shown in the above figure. First, compute the mean and standard deviation of the given sample, then use them to deduce a conclusion about the null hypothesis, if you know that the critical value of t is set at . 10%. Please approximate your mean and standard deviation to the nearest first decimal point before calculating your t-statistic. ( we reject the null hypothesis, and the firm does not need to institute an incentive program because there is strong evidence indicating that professional employees already volunteer more than 15 hours per month, on average. we fail to reject the null hypothesis. the firm should institute an incentive program because there is evidence indicating that professional employees volunteer less than 15 hours per month on average. O we reject the alternative hypothesis.Explanation / Answer
Given that,
population mean(u)=15
Given sample data:
(12,13,14,14,15,15,15,16,16,16,16,16,17,17,17,18,18,18,18,19,19,19,20,21)
from the above sample
sample mean, x =16.625
standard deviation, s =2.2227
number (n)=24
null, Ho: =15
alternate, H1: >15
level of significance, = 0.1
from standard normal table,right tailed t /2 =1.319
since our test is right-tailed
reject Ho, if to > 1.319
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =16.625-15/(2.2227/sqrt(24))
to =3.5816
| to | =3.5816
critical value
the value of |t | with n-1 = 23 d.f is 1.319
we got |to| =3.5816 & | t | =1.319
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 3.5816 ) = 0.00079
hence value of p0.1 > 0.00079,here we reject Ho
ANSWERS
---------------
null, Ho: =15
alternate, H1: >15
test statistic: 3.5816
critical value: 1.319
decision: reject Ho
p-value: 0.00079
OPTION A.
we reject the null hypothesis, and the firm does not need to institute an incentive
program because there is strong evidence indicating that professional employees already
volunteer more than 15 hours per month, on average.
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