a us based internet company offers a basic accounting course. in the first semester 315 students...

a us based internet company offers a basic accounting course. in the first semester 315 students registered. they have been sorted into 7 regions with the following enrollments in each.

test at a 10% significance and find the following:

-test statistic

-denominator of the formula

Solutions

Expert Solution

Assume, population mean(u)=43 because not given in the data.
Given that,
sample mean, x =45
standard deviation, s =10.8012
number (n)=7
null, Ho: μ=43
alternate, H1: μ!=43
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.943
since our test is two-tailed
reject Ho, if to < -1.943 OR if to > 1.943
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =45-43/(10.8012/sqrt(7))
to =0.49
| to | =0.49
critical value
the value of |t α| with n-1 = 6 d.f is 1.943
we got |to| =0.49 & | t α | =1.943
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 0.4899 ) = 0.6416
hence value of p0.1 < 0.6416,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=43
alternate, H1: μ!=43
test statistic: 0.49
critical value: -1.943 , 1.943
decision: do not reject Ho
p-value: 0.6416
we do not have enough evidence to support the claim that population mean is equal to 43

orchestra answered 1 year ago

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