Question

An article presents voltage measurements for a sample of 66 industrial networks in Estonia. Assume the rated voltage for these networks is 232 V. The sample mean voltage was 231.5 V with a standard deviation of 2.19 V. Let μ represent the population mean voltage for these networks.

a) Find the P-value for testing H0 : μ = 232 versus H1 : µ ≠ 232. Round the answer to four decimal places.

b) Either the mean voltage is not equal to 232, or the sample is in the most extreme % of its distribution. Round the answer to two decimal places.

Answer 1

Given that,
population mean(u)=232
sample mean, x =231.5
standard deviation, s =2.19
number (n)=66
null, Ho: μ=232
alternate, H1: μ!=232
level of significance, alpha = 0.05
from standard normal table, two tailed t alpha/2 =1.997
since our test is two-tailed
reject Ho, if to < -1.997 OR if to > 1.997
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =231.5-232/(2.19/sqrt(66))
to =-1.855
| to | =1.855
critical value
the value of |t alpha| with n-1 = 65 d.f is 1.997
we got |to| =1.855 & | t alpha | =1.997
make decision
hence value of |to | < | t alpha | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.8548 ) = 0.0682
hence value of p0.05 < 0.0682,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=232
alternate, H1: μ!=232
test statistic: -1.855
critical value: -1.997 , 1.997
decision: do not reject Ho
a.
p-value: 0.0682
we do not have enough evidence to support the claim that the rated voltage for these networks is 232 V
b.
Either the mean voltage is not equal to 232, or the sample is in the most extreme % of its distribution
is 6.82%