Question

Watch Corporation of Switzerland claims that its watches on average will neither gain nor lose time during a week. A sample of 18 watches provided the following gains (+) or losses (−) in seconds per week.

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−0.16	−0.15	−0.20	−0.17	+0.26	−0.19	+0.30	+0.43	−0.10
−0.31	−0.48	−0.44	−0.51	−0.67	−0.05	−0.24	−0.51	+0.05

State the null hypothesis and the alternate hypothesis

State the decision rule for 0.05 significance level. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)

Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

Answer 1

Given that,
population mean(u)=0
sample mean, x =-0.1744
standard deviation, s =0.2896
number (n)=18
null, Ho: μ=0
alternate, H1: μ!=0
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.11
since our test is two-tailed
reject Ho, if to < -2.11 OR if to > 2.11
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =-0.1744-0/(0.2896/sqrt(18))
to =-2.555
| to | =2.555
critical value
the value of |t α| with n-1 = 17 d.f is 2.11
we got |to| =2.555 & | t α | =2.11
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -2.555 ) = 0.0205
hence value of p0.05 > 0.0205,here we reject Ho
ANSWERS
---------------
null, Ho: μ=0
alternate, H1: μ!=0
test statistic: -2.555
critical value: -2.11 , 2.11
decision: reject Ho
p-value: 0.0205
we have enough evidence to support the claim that its watches on average will neither gain nor lose time during a week.