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:

         −0.45 −0.19 −0.16 −0.20 +0.28 −0.24 +0.46 +0.26 −0.14 −0.37 −0.32 −0.50 −0.51 −0.62 −0.04 −0.19 −0.56 +0.04

1. State the null hypothesis and the alternate hypothesis.

H0: ?=

H1: ?=

1. 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.)

At a level of 0.05 significance, we reject H0: ? if t < ____ or t > ___.

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

Value of the test statistic:

1. Is it reasonable to conclude that the mean gain or loss in time for the watches is 0? Use the 0.05 significance level.

___(reject) or (do not reject)__ H0. It ______ reasonable to conclude that the mean gain or loss in time for watches is 0.

1. Estimate the p-value.

Answer 1

Given that,
population mean(u)=0
sample mean, x =-0.1916
standard deviation, s =0.294
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 =-0.1916-0/(0.294/sqrt(18))
to =-2.7649
| to | =2.7649
critical value
the value of |t α| with n-1 = 17 d.f is 2.11
we got |to| =2.7649 & | 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 != -2.7649 ) = 0.0132
hence value of p0.05 > 0.0132,here we reject Ho
ANSWERS
---------------
null, Ho: μ=0
alternate, H1: μ!=0
test statistic: -2.7649
critical value: -2.11 , 2.11
decision: reject Ho
p-value: 0.0132
we have enough evidence to support the claim that the mean gain or loss in time for watches is 0.