Question

A coin-operated drink machine was designed to discharge a mean of 7 ounces of coffee per cup, In a test of the machine, the discharge amounts in 14 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 6.96 ounces and 0.12 ounces, respectively. If we assume that the discharge amounts are normally distributed, is there enough evidence, at the 0.05 level of significance, to conclude that the true mean discharge, H, differs from 7 ounces? Perform a two-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. o р The null hypothesis: IX S co The alternative hypothesis:

Answer 1

Given that,
population mean(u)=7
sample mean, x =6.96
standard deviation, s =0.12
number (n)=14
null, Ho: μ=7
alternate, H1: μ!=7
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.16
since our test is two-tailed
reject Ho, if to < -2.16 OR if to > 2.16
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =6.96-7/(0.12/sqrt(14))
to =-1.247
| to | =1.247
critical value
the value of |t α| with n-1 = 13 d.f is 2.16
we got |to| =1.247 & | t α | =2.16
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 != -1.2472 ) = 0.2343
hence value of p0.05 < 0.2343,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=7
alternate, H1: μ!=7
test statistic: -1.247
critical value: -2.16 , 2.16
decision: do not reject Ho
p-value: 0.2343
we do not have enough evidence to support the claim that true mean discharge is differ from 7.