An average computer mouse inspector can inspect 60 mice per hour with a populati

Question

An average computer mouse inspector can inspect 60 mice per hour with a population standard deviation of 8 mice per hour. The 48 computer mice inspectors at a particular factory can only inspect 55 mice per hour. Does the company have reason to believe that these inspectors are slower than average at = 0.10? ONo, because the test value -2.47 falls in the noncritical region. No, because the test value -2.47 falls in the critical region. Yes, because the test value -4.33 falls in the noncritical region. Yes, because the test value -4.33 falls in the critical region.

Explanation / Answer

Given that,
population mean(u)=60
standard deviation, =8
sample mean, x =55
number (n)=48
null, Ho: =60
alternate, H1: <60
level of significance, = 0.1
from standard normal table,left tailed z /2 =1.282
since our test is left-tailed
reject Ho, if zo < -1.282
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 55-60/(8/sqrt(48)
zo = -4.33
| zo | = 4.33
critical value
the value of |z | at los 10% is 1.282
we got |zo| =4.33 & | z | = 1.282
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : left tail - ha : ( p < -4.33 ) = 0
hence value of p0.1 > 0, here we reject Ho
ANSWERS
---------------
null, Ho: =60
alternate, H1: <60
test statistic: -4.33
critical value: -1.282
decision: reject Ho
p-value: 0
yes, because the test value -4.33 falls in the no crtical region

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