Question

In: Statistics and Probability

Keely Temp Services Agency finds it difficult to retain its employees because most of them are...

Keely Temp Services Agency finds it difficult to retain its employees because most of

them are looking for full-time positions and will leave the Keely Agency when a good

opportunity comes along. For 60 most recent Keely employees who terminated, the

average time of employment was 7.2 months. Assume that this is a random sample of

Keely employees. If the population standard deviation for the duration of employment of

Keely employees is estimated to be 10.5 months, do you think it will be reasonable to

conclude that an employee who stayed with the company for 18 months stayed

“unusually long?” To answer this question accurately, please use one of the conventional

confidence levels with the largest possible margin of error. Please show the necessary

steps and interpret your result.

Solutions

Expert Solution

Given that,
population mean(u)=18
standard deviation, σ =10.5
sample mean, x =7.2
number (n)=60
null, Ho: μ=18
alternate, H1: μ<18
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 7.2-18/(10.5/sqrt(60)
zo = -7.967
| zo | = 7.967
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =7.967 & | z α | = 1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : left tail - ha : ( p < -7.967 ) = 0
hence value of p0.05 > 0, here we reject Ho
ANSWERS
---------------
i.
null, Ho: μ=18
alternate, H1: μ<18
test statistic: -7.967
critical value: -1.645
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that employee who stayed with the company for 18 months stayed
unusually long.
ii.
given that,
standard deviation, σ =10.5
sample mean, x =7.2
population size (n)=60
I.
standard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
standard error = ( 10.5/ sqrt ( 60) )
= 1.356
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 1.356
= 2.657
Answer:
margin of error = 2.657

orchestra answered 1 year ago
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