Question

In: Statistics and Probability

Company managers at a training method improves assign 10 to each training performance is timed (...

Company managers at a training method improves assign 10 to each training performance is timed ( in large industrial complex are interested in determining if a new worker performance. The managers select 20 workers and randomly method, and upon completion of the training program the workers minutes). The following data give the completion time. Time Method1 15 20 11 23 16 21 18 16 27 24 Method2 23 31 13 19 23 17 28 26 25 28 1. Using a 90% confidence level, estimate the difference in mean job completion times using these two methods. 2. Is there a significant difference ( α = 0.10 ) between the mean job completion times using these two training methods?

Time
Method 1 15 20 11 23 16 21 18 16 27 24

Method 2 23 31 13 19 23 17 28 26 25 28

  1. Using a 90% confidence level, estimate the difference in mean job completion times using these two methods.

  2. Is there a significant difference ( α = 0.10 ) between the mean job completion times using these two training methods?

Solutions

Expert Solution

a.
TRADITIONAL METHOD
given that,
mean(x)=19.1
standard deviation , s.d1=4.8178
number(n1)=10
y(mean)=23.3
standard deviation, s.d2 =5.5588
number(n2)=10
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((23.211/10)+(30.9/10))
= 2.326
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 9 d.f is 1.833
margin of error = 1.833 * 2.326
= 4.264
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (19.1-23.3) ± 4.264 ]
= [-8.464 , 0.064]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=19.1
standard deviation , s.d1=4.8178
sample size, n1=10
y(mean)=23.3
standard deviation, s.d2 =5.5588
sample size,n2 =10
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 19.1-23.3) ± t a/2 * sqrt((23.211/10)+(30.9/10)]
= [ (-4.2) ± t a/2 * 2.326]
= [-8.464 , 0.064]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [-8.464 , 0.064] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
b.
Given that,
mean(x)=19.1
standard deviation , s.d1=4.8178
number(n1)=10
y(mean)=23.3
standard deviation, s.d2 =5.5588
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =1.833
since our test is two-tailed
reject Ho, if to < -1.833 OR if to > 1.833
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =19.1-23.3/sqrt((23.2112/10)+(30.90026/10))
to =-1.806
| to | =1.806
critical value
the value of |t α| with min (n1-1, n2-1) i.e 9 d.f is 1.833
we got |to| = 1.80553 & | t α | = 1.833
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 != -1.8055 ) = 0.104
hence value of p0.1 < 0.104,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.806
critical value: -1.833 , 1.833
decision: do not reject Ho
p-value: 0.104
we do not have enough evidence to support the claim that difference ( α = 0.10 ) between the mean job completion times using these two training methods


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