Question

Referring back to the process map that I worked in class that involved “getting to work on time”. Suppose that during the construction of that process map, we came up with many potential causes for why we might be getting to work late. We placed those causes in a FMEA and one of the causes that had a high RPN was “getting to bed too late.” Up until that point, it was just a feeling that getting to bed late was causing us to get to work late. However, we decided to do an experiment. For 8 weeks (5 Days per week) we tried to go to bed earlier. For these 8 weeks, we recorded our arrival time at work. The following is the arrival times. Note: Our specification for arrival is 8:00 am, which is 480 minutes.) Using the same notation, we put the arrival times in minutes. The mean and standard deviation for the 40 arrival times above is 447.38 and 11.21, respectively, if you want to work the following two questions by hand.

1) Calculate the 95% confidence interval for the mean arrival time using the data above.

2) If our previous average arrival time was 452 minutes, did we significantly change the process of getting to work on time? Or in other words, is the arrival time different? Use your confidence interval from question 1 to answer this question.

3) If our previous average arrival time was 452 minutes, did we significantly change the process of getting to work on time? Or in other words, is the arrival time different? Perform a hypothesis test to answer this question at the 5% level of significance. (Use a t-statistic)

Answer 1

1.

TRADITIONAL METHOD
given that,
sample mean, x =447.38
standard deviation, s =11.21
sample size, n =40
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 11.21/ sqrt ( 40) )
= 1.772
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 39 d.f is 2.023
margin of error = 2.023 * 1.772
= 3.586
III.
CI = x ± margin of error
confidence interval = [ 447.38 ± 3.586 ]
= [ 443.794 , 450.966 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =447.38
standard deviation, s =11.21
sample size, n =40
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 39 d.f is 2.023
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 447.38 ± t a/2 ( 11.21/ Sqrt ( 40) ]
= [ 447.38-(2.023 * 1.772) , 447.38+(2.023 * 1.772) ]
= [ 443.794 , 450.966 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 443.794 , 450.966 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
2.
TRADITIONAL METHOD
given that,
sample mean, x =452
standard deviation, s =11.21
sample size, n =40
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 11.21/ sqrt ( 40) )
= 1.772
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 39 d.f is 2.023
margin of error = 2.023 * 1.772
= 3.586
III.
CI = x ± margin of error
confidence interval = [ 452 ± 3.586 ]
= [ 448.414 , 455.586 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =452
standard deviation, s =11.21
sample size, n =40
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 39 d.f is 2.023
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 452 ± t a/2 ( 11.21/ Sqrt ( 40) ]
= [ 452-(2.023 * 1.772) , 452+(2.023 * 1.772) ]
= [ 448.414 , 455.586 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 448.414 , 455.586 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
3.
Given that,
population mean(u)=452
sample mean, x =447.38
standard deviation, s =11.21
number (n)=40
null, Ho: μ=452
alternate, H1: μ!=452
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.02
since our test is two-tailed
reject Ho, if to < -2.02 OR if to > 2.02
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =447.38-452/(11.21/sqrt(40))
to =-2.607
| to | =2.607
critical value
the value of |t α| with n-1 = 39 d.f is 2.02
we got |to| =2.607 & | t α | =2.02
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -2.6066 ) = 0.0129
hence value of p0.05 > 0.0129,here we reject Ho
ANSWERS
---------------
null, Ho: μ=452
alternate, H1: μ!=452
test statistic: -2.607
critical value: -2.02 , 2.02
decision: reject Ho
p-value: 0.0129
we have enough evidence to support the claim that the arrival time different.