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.

451 438 430 471 434 449 460 470 449 452 448 445 454 452 442 436 457 446 462 454 414 453 457 442 442 447 446 447 456 447 431 452 437 446 466 454 443 439 443 433

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

Q1.
given that,
sample mean, x =447.38
standard deviation, s =11.21
sample size, n =40
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 11.21/ sqrt ( 40) )
= 1.772
II.
margin of error = t alpha/2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, alpha = 0.05
from standard normal table, two tailed value of |t alpha/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 ]
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Q2.
since 452 minutes is lies outside the confidence interval in above calculated, hence evidence that it significantly change the process of getting to work on time
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Q3.
null, Ho: μ=452
alternate, H1: μ!=452
level of significance, alpha = 0.05
from standard normal table, two tailed t alpha/2 =2.023
since our test is two-tailed
reject Ho, if to < -2.023 OR if to > 2.023
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =447.38-452/(11.21/sqrt(40))
to =-2.6066
| to | =2.6066
critical value
the value of |t alpha| with n-1 = 39 d.f is 2.023
we got |to| =2.6066 & | t alpha | =2.023
make decision
hence value of | to | > | t alpha| 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
hence evidence that it significantly change the process of getting to work on time