Plaginicj

2022-10-08

Supervisor Tom Choi at Tempe Equipment Company is concerned that material is not arriving as promptly as needed at work cells. A new kanban system has been installed, but there seems to be some delay in getting the material moved to the work cells so that the job can begin promptly. Choi is interested in determining how much delay there is on the part of his highly paid machinists. Ideally, the delay would be close to zero. He has asked his assistant to determine the delay factor among his 10 work cells. The assistant collects the data on a random basis over the next 2 weeks and determines that of the 1,200 observations, 100 were made while the operators were waiting for materials. Use a 99% confidence level and a 3% acceptable error.

In your report, the percent of time a delay occurs =________ % (enter your response as a percentage rounded to two decimal places). Based on the required confidence level and the acceptable error rate, the sample size of 1,200 is

In your report, the percent of time a delay occurs =________ % (enter your response as a percentage rounded to two decimal places). Based on the required confidence level and the acceptable error rate, the sample size of 1,200 is

Tanya Anthony

Beginner2022-10-09Added 5 answers

For the given work sampling plan:

Sample Size = 1200 observations

Number of observations with material waiting or time delay = 100 observations

Confidence interval = 99%

z-score for 99% = 2.58

e = error level = 3% = ± 0.03

To determine sample number of observations required for the work study following formula is used:

$n=(\frac{z}{e}{)}^{2}(p)(1-p)=(\frac{2.053}{0.06}{)}^{2}\xad(0.18)(1\u20130.18)=173\text{samples}$

where,

$p=\text{proportion of time delay occurs}=\frac{(\text{number of time delay})}{(\text{total no. of obs.})}=\frac{100}{1200}=0.08333$

$\text{percentage of time delay}=\text{proportion of time delay occurs}\times 100=0.08333\times 100=8.33\text{\%}$

Confidence interval = 99.73%

z-score for 99% confidence interval = 2.58

$e=\text{error level}=\pm 0.03$

Number of sample observation required $=n=(\frac{2.58}{0.03}{)}^{2}\xad(0.08333)(1\u20130.08333)$

$n=7396\times 0.0764$

n = 565 observations

Since the estimate of number of observations (565) required is less than the actual number of observations, thus based on the required confidence level and the acceptable error rate, the sample size of 1,200 is adequate.

Sample Size = 1200 observations

Number of observations with material waiting or time delay = 100 observations

Confidence interval = 99%

z-score for 99% = 2.58

e = error level = 3% = ± 0.03

To determine sample number of observations required for the work study following formula is used:

$n=(\frac{z}{e}{)}^{2}(p)(1-p)=(\frac{2.053}{0.06}{)}^{2}\xad(0.18)(1\u20130.18)=173\text{samples}$

where,

$p=\text{proportion of time delay occurs}=\frac{(\text{number of time delay})}{(\text{total no. of obs.})}=\frac{100}{1200}=0.08333$

$\text{percentage of time delay}=\text{proportion of time delay occurs}\times 100=0.08333\times 100=8.33\text{\%}$

Confidence interval = 99.73%

z-score for 99% confidence interval = 2.58

$e=\text{error level}=\pm 0.03$

Number of sample observation required $=n=(\frac{2.58}{0.03}{)}^{2}\xad(0.08333)(1\u20130.08333)$

$n=7396\times 0.0764$

n = 565 observations

Since the estimate of number of observations (565) required is less than the actual number of observations, thus based on the required confidence level and the acceptable error rate, the sample size of 1,200 is adequate.

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