Sample size The operations manager from Q4 wants to make sure the right sample s

Question

Sample size

The operations manager from Q4 wants to make sure the right sample size was used to collect the observations and calculate the standard time.

The company’s owner will be satisfied within an accuracy 2% in the time estimates but wants to be 99% confident.

Determine of the sample size of 10 that was used was sufficient to meet the owner’s expectations. Describe your findings. Hint: You will have to recall how to calculate the standard deviation from the observations in Q4.

The follwing is Q4 with it answer.

A large cable company received a shipment of audio video cables with the wrong connectors. The error was made by the cable company’s specifications provided to the overseas supplier. The operations manager calculates that it would be cheaper to manually change the connectors rather than the supplier re-doing or fixing the cables. The manager observes the first set of workers pulled aside to change the cables and from these observations will set a standard time. The observations are as follows for workers to each change the connectors on sets of 20 cables.

The manager will use a performance rating factor of .95 and an allowance factor of .04.

Observation

Time (min)

1

7.17

2

7.23

3

6.98

4

7.05

5

7.54

6

7.33

7

7.11

8

7.27

9

7.09

10

6.99

Determine the standard time the manager will use to change the connectors.

Performance rating factor = .95, Allowance factor = .04

Average observation time from above table = 7.176 minutes

Normal cycle time = Average observation time* Rating factor = 7.176*0.95 = 6.8172 minutes

Standard time = Normal cycle time*(1+allowance factor) = 6.8172*1.04=7.089 minutes

So, the standard time the manager will use to change the connectors = 7.089 minutes

Observation

Time (min)

1

7.17

2

7.23

3

6.98

4

7.05

5

7.54

6

7.33

7

7.11

8

7.27

9

7.09

10

6.99

Explanation / Answer

Sample size = (zs/ey) 2

                        z- Confidence level

                        y- The sample mean

                        e- Error term

                        s- Sample standard deviation

Standard deviation = (0.2688/9)1/2

                        = 0.1728

Therefore; sample size

                        = 2.58x 0.1728

                                  0.02 x 7.18

                        = 9.63

                        = 10 observations

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