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the production manager for the xyz manufacturing company is concerned that the customer orders are being shipped late. he asked one of his planners to check the timeliness of shipments. the planner randomly selected 1000 orders and found that 120 orders were shipped late. construct the 95% confidence interval for the proportion of orders shipped late.

Answer :

The confidence interval for the proportion of orders shipped late is 2.763 and 2.523

We are to going to construct a 95% confidence interval, so here the value is x, divided by n.

That is 120 divided by 1000

120/1000= 0.12

Capital value = 1 - 0.12

                      = 0.88

So definitely we are going to right here for this confidence interval must be written as on.

Confidence interval must be written as a cap plus minus go alpha by 2 multiplied by under root of 0.12 multiplied by 0.88 and this must be divisible by n, which is 1000 points.

So if I write here for this, this is 0.12 plus minus

x = 0.12

n = 1000

s = 0.88 = 0.8

z = 95 = 99

Formula to calculate the confidence interval

CI = x ± zs/ √n

    =0.12 ± 95 × 0.88/√1000

    = 0.12 ± 2.643

     = 2.763 and 2.523

Therefore the confidence interval is 2.763 and 2.523.

To know more about the confidence interval refer to the link given below:

https://brainly.com/question/17212516

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