We have to accept or reject a large shipment of items. For quality control purposes, we collect a sample of 200 items and find 24 defective items. Construct a 95% percent confidence interval for the proportion of defective items in the whole shipment.

A 95% confidence for the population proportion of defective items in the whole shipment is [0.075, 0.165] .

Step-by-step explanation:

We are given that for quality control purposes, we collect a sample of 200 items and find 24 defective items.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

                             P.Q.  =    ~  N(0,1)

where, = sample proportion of defective items = = 0.12

            n = sample of items = 200

            p = population proportion  of defective items

Here for constructing a 95% confidence interval we have used a One-sample z-test statistics for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                                      of significance are -1.96 & 1.96}  

P(-1.96 < < 1.96) = 0.95

P( < < ) = 0.95

P( < p < ) = 0.95

95% confidence interval for p = [ , ]

  = [ , ]

 = [0.075, 0.165]

Therefore, a 95% confidence for the population proportion of defective items in the whole shipment is [0.075, 0.165] .

step-by-step explanation:

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answer: it's d

step-by-step explanation: