In a sample of 200 books, 26 books contained at least one punctuation error. the sampling method had a margin of error of 0.03. what is the interval estimate for the proportion that would have at least one punctuation error in the form (lower limit, upper limit) ?

confidence interval =(0.10,0.16)

Step-by-step explanation:

Given that sample proportion was

26/200 =0.13 for a sample size of 200

Margin of error is given as 0.03

We know that confidence interval is one which is calculated as sample proportion ±Margin of error

Here sample proportion=0.13

Margin of error = 0.03

Hence

confidence interval lower limit = 0.13-0.03 = 0.10

Confidence interval upper limit = 0.13+0.03 = 0.16

Hence confidence interval =(0.10,0.16)

8/k

step-by-step explanation:

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