Question

Of 346 items tested, 12 are found to be defective. Construct a 98% confidence interval to estimate the proportion of all such items that are defective.
A. (0.012, 0.058)
B. (0.093, 0.600)
C. (0.014, 0.055)
D. (0.015, 0.054)
E. (0.013, 0.680)

Answer 1

A 98% confidence interval to estimate the proportion of all such items that are defective is A . ( 0.012 , 0.058 ) .

Given :

Of 346 items tested, 12 are found to be defective. Construct a 98% confidence interval to estimate the proportion of all such items that are defective.

Middle = 0.02 / 2

= 0.01

At 98 % z score ;

z = -2.33

z* = 2.33

CI = p ± z $\sqrt{p ( 1 - p) / n}$

= 12 / 346 + 2.33 * $\sqrt{12/346 * 334/346 / 346}$

= 0.347 ± 0.229

= 0.347 - 0.229 to 0.347 + 0.229

= 0.118 to 0.576

≈ 0.012 to 0.058

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