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itz64
04.12.2019 •
Mathematics
The workers at sandbachian, inc. took a random sample of 800 manhole covers and found that 40 of them were defective. what is the 95% ci for p, the true proportion of defective manhole covers, based on this sample? a) (37.26, 42.74)b) (.035, .065)c) (.047, .053)d) (.015, .085)
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Ответ:
b)![(0.035,\ 0.065)](/tpl/images/0402/2772/9259f.png)
Step-by-step explanation:
The confidence interval for proportion (p) is given by :-
, where
= Sample proportion
n= sample size.
z* = Critical z-value.
Let p be the true proportion of defective manhole covers, based on this sample.
Given : The workers at Sandbachian, Inc. took a random sample of 800 manhole covers and found that 40 of them were defective.
Then , n= 800
Confidence interval = 95%
We know that the critical value for 95% Confidence interval : z*=1.96
Then, the 95% CI for p, the true proportion of defective manhole covers will be :-
Hence, the required confidence interval : b)![(0.035,\ 0.065)](/tpl/images/0402/2772/9259f.png)
Ответ:
The pic? For the question for me to answer it