Question 1042675
probability of 0 broken mirrors is
0.99^10=0.904
The probability of one or more broken mirrors is 0.0996. At a 5% level of significance, this meets the criteria, assuming independence of one mirror from another.
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Expected value is 0.01*2000=20
use alpha=0.05
reject if |z|(can use z because t df=1999) is >1.96
probability of 30 or more mirrors broken
Can use normal probability distribution as well as 1-sample proportion.
First: np=mean=20
sd=sqrt (np(1-p)=sqrt(2000*0.99*0.01)=sqrt(19.8)=4.45
z=(x-mean)/sd=(30-20)/4.45=2.26
p-value=0.0123.
This value is highly significant, the likelihood of this result happening by chance is <2%, and the company is not living up to its guarantee if we set the standard at 5% significance.
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The proportion approach is z=(0.015-0.01)/sqrt{(0.01)(0.99)/2000}
This is 0.005/0.00222=2.245, which is almost the same result.