Question

A certain large shipment comes with a guarantee that it contains no more than 20% defective items. If the proportion of items in the shipment is greater than 20%, the shipment may be returned. You draw a random sample of 10 items and test each one to determine whether it is defective.

1.             If in fact 20% of the items in the shipment are defective (so that the shipment is good, but just barely) what is the probability that 7 or more of the 10 sampled items are defective?
2.             Based on the answer to part (a), if 20% of the items in the shipment are defective would 7 defectives in a sample of size 10 be an unusually large number?
3.              If you found that 7 of the 10 sample items were defective, would this be convincing evidence that the shipment should be returned?
4.             If in fact 20% of the items in the shipment are defective, what is the probability that 2 or more of the 10 sampled items are defective?

Answer 1

n =10

p(defective) = 0.2

P ( X = 7) = C (10,7) * 0.2^7 * ( 1 - 0.2)^3=	7	0.0008
P ( X = 8) = C (10,8) * 0.2^8 * ( 1 - 0.2)^2=	8	0.0001
P ( X = 9) = C (10,9) * 0.2^9 * ( 1 - 0.2)^1=	9	0.0000
P ( X = 10) = C (10,10) * 0.2^10 * ( 1 - 0.2)^0=	10	0.0000

p(x>=7) = p(7)+p(8)+p(9) +p(10)

= 0.0009

.............

b)

from part a , if 20% of the items in the shipment are defective,   7 out of 10 is a large number

...............

c)

Ho :   p =    0.2                  
H1 :   p >   0.2       (Right tail test)          
                          
Level of Significance,   α =    0.05                  
Number of Items of Interest,   x =   7                  
Sample Size,   n =    10                  
                          
Sample Proportion ,    p̂ = x/n =    0.7000                  
                          
Standard Error ,    SE = √( p(1-p)/n ) =    0.1265                  
Z Test Statistic = ( p̂-p)/SE = (   0.7000   -   0.2   ) /   0.1265   =   3.9528
                                
                          
p-Value   =   3.86134E-05   [Excel function =NORMSDIST(-z)              
Decision:   p-value<α , reject null hypothesis                      

There is enough evidence to say that shipment should be returned

..................

d)

	X	P(X)
P ( X = 0) = C (10,0) * 0.2^0 * ( 1 - 0.2)^10=	0	0.1074
P ( X = 1) = C (10,1) * 0.2^1 * ( 1 - 0.2)^9=	1	0.2684

p(x >=2) = 1- (p(0) +p(1))

= 1- (0.1074 + 0.2684)

= 0.6242

...................

Please revert back in case of any doubt.

Please upvote. Thanks in advance.