SOLUTION: A sales clerk in a departmental store claims that 45% of the shoppers entering the store leave without making a purchase. A random sample of 60 shoppers showed that 40 of them left
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Question 1185090: A sales clerk in a departmental store claims that 45% of the shoppers entering the store leave without making a purchase. A random sample of 60 shoppers showed that 40 of them left without buying anything. Can we accept the clerk's claim at α=5% level of significance?
The test value is: Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! p hat is 0.667 because 40/60 shoppers left without buying
Ho: p is =0.45
Ha: p is NE 0.45
alpha is 0.05 p{reject Ho|Ho true}
test stat is a z, 2-way 1-proportion test
reject Ho for |z|> 1.96
z=(p hat-p)/sqrt (p(1-p)/n)
=-0.667/sqrt (0.45*0.55/60)
=-0.217/0.0642
=-3.37
Reject Ho; the true proportion is not 0.45.
