SOLUTION: A sales clerk in a departmental store claims that 60%of the shoppers entering the store without making a purchase .A random sample of 50 shoppers show that 40 of them leave without

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Question 1184381: A sales clerk in a departmental store claims that 60%of the shoppers entering the store without making a purchase .A random sample of 50 shoppers show that 40 of them leave without buying anything .Can we accept the clerk claim at0.05 level of significance.
Find the test value and critical value
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
population proportion = .6
sample size = 50
sample proportion = 40/50 = .8
test standard error = sqrt(.6 * .4 / 50) = .069282.
test z-score = (.8 - .6) / .069282 = 2.88675
critical z-score at two-tail level of significance of .05 = plus or minus 1.96.
test z-score is greater than this, therefore the results are significant and it can be concluded that more than 60% of the shoppers entering the store leave without making a purchase.