SOLUTION: A survey claims that 9 out of 10 doctors (i.e., 90%) recommend brand Z for their patients who have children. To test this claim against the alternative that the actual proportion o

Question 1180570: A survey claims that 9 out of 10 doctors (i.e., 90%) recommend brand Z for their patients who have children. To test this claim against the alternative that the actual proportion of doctors who recommend brand Z is less than 90%, a random sample of 100 doctors results in 87 who indicate that they recommend brand Z. The test statistic in this problem is approximately (round to the nearest hundredth):

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
the test statistic is z=(0.87-0.9)/sqrt(0.9*0.1/100), substituting in formula z=(p hat-p)/sqrt(p*(1-p)/n)
=-0.03/sqrt(0.09/100)
=-1.00
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