CFA Practice Question

There are 434 practice questions for this study session.

CFA Practice Question

A researcher randomly samples 100 residents of Florida and finds that 95 of the citizens are literate. The researcher then finds a 90% confidence interval. Which of the following is false?
A. The researcher is 90% sure that the percentage of literate citizens is between 91.4% and 98.6%.
B. The sample proportion p' is 0.95 and the margin of error is 3.6%.
C. The confidence interval procedure is invalid because Np' and N(1 - p') are not both greater than 5.
Explanation: The procedure used for confidence intervals assumes that Np' > 5 and that N(1 -p') > 5. Because this is not the case and 100(1 - 0.95) = 5, the researcher cannot conclude with 90% confidence that the percentage of Florida citizens that are literate is between 91.4% and 98.6%.

User Contributed Comments 7

User Comment
haarlemmer How to caculate the margin of error though?
tengo variance = np(1-p) 100*.95*.05 =4.75
MUSK What is this "NP, N(1-P) both not being higher than 5?" Is there a rule like that for determining the intervals? Please reply.
aggabad margin of error= 1.645*(.05*.95/100)^1/2
Epix Where in the book does it talk about the rule for C. I cannot find it.
Mariecfa Margin of Error
E = Test Statistic*(P*(1-P)/(SD)/n^1/2))
aggabad forgot the standard deviation 2.179.
E = 1.645*(.95*.05)/(2.179/(100^1/2))
or you use the variance from tengo's suggestion.
E = 1.645*(.95*.05)/(4.75/100)^1/2
GBolt93 Mariecfa, those give you the wrong answer. you end up with a error of .358 when it should be .0358 which you get from aggabad's calculation. .95*.05 is the variance, so not sure why you're dividing that by itself.
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