In 2005, about 31% of the citizens of the U.S. were Republicans. (a) If 435 citi

Question

In 2005, about 31% of the citizens of the U.S. were Republicans.

(a) If 435 citizens were selected at random, what is the probability of getting 53% or more who are Republicans. Highlight your final answer. Assume that 435 is large enough for you to use the normal density to make your calculations.

(b) In 2006, the U.S. House of Representatives had 435 members, of whom 231 ( 53%) were Republicans. Is this % higher than can reasonably be seen by chance. Why? If so, what is a possible reason for this unusually large sample percentage?

Explanation / Answer

here p=0.31 and n=435

therefore std error =(p(1-p)/n)1/2 =0.0222

a) hence probability of getting 53% or more who are Republicans=P(X>0.53)=1-P(X<0.53)

=1-P(Z<(0.53-0.31)/0.0222)=1-P(Z<9.9211)=1-(~1.000) =0.0000

b) yes the chance of happening that is almost nil. possible reason for this unusually large sample percentage is that populaton % of republicans has been increased in US,

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