A political strategist claims that 55% of voters in Madison County support his candidate. In a poll of 200 ran?

Posted on May 31st, 2011 by admin1 in political strategist

Can someone point me in the right direction as to how to go about solving this – I am stuck and cannot figure out what to do – Thanks for your time!

A political strategist claims that 55% of voters in Madison County support his candidate. In a poll of 200 randomly selected voters, 98 of them support the strategist’s candidate. At α= .05, is the political strategist’s claim warranted?
A) No, because the test value –1.00 is in the noncritical region.
B) No, because the test value –2.22 is in the critical region.
C) Yes, because the test value –2.22 is in the noncritical region.
D) Yes, because the test value –1.71 is in the noncritical region.

p= 0.55
sqrt (p(1-p)/n) = 0.035
test value = -0.06/0.035 = -1.71
Answer D

2 Comments on “A political strategist claims that 55% of voters in Madison County support his candidate. In a poll of 200 ran?”

  1. Justin

    null hypothesis: p = .55 alternate hypotheisi: p does not equal .55

    Estimator = p_1 = 98/200 = .49

    With α= .05, we reject the null hypothesis if Z <= -1.96 or Z >= 1.96

    Z =(p_1 – p)/ sq. root(p * (1-p) / n) = (.49 – .55) / (sq. root( .55* .49/ 200)) = -1.71

    Since -1.71 is in the noncritical region, we can conclude that Yes the strategist’s claim is warranted.

    So D) is the answer.

    hope this helpsReferences :

  2. Bail Out

    p= 0.55
    sqrt (p(1-p)/n) = 0.035
    test value = -0.06/0.035 = -1.71
    Answer DReferences :

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