Question

In a recent poll, 134 registered voters who planned to vote in the next election were asked if they would vote for a particular candidate and 80 of these people responded that they would. We wish to predict the proportion of people who will vote for this candidate in the election.

a) Find a point estimator of the proportion who would vote for this candidate.

b) Construct a 90% confidence interval for the true proportion who would vote for this candidate.

c) If we wanted the margin of error in the previous problem to be less than 2%, what how many people should we sample?

If the candidate receives more than 50% of the votes, a win is guaranteed.

d) Construct the null and alternative hypotheses needed to test the claim that the candidate will have more than 50% of the vote and thus win the election.

e) Calculate the test-statistic and p-value for this hypothesis test.

f) At the .05 level, should the null hypothesis be rejected? Do we have evidence to suggest that the candidate will win the election?

Answer 1

Solution :

Given,

n=134 , x=80

a)

Point estimator for the proportion. P= 80/134 =   0.5970

b)

90% confidence interval for true proportion is given by

p ± Z0.10 * sqrt{p(1-p)/ n}

0.5970 ± 1.645 *sqrt{0.5970 *0.4030/ 134}

0.5970 ± 0.06970

{0.5273, 0.6667}

c)

If Margin of error < 0.02

let sample size is n

1.645 *sqrt{0.5970 *0.4030/ n} = 0.02

sqrt( n) = 1.645 * sqrt(0.5970 *0.4030)/0.02

n= 1627.61 =1628

that is for Margin of error < 0.02

n > 1628 Answer

d)

H0: P= 0.50

H1: P > 0.50

Level =0.05

e)

Test statistics Z = (0.5970- 0.5)/ sqrt{0.5970 *0.4030/ 134}

Z= 0.097/ 0.042373

Z = 2.289

p value = 0.0110

f)

Conclusion : p value ( 0.0110) < 0.05

Reject Ho.

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