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.**

**Let me know in the comment section if anything is not
clear. I will reply ASAP!**

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will be quite encouraging for me.Thank-you!**

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