Question

1. Voting records show that 61% of eligible voters actually did vote in a recent presidential election. In a survey of 1002 people, 70% said that they voted in that election. Show work:

1. Use the survey results to test the claim that the percentage of all voters who say that they voted is equal to 61%.
2. Test the claim by constructing an appropriate confidence interval.
3. What are the null and alternative hypotheses?
4. What is the value of α= significance level?
5. Is the test two-tailed, left-tailed or right-tailed?
6. What is the test statistic and what is its distribution?
7. What is the value of the test statistic?
8. What is the P-value?
9. What is (are) the critical value(s)?
10. How would you state a conclusion that addresses the original claim?

Answer 1

Given that,
possibile chances (x)=701.4
sample size(n)=1002
success rate ( p )= x/n = 0.7
success probability,( po )=0.61
failure probability,( qo) = 0.39
null, Ho:p=0.61
alternate, H1: p!=0.61
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.7-0.61/(sqrt(0.2379)/1002)
zo =5.841
| zo | =5.841
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =5.841 & | z α | =1.96
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 5.84089 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
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a.
null, Ho:p=0.61
alternate, H1: p!=0.61
test statistic: 5.841
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that the percentage of all voters who say that they voted is equal to 61%.
b.
TRADITIONAL METHOD
given that,
possible chances (x)=701.4
sample size(n)=1002
success rate ( p )= x/n = 0.7
I.
sample proportion = 0.7
standard error = Sqrt ( (0.7*0.3) /1002) )
= 0.014
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.014
= 0.028
III.
CI = [ p ± margin of error ]
confidence interval = [0.7 ± 0.028]
= [ 0.672 , 0.728]
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DIRECT METHOD
given that,
possible chances (x)=701.4
sample size(n)=1002
success rate ( p )= x/n = 0.7
CI = confidence interval
confidence interval = [ 0.7 ± 1.96 * Sqrt ( (0.7*0.3) /1002) ) ]
= [0.7 - 1.96 * Sqrt ( (0.7*0.3) /1002) , 0.7 + 1.96 * Sqrt ( (0.7*0.3) /1002) ]
= [0.672 , 0.728]
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interpretations:
1. We are 95% sure that the interval [ 0.672 , 0.728] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
c.
null, Ho:p=0.61
alternate, H1: p!=0.61
d.
level of significance =0.05
e.
two tailed test
f.
standard normal distribution
g.
test statistic: 5.841
h.
p-value: 0
i.
critical value: -1.96 , 1.96
j.
we have enough evidence to support the claim that the percentage of all voters who say that they voted is equal to 61%.