Consider an election with generic candidates X vs Y (which we could interpret as
ID: 2927065 • Letter: C
Question
Consider an election with generic candidates X vs Y (which we could interpret as chromosomes but not necessarily!).
a. Candidate X had the support of 54% of the voters in the latest poll, which had 200 people. What is a 90% confidence interval for the actual level of support?
b. If the polling organization were to add 100 more people to the sample size, how much would the 90% confidence interval change?
c. Last month's poll had the candidate supported by 51% -- both the previous poll and the current poll had 200 people. Is the difference in polling statistically significant?
d. Candidate X must win 2 particular states in order to win the election; the forecast says she has a 60% chance of winning each state individually. Your friend, a wannabe
statistician, explains that a 0.6 chance of winning one state and a 0.6 chance of winning the other means only a 0.6*0.6= 0.36 chance of winning both – so the "favorite" is actually not the favorite! Explain why your friend is wrong.
Please explain to me clearly and little by little .thanks
Explanation / Answer
PART A.
TRADITIONAL METHOD
given that,
sample size(n)=200
success rate ( p )= x/n = 0.54
I.
sample proportion = 0.54
standard error = Sqrt ( (0.54*0.46) /200) )
= 0.035
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
margin of error = 1.645 * 0.035
= 0.058
III.
CI = [ p ± margin of error ]
confidence interval = [0.54 ± 0.058]
= [ 0.482 , 0.598]
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DIRECT METHOD
given that,
possibile chances (x)=108
sample size(n)=200
success rate ( p )= x/n = 0.54
CI = confidence interval
confidence interval = [ 0.54 ± 1.645 * Sqrt ( (0.54*0.46) /200) ) ]
= [0.54 - 1.645 * Sqrt ( (0.54*0.46) /200) , 0.54 + 1.645 * Sqrt ( (0.54*0.46) /200) ]
= [0.482 , 0.598]
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interpretations:
1. We are 90% sure that the interval [ 0.482 , 0.598] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
PART B.
sample size(n)=300
success rate ( p )= x/n = 0.54
CI = confidence interval
confidence interval = [ 0.54 ± 1.645 * Sqrt ( (0.54*0.46) /300) ) ]
= [0.54 - 1.645 * Sqrt ( (0.54*0.46) /300) , 0.54 + 1.645 * Sqrt ( (0.54*0.46) /300) ]
= [0.493 , 0.587]
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PART C.
Given that,
sample one,n1 =200, p1= x1/n1=0.54
sample two,n2 =200, p2= x2/n2=0.51
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.05
from standard normal table, two tailed z /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) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.54-0.51)/sqrt((0.525*0.475(1/200+1/200))
zo =0.601
| zo | =0.601
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =0.601 & | z | =1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.6008 ) = 0.548
hence value of p0.05 < 0.548,here we do not reject Ho
ANSWERS
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null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 0.601
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.548
there is no significnace diffrence b/w them
PART D.
both probability chances are independent to each other and the
combined probability will be average of both = (0.60+0.60)/2 = .60