Question

In a survey of 1145 ​people, 862 people said they voted in a recent presidential election. Voting records show that 73​% of eligible voters actually did vote. Given that 73​% of eligible voters actually did​ vote, (a) find the probability that among 1145 randomly selected​ voters, at least 862 actually did vote.​ (b) What do the results from part​ (a) suggest?

​(a) P(X>862)=

Answer 1

Solution:
Given in the question
P(Eligible voters actually did vote) = 0.73
Number of sample N = 1145
No. of people said they voted in recent presidential election X = 862
Sample proportion p^ = X/N = 862/1145
We need to calculate the probability that among 1145 randomly selected voters' at least 862 actually did vote
P(X>862)=? can be calculated as
Here we will use one proportion Z test as the sample size is large enough because np(0.73*1145) =836  and nq = (0.27*1145) = 309 are greater than 10
So first we need to calculate Z-score which can be calculated as
Z-score = (p^ - p)/sqrt(p*(1-p)/n) = ((862/1145)-0.73)/sqrt(0.73*(1-0.73)/1145)) = 0.0228/0.0131 = 1.74
From Z score we found p-value
P(X>862)= 1 -P(X<=862) = 1 - 0.9591 = 0.0409
Solution(b)
from the above probability we can conclude that 4.09% of selected voters at least 862 actually did vote.