Complete Question
An analysis of several polls suggests that 60% of all Florida voters plan to vote for Anderson. A poll of 250 randomly selected Florida voters shows that 144 plan to vote for Anderson.
Required:
a. What is the probability of this result (i.e. 144 voters or less out of 250) happening by chance, assuming the aggregate poll model proportion of 60% is correct?
b. Does your result from part I indicate that the number of voters who plan to vote for Anderson has decreased? In other words, is this outcome unusual?(Recall that an unusual event has a probability of 0.05 or less of occurring )
Answer:
a
[tex]P(p < \^p) = 0.2206[/tex]
b
It is not an unusual event
Step-by-step explanation:
From the question we are told that
The population proportion is p = 0.60
The sample size is n = 250
The number that plans to vote for Anderson is k = 144
Generally the mean of the sampling distribution is
[tex]\mu _{x} = p = 0.60[/tex]
Generally the standard deviation is
[tex]\sigma = \sqrt{ \frac{p(1 - p )}{n} }[/tex]
=> [tex]\sigma = \sqrt{ \frac{0.60(1 - 0.60 )}{250} }[/tex]
=> [tex]\sigma = 0.0310[/tex]
Generally the sample proportion is mathematically represented as
[tex]\^{p} = \frac{k}{n}[/tex]
=> [tex]\^{p} = \frac{144}{250}[/tex]
=> [tex]\^{p} = 0.576 [/tex]
Gnerally the probability of this result (i.e. 144 voters or less out of 250) happening by chance, assuming the aggregate poll model proportion of 60% is correct is mathematically represented as
[tex]P(p < \^p) = P(\frac{ p - \mu_{x}}{ \sigma } < \frac{ \^ p - \mu_{x}}{ \sigma } )[/tex]
=> [tex]P(p < \^p) = P(Z < \frac{ 0.576 - 0,60}{ 0.0310} )[/tex]
=> [tex]P(p < \^p) = P(Z < -0.77 )[/tex]
From the z-table the probability of (Z < -0.77) is
[tex]P(Z < -0.77 ) = 0.2206[/tex]
So
[tex]P(p < \^p) = 0.2206[/tex]
Since
[tex]0.2206 > 0.05[/tex] it implies that it is not an unusual event
