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?

Answer 1

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

`P(p  <  \^p) =   0.2206`

b

It is not an unusual event

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

`\mu _(x) =  p =  0.60`

Generally the standard deviation is

`\sigma  =  \sqrt{ (p(1 - p ))/(n) }`

=>
`\sigma  =  \sqrt{ (0.60(1 - 0.60 ))/(250) }`

=>
`\sigma  =   0.0310`

Generally the sample proportion is mathematically represented as

`\^(p) =  (k)/(n)`

=>
`\^(p) =  (144)/(250)`

=>
`\^(p) = 0.576`

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

`P(p  <  \^p) =  P(( p - \mu_(x))/( \sigma )  <  ( \^ p - \mu_(x))/( \sigma ) )`

=>
`P(p  <  \^p) =  P(Z <  ( 0.576 - 0,60)/( 0.0310) )`

=>
`P(p  <  \^p) =  P(Z <  -0.77 )`

From the z-table the probability of (Z < -0.77) is

`P(Z <  -0.77 )  =  0.2206`

So

`P(p  <  \^p) =   0.2206`

Since

`0.2206 > 0.05` it implies that it is not an unusual event