Question

A professor is interested in comparing the characteristics of students who do and do not vote in elections. For a random sample of 114 students who claimed to have voted in the last presidential election, she found a mean CPA of 2.71 and a standard deviation of 0.64. For an independent sample of 123 students who did not vote, the mean CPA was 2.79 and the standard deviation was 0.56. Test at the 5% level of significance if the mean CPA of all students who vote is lower than the CPA of students who do not vote.

Answer 1

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to conclude that the mean CPA of all students who vote is lower than the CPA of students who do not vote

Explanation:

From the question we are told that

The first sample size is
`n_1 =  114`

The sample mean is
`\= x_1 =  2.71`

The standard deviation is
`s_1 =  0.64`

The second sample size is
`n_2 =  123`

The sample mean is
`\= x_2 =  2.79`

The level of significance is
`\alpha  =  0.05`

The standard deviation is
`s_2 =  0.56`

The null hypothesis is
`H_o  :  \mu_1 = \mu _2`

The alternative hypothesis is
`H_a :  \mu_1 <  \mu_2`

Generally the test hypothesis is mathematically represented as

`t =  \frac{\= x_1 - \= x_2}{ \sqrt{ (s_1^2 )/(n_1) + (s_2^2)/(n_2)  } }`

=>
`t =  \frac{2.71 - 2.79}{ \sqrt{ (0.64^2 )/( 114) + (0.56^2)/(123)  } }`

=>
`t =  -1.01`

Generally given that variance are not equal (the standard deviation squared of both population are not equal ) the degree of freedom is mathematically represented as

`df  =  ([(s_1^2)/(n_1 ) + (s_2^2)/(n_2)  ])/(([(s_1^2)/(n_1)]^2 )/(n_1 - 1) + ([(s_2^2)/(n_2)]^2 )/(n_2 - 1))`

=>
`df  =  ([(0.64^2)/(114) + (0.56^2)/(123)  ])/(([(0.64^2)/(114)]^2 )/(114 - 1) + ([(0.56^2)/(123)]^2 )/(123 - 1))`

=>
`df  =  225`

Generally from the t distribution table the the probability of (t < -1.01) at a degree of freedom of
`df  =  225`is

`p-value  = P(t <  -1.01 ) =  0.15679`

Generally the value obtained we see that
`p-value  > \alpha` so

The decision rule is

Fail to reject the null hypothesis

The conclusion is

There is no sufficient evidence to conclude that the mean CPA of all students who vote is lower than the CPA of students who do not vote