Answer:
The decision rule is
Fail to reject the null hypothesis
The conclusion is Â
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
Step-by-step explanation:
From the question we are told that
    The first sample size  is  [tex]n_1 =  114[/tex]
    The sample mean is  [tex]\= x_1 =  2.71[/tex]
    The standard deviation is  [tex]s_1 =  0.64[/tex]
   The second sample size is  [tex]n_2 =  123 [/tex]
   The sample mean is  [tex]\= x_2 =  2.79[/tex]
   The level of significance is  [tex]\alpha  =  0.05[/tex]
   The standard deviation is  [tex]s_2 =  0.56 [/tex]
The null hypothesis is  [tex]H_o  :  \mu_1 = \mu _2[/tex]
The alternative hypothesis is  [tex]H_a :  \mu_1 <  \mu_2[/tex]
Generally the test hypothesis is mathematically represented as
   [tex]t =  \frac{\= x_1 - \= x_2}{ \sqrt{ \frac{s_1^2 }{n_1} + \frac{s_2^2}{n_2}  } }[/tex]
=> [tex]t = Â \frac{2.71 - 2.79}{ \sqrt{ \frac{0.64^2 }{ 114} + \frac{0.56^2}{123} Â } }[/tex]
=> [tex]t = Â -1.01 [/tex]
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
[tex]df  =  \frac{[\frac{s_1^2}{n_1 } + \frac{s_2^2}{n_2}  ]}{\frac{[\frac{s_1^2}{n_1}]^2 }{n_1 - 1} + \frac{[\frac{s_2^2}{n_2}]^2 }{n_2 - 1}}[/tex]
=>  [tex]df  =  \frac{[\frac{0.64^2}{114} + \frac{0.56^2}{123}  ]}{\frac{[\frac{0.64^2}{114}]^2 }{114 - 1} + \frac{[\frac{0.56^2}{123}]^2 }{123 - 1}}[/tex]
=>  [tex]df  =  225[/tex]
Generally from the t distribution table the the probability of  (t < -1.01) at a degree of freedom of [tex]df  =  225[/tex]is Â
   [tex] p-value  = P(t <  -1.01 ) =  0.15679[/tex]
Generally the value obtained we see that  [tex]p-value  > \alpha[/tex] so
The decision rule is
Fail to reject the null hypothesis
The conclusion is Â
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
