Answer :
Answer:
Decision rule is
Fail to reject the null hypothesis
The conclusion is
There is sufficient evidence that indicate that there is a significant difference between the two treatments
Step-by-step explanation:
From the question we are told that
The sample size is n = 12
The first mean is [tex]M_1 = 55[/tex]
The variance is [tex]s^2_1 = 8[/tex]
The second mean is [tex]M_2 = 52[/tex]
The second variance is [tex]s^2_2 = 4[/tex]
Let the level of significance be [tex]\alpha = 0.05[/tex]
The null hypothesis is [tex]H_o : M_1 = M_2[/tex]
The alternative hypothesis is [tex]H_a : M_1 \ne M_2[/tex]
Generally the test statistics is mathematically
[tex]t = \frac{M_1 - M_2 }{ \sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2}} }[/tex]
=> [tex]t = \frac{55 - 52 }{ \sqrt{\frac{8}{12} +\frac{4}{12 }} }[/tex]
=> [tex]t = 1.1619 [/tex]
Generally the degree of freedom is mathematically represented as
[tex]df = 12 -1[/tex]
=> [tex]df = 11 [/tex]
Generally from the t- distriibution table the critical value of [tex]\alpha[/tex] at a degree of freedom of [tex]df = 11 [/tex] is
[tex]t_{0.05 , 11} = 2.20098516[/tex]
Here given that the critical value is greater than the t statistics value the
Decision rule is
Fail to reject the null hypothesis
The conclusion is
There is sufficient evidence that indicate that there is a significant difference between the two treatments