Respuesta :
Answer:
The decision Rule is
Reject the null hypothesis
The conclusion is
There is sufficient evidence to states that tim and per's mutations lead to different mean copulation duration.
The test statistics is [tex]z =-2.2187[/tex]
Step-by-step explanation:
From the question we are told that
The Mutation Mean copulation duration is for per [tex]\= x_1 = 17.5[/tex]
The Mutation Mean copulation duration is for tim [tex]\= x_2 = 19.9[/tex]
The Mutation Standard deviation copulation duration for is per [tex]s _1 = 3.37[/tex]
The Mutation Standard deviation copulation duration for is tim [tex]s _2 = 2.47[/tex]
The sample size for per is [tex]n_1 = 14[/tex]
The sample size for tim is [tex]n_2 = 17[/tex]
The null hypothesis is [tex]H_o : \mu_1 = \mu_2[/tex]
The alternative hypothesis is [tex]H_a : \mu_1 \ne \mu_2[/tex]
Generally the test statistics is mathematically represented as
[tex]z = \frac{\=x_1 - \= x_2 }{ \sqrt{\frac{s_1^2 }{n_1} + \frac{s_2^2 }{n_2} } }[/tex]
=> [tex]z = \frac{17.5 - 19.9 }{ \sqrt{\frac{3.37^2 }{14} + \frac{2.47^2 }{17} } }[/tex]
=> [tex]z =-2.2187[/tex]
From the z table the area under the normal curve to the left corresponding to -2.2187 is
[tex]P(Z < -2.2187) = 0.013254[/tex]
Generally the p-value is mathematically represented as
[tex]p-value = 2 * P(Z <-2.2187 )[/tex]
=> [tex]p-value = 2 * 0.013254[/tex]
=> [tex]p-value = 0.02651[/tex]
Let assume that the level of significance is [tex]\alpha = 0.05[/tex]
So
From the value obtained we see that [tex]p-value < \alpha[/tex] hence
The decision Rule is
Reject the null hypothesis
The conclusion is
There is sufficient evidence to states that tim and per's mutations lead to different mean copulation duration.