Answer:\
a
The null hypothesis is [tex]H_o : \mu_1 - \mu_2 \le 0[/tex]
The alternative hypothesis is [tex]H_a : \mu_1 - \mu_2 > 0[/tex]
b
[tex]t = -0.39[/tex]
c
The conclusion
There is sufficient evidence to conclude that the female outscores the male.
Step-by-step explanation:
From the question we are told that
The sample size for male is [tex]n_1 = 30[/tex]
The sample size for female is [tex]n_2 = 30[/tex]
The level of significance is [tex]\alpha = 0.01[/tex]
The null hypothesis is [tex]H_o : \mu_1 - \mu_2 \le 0[/tex]
The alternative hypothesis is [tex]H_a : \mu_1 - \mu_2 > 0[/tex]
Generally the sample mean for male is
[tex]\= x_1 = \frac{\sum x_i}{n_1}[/tex]
=> [tex]\= x_1 = \frac{620 + 570 +540 + \cdots + 580 }{30}[/tex]
=> [tex]\= x_1 = 574.33 [/tex]
Generally the standard deviation of male is
[tex]s_1 = \sqrt{\frac{\sum (x_1 - \= x)^2}{n_1} }[/tex]
=> [tex]s_1 = \sqrt{\frac{ (620 - 574.33)^2 + (570 - 574.33)^2 + (540 - 574.33)^2 + \cdots +(580 - 574.33)^2 }{30} }[/tex]
=> [tex]s_1 =206.24 [/tex]
Generally the sample mean for female is
[tex]\= x_2 = \frac{\sum x_i}{n_2}[/tex]
=> [tex]\= x_2 = \frac{660 + 590 +560 + \cdots + 560 }{30}[/tex]
=> [tex]\= x_2 = 593.33 [/tex]
Generally the standard deviation of male is
[tex]s_2 = \sqrt{\frac{\sum (x_1 - \= x)^2}{n_2} }[/tex]
=> [tex]\sigma_1 = \sqrt{\frac{ (660 - 593.33)^2 + (590 - 593.33)^2 + (560 - 593.33 )^2 + \cdots +(560 - 593.33)^2 }{30} }[/tex]
=> [tex]s_2 =169.31 [/tex]
Generally the degree of freedom for unequal variance is mathematically represented as
[tex]df = \frac{[\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} ]^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{206.24^2}{30} +\frac{169.31^2}{30} ]^2}{ \frac{\frac{206.24^2}{30}}{30 -1} +\frac{\frac{169.31^2}{30}}{30 -1} }[/tex]
=>[tex]df = 56[/tex]
Generally the test statistics 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 = \frac{574.33- 593.33 }{\sqrt{\frac{206.24^2}{30} + \frac{169.31^2}{30}} }[/tex]
=> [tex]t = -0.39[/tex]
From the t distribution table the value of [tex]P(t < -0.39)[/tex] at a degree of freedom of [tex]df = 56[/tex] is
[tex]P(t < -0.39) = 0.3490080[/tex]
Hence the p-value is [tex]p-value = 0.3490080[/tex]
From the values obtained we see that the p-value is >[tex]\alpha[/tex]
Hence we fail to reject the null hypothesis.
The conclusion is
There is sufficient evidence to conclude that the female outscores the male
