Answer:
The decision rule is
Reject the null hypothesis
The test statistics is [tex]t = -2.699[/tex]
Step-by-step explanation:
From the question we are told that
The first sample size is [tex]n_1 = 15[/tex]
The mean at first deployment is [tex]\= x _1 = \$ 150 000[/tex]
The standard deviation is [tex]s_1 = \$ 40000[/tex]
The second sample size is [tex]n_2 = 25[/tex]
The mean at second deployment is [tex]\= x_2 = \$ 180000[/tex]
The standard deviation is [tex]s_2 = \$ 30 000[/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]
The level of significance is [tex]\alpha = 0.05[/tex]
Generally the degree of freedom is mathematically represented as
[tex]df =n_1 + n_2 -2[/tex]
=> [tex]df =40 -2[/tex]
=> [tex]df =38[/tex]
Generally the pooled variance is mathematically represented as
[tex]s_p^2 = \frac{x(n_1 -1) s_1^2 + (n_2 - 1)s_2^2}{df}[/tex]
=> [tex]s_p^2 = \frac{(15 -1) 40000^2 + (25 - 1)30000^2}{38}[/tex]
=> [tex]s_p^2 = 1.1579 * 10^{9}[/tex]
Generally the test statistics is mathematically represented as
[tex]t = \frac{ \= x_1 - \= x_2 }{\sqrt{s_p [\frac{1}{n_1} + \frac{1}{n_2} ]} }[/tex]
=> [tex]t = \frac{ 150000 - 180000 }{\sqrt{1.11579 *10^{9} [\frac{1}{15} + \frac{1}{25} ]} }[/tex]
=> [tex]t = -2.699[/tex]
Generally from the t distribution table the probability corresponding to the t statistics value to the left is
[tex]t_{-2.699 , 38} = 0.00516046[/tex]
Generally the p -value is mathematically represented as
[tex]p-value = 2* t_{-2.699, 38}[/tex]
=> [tex]p-value = 2* 0.00516046[/tex]
=> [tex]p-value = 0.01032[/tex]
From the obtained value we see that the [tex]p-value < \alpha[/tex] hence
The decision rule is
Reject the null hypothesis
