Answer :
Answer
test statistic [tex]t = -2.27[/tex]
P-value [tex]p- value = 0.0198[/tex]
Step-by-step explanation:
From the question we are told that
The first sample size is [tex]n_1 = 5[/tex]
The first sample mean is [tex]\= x_1 = 1.76 \ mm[/tex]
The first standard is [tex]s_1 = 0.55[/tex]
The second is [tex]sample \ size\ is\ n_2 = 11[/tex]
The second sample mean is [tex]\= x_2 = 2.56[/tex]
The second standard deviation is [tex]\sigma = 0.84[/tex]
The level of significance is [tex]\alpha = 0.01[/tex]
The null hypothesis is [tex]H_o : \mu_1 - \mu_2 = 0[/tex]
The alternative hypothesis is [tex]H_a: \mu_1 - \mu_2 < 0[/tex]
Generally the test statistics is mathematically represented as
[tex]t = \frac{ (\= x _1 - \= x_2) - ( \mu_1 - \mu_2)}{ \sqrt{\frac{s_1 ^2 }{ n_1 } + \frac{s_1 ^2 }{ n_1 } } }[/tex]
=> [tex]t = \frac{ (1.76 - 2.56 ) - 0 }{ \sqrt{\frac{0.55^2 }{ 5 } + \frac{0.84^2 }{11 } } }[/tex]
=> [tex]t = -2.27[/tex]
Generally degree of freedom is mathematically represented as
[tex]df = n_1 + n_2 - 2[/tex]
=> [tex]df = 5 + 11 - 2[/tex]
=> [tex]df = 14[/tex]
From the student t distribution table the probability value to the left that corresponds to [tex]t = -2.27[/tex] at a degree of freedom of [tex]df = 14[/tex] is
[tex]p- value = 0.0198[/tex]