Answer:
The 95% confidence interval is
[tex] 0.45 < \mu_1 - \mu_2 < 5.35 [/tex]
Step-by-step explanation:
From the question we are told that
The first sample size is [tex]n_1 = 36[/tex]
The first sample mean is [tex]\= x_1 = 3.5[/tex]
The first standard deviation is [tex]\sigma_1 = 5.9 \ pounds[/tex]
The second sample size is [tex]n_2 = 36[/tex]
The second sample mean is [tex]\= x_2 = 0.6[/tex]
The second standard deviation is [tex]\sigma = 4.4[/tex]
Generally the degree of freedom is mathematically represented as
[tex]df = \frac{ [ \frac{s_1^2 }{n_1 } + \frac{s_2^2 }{n_2} ]^2 }{ \frac{1}{(n_1 - 1 )} [ \frac{s_1^2}{n_1} ]^2 + \frac{1}{(n_2 - 1 )} [ \frac{s_2^2}{n_2} ]^2 }[/tex]
=> [tex]df = \frac{ [ \frac{5.9^2 }{34 } + \frac{4.4^2 }{34} ]^2 }{ \frac{1}{(34 - 1 )} [ \frac{5.9^2}{34} ]^2 + \frac{1}{(34- 1 )} [ \frac{4.4^2}{ 34} ]^2 }[/tex]
=> [tex]df =63[/tex]
Generally the standard error is mathematically represented as
[tex]SE = \sqrt{ \frac{s_1 ^2 }{n_1} + \frac{s_2^2 }{ n_2 } }[/tex]
=> [tex]SE = \sqrt{ \frac{ 5.9 ^2 }{ 36 } + \frac{ 4.4^2 }{36} }[/tex]
=> [tex]SE = 1.227[/tex]
From the question we are told the confidence level is 95% , hence the level of significance is
[tex]\alpha = (100 - 95 ) \%[/tex]
=> [tex]\alpha = 0.05[/tex]
Generally from the t distribution table the critical value of at a degree of freedom of is
[tex]t_{\frac{\alpha }{2}, 63 } = 1.998 [/tex]
Generally the margin of error is mathematically represented as
[tex]E = t_{\frac{\alpha }{2}, 63 } * SE[/tex]
=> [tex]E = 1.998 * 1.227[/tex]
=> [tex]E = 2.45[/tex]
Generally 95% confidence interval is mathematically represented as
[tex](\= x_1 - \x_2) -E < \mu <(\= x_1 - \x_2) + E[/tex]
=> [tex](3.5 - 0.6) - 2.45 < \mu_1 - \mu_2 < ( 3.5 - 0.6) + 2.45 [/tex]
=> [tex] 0.45 < \mu_1 - \mu_2 < 5.35 [/tex]
