Question

A randomized trial tested the effectiveness of diets on adults. Among 36 subjects using Diet 1, the mean weight loss after a year was 3.5 pounds with a standard deviation of 5.9 pounds. Among 36 subjects using Diet 2, the mean weight loss after a year was 0.6 pounds with a standard deviation of 4.4 pounds. Construct a 95% confidence interval estimate of the difference between the population means, assuming the population standard deviations are equal.

Answer 1

Answer:

The 95%  confidence interval is

           $0.45 < \mu_1 - \mu_2 < 5.35$

Step-by-step explanation:

From the question we are told that

   The first sample size is  $n_1 = 36$

   The first  sample mean is  $\= x_1 = 3.5$

   The first standard deviation is  $\sigma_1 = 5.9 \ pounds$

   The second  sample size is $n_2 = 36$

    The second  sample mean is  $\= x_2 = 0.6$

    The second  standard deviation is $\sigma = 4.4$

Generally the degree of freedom is mathematically represented as

     $df = \frac{ [ \frac{s_1^2 }{n_1 } + \frac{s_2^2 }{n_2} ]^2 }{ \frac{1}{(n_1 - 1 )} [ \frac{s_1^2}{n_1} ]^2 + \frac{1}{(n_2 - 1 )} [ \frac{s_2^2}{n_2} ]^2 }$

=>  $df = \frac{ [ \frac{5.9^2 }{34 } + \frac{4.4^2 }{34} ]^2 }{ \frac{1}{(34 - 1 )} [ \frac{5.9^2}{34} ]^2 + \frac{1}{(34- 1 )} [ \frac{4.4^2}{ 34} ]^2 }$

=>  $df =63$

Generally the standard error is mathematically represented as

      $SE = \sqrt{ \frac{s_1 ^2 }{n_1} + \frac{s_2^2 }{ n_2 } }$

=>  $SE = \sqrt{ \frac{ 5.9 ^2 }{ 36 } + \frac{ 4.4^2 }{36} }$

=>  $SE = 1.227$

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

Generally from the t distribution table the critical value  of   at a degree of freedom of is  

     $t_{\frac{\alpha }{2}, 63 } = 1.998$

Generally the margin of error is mathematically represented as

        $E = t_{\frac{\alpha }{2}, 63 } * SE$

=>    $E = 1.998 * 1.227$

=>    $E = 2.45$

Generally 95% confidence interval is mathematically represented as  

      $(\= x_1 - \x_2) -E < \mu$

 => $(3.5 - 0.6) - 2.45 < \mu_1 - \mu_2 < ( 3.5 - 0.6) + 2.45$

=>   $0.45 < \mu_1 - \mu_2 < 5.35$