Question

Two​ shooters, Rodney and​ Philip, practice at a shooting range. They fire rounds each at separate targets. The targets are marked with circles and each bullet hitting a particular circle gets them a particular number of points. rounds are selected at random. The sample mean scores of Rodney and Philip are and ​, respectively.​ And, their variances are and ​, respectively. The standard error of the difference between their mean scores is . 13. ​(Round your answer to two decimal places​.) The​ 95% confidence interval for the difference between the mean scores of Rodney and Philip​ is

Answer 1

Complete Question

Answer:

a

  $SE = 0.66}$

b

$-3.29 < \mu_1 - \mu_2 < -0.70$  

Step-by-step explanation:

From the question we are told that

  The sample size is  n  = 60

   The first sample mean is  $\= x _1 = 8$

    The second sample mean is   $\= x _2 = 10$

    The first variance is  $v_1 = 0.25$

    The first variance is  $v_2 = 0.55$

Given that  the confidence level is 95% then the level of significance is 5% =  0.05

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the first standard deviation is  

     $\sigma_1 = \sqrt{v_1}$

=>   $\sigma_1 = \sqrt{0.25}$

=>   $\sigma_1 = 0.5$

Generally the second standard deviation is

     $\sigma_2 = \sqrt{v_2}$

=>   $\sigma_2 = \sqrt{0.55}$

=>   $\sigma_2 = 0.742$    

Generally the first standard error is

     $SE_1 = \frac{\sigma_1}{\sqrt{n} }$

      $SE_1 = \frac{0.5}{\sqrt{60} }$

     $SE_1 = 0.06$

Generally the second standard error is

     $SE_2 = \frac{\sigma_2}{\sqrt{n} }$

      $SE_2 = \frac{0.742}{\sqrt{60} }$

     $SE_2 = 0.09$

Generally the standard error of the difference between their mean scores is mathematically represented as    

      $SE = \sqrt{SE_1^2 + SE_2^2 }$

=>     $SE = \sqrt{0.06^2 +0.09^2 }$

=>     $SE = 0.66}$

Generally 95% confidence interval is mathematically represented as  

      $(\= x_1 -\= x_2) -(Z_{\frac{\alpha }{2} } * SE) < \mu_1 - \mu_2 < (\= x_1 -\= x_2) +(Z_{\frac{\alpha }{2} } * SE)$

=> $(8 -10) -(1.96 * 0.66) < \mu_1 - \mu_2 < (8-10) +(Z_{\frac{\alpha }{2} } * 0.66)$  

=>  $-3.29 < \mu_1 - \mu_2 < -0.70$