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`

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
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.96`

Generally the first standard deviation is

`\sigma_1 =  √(v_1)`

=>
`\sigma_1 =  √(0.25)`

=>
`\sigma_1 =  0.5`

Generally the second standard deviation is

`\sigma_2 =  √(v_2)`

=>
`\sigma_2 =  √(0.55)`

=>
`\sigma_2 =  0.742`

Generally the first standard error is

`SE_1  =  (\sigma_1)/(√(n) )`

`SE_1  =  (0.5)/(√(60) )`

`SE_1  =  0.06`

Generally the second standard error is

`SE_2  =  (\sigma_2)/(√(n) )`

`SE_2  =  (0.742)/(√(60) )`

`SE_2  =  0.09`

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

`SE  =  √(SE_1^2 + SE_2^2 )`

=>
`SE  =  √(0.06^2 +0.09^2 )`

=>
`SE  = 0.66}`

Generally 95% confidence interval is mathematically represented as

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

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

=>
`-3.29 <  \mu_1 - \mu_2 <  -0.70`