Question

An avid fan of early 2000's baseball wanted to test whether corked bats actually improved a baseball bat's hitting power to see if Mark McGuire was wrongfully removed from Backyard Baseball 2003. To do so, the researcher gathered 25 men together from his college campus and divided into two groups, Group A with 13 people and Group B with 12. Group A was given batting practice with corked bats whereas Group B used standard wood bats. Group A hit an average of 11.4 balls into the outfield with s = 2.2. Group B averaged hitting 9.7 balls into the outfield with s = 3.5. Assume the two groups are independent.

Rounding to three units, the lower bound, for a 95% confidence interval for ?μ_{1 -} ?μ₂ using conservative degrees of freedom would be:

Rounding to three units, the upper bound, for a 95% confidence interval for ?μ₁ - ?μ₂ using conservative degrees of freedom would be:

Answer 1

Given

Group A with 13 people and Group B with 12. Group A was given batting practice with corked bats whereas Group B used standard wood bats. Group A hit an average of 11.4 balls into the outfield with s = 2.2. Group B averaged hitting 9.7 balls into the outfield with s = 3.5.

Assume the two groups are independent.

Thus

n1 = 13           ;       n2 = 12

= 11.4         ;       = 9.7

s1 = 2.2          ;      s2 = 3.5

95% confidence interval for μ_{1 -} μ₂ is given by

C.I = { ( - ) * S.E }

Calculation -

Standard Error S.E =

where sd² =

              =

          sd² = 192.83 / 23 = 8.383913

sd² = 8.383913

Hence

S.E = = = 1.159127

Hence S.E = 1.159127

now is t-distributed with =0.05

Since we need to use Conservative approximation, to estimate the degrees of freedom. This is simply the smaller of the two numbers n1-1 and n2-1.

Thus degree of freedom = min( n1-1 , n2 -1 ) = min( 13-1 , 12 -1 ) =min(12,11) = 11 degree of freedom

i.e is t-distributed with degree of freedom = 11 and =0.05

It can be computed from statistical book or more accurately from any software like R,Excel

From R

> qt(1-0.05/2,df=11)
[1] 2.200985

Thus = 2.200985               with df = 11 at =0.05

Hence

95% confidence interval for μ_{1 -} μ₂ is given by

C.I = { ( - ) * S.E }

C.I = { (11.4   - 9.7 ) 2.200985 * 1.159127 }

The lower bound ( LB ), for a 95% confidence interval for μ_{1 -} μ₂ is

LB = { (11.4   - 9.7 ) - 2.200985 * 1.159127 } =   -0.8512211

lower bound = -0.851                       ( Rounding to three units )

And

The upper bound ( UB ), for a 95% confidence interval for μ_{1 -} μ₂ is

UB = { (11.4   - 9.7 ) + 2.200985 * 1.159127 } = 4.251221

upper bound = 4.251                      ( Rounding to three units )

Hence 95% confidence interval for μ_{1 -} μ₂ is ( -0.851   , 4.251      )

Conclusion - Since our 95% confidence interval include the interger zero { 0 } , hence we conclude that at =0.05 , the averaged hitting by two bats may be same .