Question

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Answer 1

a) This is a good idea in designing the experiment because it controls for any “learning” that may occur in using the simulators.

b) 95% confidence interval :

First we need to find the differences (d) = impaired - Normal

Then we have to find the mean of differences $\bar{d}$ and standard deviation of differences S_d using the excel function =AVERAGE() and STDEV.S() respectively.

A B C D Impaired Normal d 1 2 5.86 4.49 1.37 3 5.67 4.24 1.43 4 5.45 .58 -AVERAGE (C2 :C10) 0.9689 -STDEV.S(C2:C10) 5 5.29 4.56 0.73 5.83 4.31 1.52 0.3643 7 5.49 4.8 0.69 5.23 4.59 0.64 9 5.63 5 0.63 10 5.63 4.79 0.84 11 co.

$\bar{d}$ = 0.9689 and S_d = 0.3643

Lower bound =   - E

Upper bound =   + E

E = ; t is critical value follows t distribution with degrees of freedom (df) = n-1

t* S Vn

We are given confidence level = 0.95

We can find critical value using excel function =TINV( α , d.f )

α = 1 - confidence level = 1 - 0.95 = 0.05

degrees of freedom (d.f) = n - 1 = 9-1 = 8

t =TINV(0.05,8)

t = 2.306

Therefore,

E =

2.3060.3643

E = 0.28

Lower bound =   - E = 0.9689 - 0.28 = 0.689

Upper bound =   + E = 0.9689 + 0.28 = 1.249

95% Confidence interval is ( 0.689,1.249 )

Conclusion : As confidence interval does not contain 0 ;

There is sufficient evidence to conclude that there is difference in braking time with impaired vision and normal vision .