Question

Suppose you ran your elevator simulation program N times with different random number seeds to generate N independent measurements of the average travel time
X₁, X₂,⋅⋅⋅, X_N

a. State the correct formula for the 95% confidence interval for the global average travel time across all N runs assuming that each run of your simulation program was long enough to assume that the average travel times across different runs have an i.i.d. normal distribution. [HINT: Remember to use the t-distribution, because your formula only has access to the sample variance.]

b. Suppose you increase the total number of runs, N, from 9 to 36. How does that affect the confidence interval in part (a), assuming the sample mean and sample variance do not change significantly.

c. Now suppose you must reduce the width of the confidence interval to 1/10th of its size when N = 36. Estimate how many additional runs of your simulation will be needed to satisfy this requirement.

Answer 1

Solution

Back-up Theory

Given X ~ N(μ, σ²),

100(1 - α) % Confidence Interval for μ, when σ is not known is: Xbar ± (t_{n- 1, α /2})s/√n ………… (1)

where

Xbar = sample mean,

t_{n – 1, α /2} = upper (α /2)% point of t-distribution with (n - 1) degrees of freedom,

s = sample standard deviation and

n = sample size.

Now to work out the solution,

Part (a)

Vide (1),

the correct formula for the 95% confidence interval for the global average travel time across all N runs assuming that each run of your simulation program was long enough to assume that the average travel times across different runs have an i.i.d. normal distribution

= Xbar ± (t_{n- 1, 0.025})s/√n   Answer

Part (b)

Vide (1), width of the Confidence Interval = 2(t_{n- 1, α /2})s/√n

When n = 9, the width, W1 = 2 x 2.306s/3 [t_{8, 0.025} from standard t-distribution table]

When n = 36, the width, W2 = 2 x 2.030s/6 [t_{35, 0.025} Using Excel Function: Statistical TINV]

So, W2/W1 = 0.4402

=> when the sample size increases from 9 to 36,

the width of the confidence interval reduces by 56%. Answer

Part (c)

Continuing the same analysis as in Part (b), we want n such that:

{2(t_{n- 1, 0.025})s/√n}/(2 x 2.030s/6) = 1/10 = 0.1

Or, (t_{n- 1, 0.025})/√n = 0.0338

Solution to the above is obtained using Excel calculations and the value of n is 3350 Answer

Excel Calculations Details

α = 0.05		t = t(n - 1, α/2)
n	n - 1	√n	t	t/√n
36	35	6	2.030107915	0.338351
100	99	10	1.9842169	0.198422
500	499	22.36068	1.964729307	0.087865
1000	999	31.6227766	1.962341416	0.0620547
2000	1999	44.7213595	1.96115137	0.0438527
3000	2999	54.7722558	1.960755267	0.0357983
3500	3499	59.1607978	1.960642148	0.0331409
3400	3399	58.30952	1.960662108	0.033625
3,350	3,349	57.87918	2	0.033875

DONE