Given a sample {X1,X2 ,X,), let be a sample range: = max(X1, . . . , xn)-min(X1,

Question

Given a sample {X1,X2 ,X,), let be a sample range: = max(X1, . . . , xn)-min(X1, . . . ,%) Suppose that the statistic from 0 to 100. is known to be uniformly distributed 1. Find two numbers u and u2 such that (HINT: These numbers are such that the probability of the uniform random variable U(0, 100) taking a value between them is 0.90) 2. Transform the equation P(u1 s U2) 0.90 to the equivalent form, where is in the middle of the inequality: 3. If you take a sample and equals to 1050 for that sample, what is the 90% confidence interval on ?

Explanation / Answer

1. Given that (-)/10 follows a uniform distribution from 0 to 10,

using P( u1 <= (-)/10 <= u2) = 1 - (P((-)/10 > u2) + P((-)/10 < u1)) = 0.9

let, P((-)/10 > u2) = k & P((-)/10 < u1) = 0.1-k where 0 <= k <= 0.1

Using the CDF of uniform distribution,

P((-)/10 > u2) = 1 - u2/100

P((-)/10 < u1) = u1/100

Solving by comparing the two values of the probabilities,

1 - u2/100 = k, u1/100 = 0.1 - k

-> u2 = 100*(1-k)

-> u1 = 10 - 100*k for k in [0,0.1]

Using k=0.05 (Arbitrary choice)

u2 = 95, u1 = 5

2.

Taking the expression from the probability P( u1 <= (-)/10 <= u2),

u1 <= (-)/10 <= u2

Multiplying the equation with 10,

10*u1 <= (-) <= 10*u2

Subtracting from all sides of the equation,

10*u1- <= - <= 10*u2-

Multiplying by -1 on all sides of the equation,

- 10*u1 <= <= - 10*u2

3.

A 90% confidence interval for (-)/10 is given by P( u1 <= (-)/10 <= u2) = 0.9. Using equal split on both the tails, we get the values u1 = 5 and u2 = 95 as seen in the first part.

Hence using the second part to convert the confidence interval of  (-)/10 into that of ,

- 10*u1 <= <= - 10*u2

Putting the values = 1050, u2 = 95 and u1 = 5,

1050 - 10*5 <= <= 1050 - 10*95

1000 <= <= 100

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