SOLUTION FOR HOMEWORK 4, STAT 4352

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1 SOLUTION FOR HOMEWORK 4, STAT 4352 Welcome to your fourth homework. Here we begin the study of confidence intervals, Errors, etc. Recall that X n := (X 1,...,X n ) denotes the vector of n observations. Try to find mistakes (and get extra points) in my solutions. Typically they are silly arithmetic mistakes (not methodological ones). They allow me to check that you did your HW on your own. Please do not me about your findings just mention them on the first page of your solution and count extra points. Now let us look at your problems. 1. Problem Let X be distributed according to the exponential distribution Expon(θ). Please note that θ is the scale parameter, as a result here X/θ = Y Expon(1). As a result, Y is the pivot, and its distribution, which does not depend on θ, will be used to define the confidence interval. But now let us return to the problem at hand. We need to find k such that Recall the pivot Y = X/θ and write, P θ (0 < θ < kx) = 1 α. P θ (0 < θ < kx) = P θ (0 < 1 < kx/θ) = P(Y > k 1 ). Please note that in the last probability I skipped the subscript θ because the distribution of Y does not depend on θ. Now we can find k: which implies the answer 1 α = P(Y > k 1 ) = k = 1/ ln(1 α). e y dy = e k 1, k 1 2. Problem 11.2(a). This problem is a good review of the probability technique on how to work with two RVs. Let X 1 and X 2 be two independent RVs uniformly distributed on (0, θ). Then we need to find k such that P θ (0 < θ < k(x 1 + X 2 )) = 1 α. Again, you may note that the pivot here is (X 1 + X 2 )/θ = Y 1 + Y 2 where Y 1 and Y 2 are independent and uniformly distributed on (0, 1). [In other words, θ is the scale parameter]. Then we can write P θ (0 < θ < k(x 1 + X 2 )) = P(Y 1 + Y 2 > k 1 ) = 1 α. Now the problem is converted into finding αth quantile of the distribution of the sum Z = Y 1 + Y 2 (well, we did this problem in the probability class, by the way.) Let us solve it directly (denote a := 1/k): α = P(Y 1 + Y 2 a) = I(0 < y 1 < 1)I(0 < y 2 < 1)dy 1 dy 2 =... 0 y 1 +y 2 a 1

2 By definition of this particular integral, this integral is equal to the area in the square [0, 1] 2 under the straight line y 2 = a y 1 [please draw a nice diagram of the square and the line to see this set and its area; pay attention to the fact that shape of the set changes depending on a < 1 and a > 1. For the considered case α < 1/2 the set is a lower triangle with the area equal to a 2 /2. Then α = (1/k) 2 /2. This yields the wished k = (1/2α) 1/2. Remark 1: It is possible that I (or you) made a mistake, and it is difficult to check numbers, but this verification may be helpful. If α decreases (the confidence coefficient increases) then the confidence interval must increase (you wish a larger probability of covering the interval must grow!). In our example this means that k must increase when α decreases. Our answer passes this test. [At the same time, to check the exact formulae you need to repeat your solution/check all steps.] Remark2. Think about the answer for the case α > 1/2 where a different set defines the quantile. 3. Problem Here we are considering the classical normal case with z-scoring being the pivot. We need to satisfy Using z-scoring we get P µ ( X z kα σn 1/2 < µ < X + z (1 k)α σn 1/2 ) 1 α. where Z N(0, 1). The latter holds iff P( z kα < Z < z (1 k)α ) 1 α kα + (1 k)α α We conclude that any k [0, 1] implies the desired confidence interval. But which k yields the minimal (most accurate) confidence interval? The answer is k = 1/2, and to prove it, you need to draw the bell-shaped normal density, look at the case k = 1/2 which implies so-called equal-tails confidence interval. Then shift the left side of the interval and look at the required length of a new interval to get the area 1 α under the normal density. By taking into account bell-shaped curve of the density, you prove that the interval becomes larger. 4. Problem Let us assume that ( X µ)/(σ/n 1/2 ) = Z N(0, 1). [This is the case if X is normal or n > 30]. Then we need to find a minimal sample size n 0 such that for the given confidence error E P µ X µ E) = 1 α. Using z-scoring we find that P µ ( X µ E) = P µ ( X µ σ/n 1/2 E/(σ/n1/2 ) ) 2

3 = P( Z E/(σ/n 1/2 )) = 1 α. Solving the last equality we find that n 0 should be equal to the rounded up [z α/2 σ/e] 2. Note that it is increasing in σ and decreasing in 1 α and E. 5. Problem Let X 1 N(µ 1, σ 2 1/n 1 ) and X 2 N(µ 2, σ 2 2/n 2 ). The parameter of interest is µ = µ 1 µ 2. Please note that the sufficient statistic here is Y = X 1 X 2 (check this using the Factorization Theorem). Then we get As a result, This yields a formula for the error Y N(µ, σ 2 ), σ 2 := σ 2 1/n 1 + σ 2 2/n 2. P µ ( Y µ /σ zα/2 ) = 1 α. E α = z α/2 [σ 2 1 /n 1 + σ 2 2 /n 2] 1/2. 6. Problem Suppose that we have X 1 N(θ 1, σ 2 /n 1 ) and independent X 2 N(θ 2, σ 2 /n 2 ) calculated from the samples X 11, X 12,..., X 1n1 and X 21, X 22,..., X 2n2, respectively. Then a direct calculation yields X 1 X 2 N(θ 1 θ 2, σ 2 (1/n 1 + 1/n 2 )). If σ 2 is known then this allows us to construct a confidence interval using the Z-pivot [ X 1 X 2 ]/[σ 2 (1/n 1 + 1/n 2 )] 1/2. Note that the pivot has standard normal distribution, so we can use the usual z α/2 in our formulae. However, if σ 2 is unknown, we need to estimate it. Because σ 2 is the same in both samples, we can pool them together and calculate n 1 ˆV := (X 1l X 1 ) 2 + (X 2l X 2 ) 2. Now: the next step is important. Consider a new random variable n 1 W := ˆV /σ 2 = (X 1l X 1 ) 2 /σ 2 + (X 2l X 2 ) 2 /σ 2. As we know from Section 8.4 (and our first homework), the first sum in the last equation has Chisq(n 1 1) distribution and the second has Chisq(n 2 1) distribution. Because these sums are independent, their total W has Chisq(n 1 + n 2 2) distribution. n 2 n 2 3

4 Further, we can introduce another random variable: ˆT := [( X 1 X 2 ) (µ 1 µ 2 )]/[σ 2 (n n 1 2 )] 1/2 [ˆV /σ 2 (n 1 + n 2 2)] 1/2 =: ξ 0 [χ 2 n 1 +n 2 2/(n 1 + n 2 2)] 1/2, where ξ 0 N(0, 1) is independent of χ 2 n 1 +n 2 2, which is a Chisq(n 1 + n 2 2). The latter is true because X 1 is independent from n 1 (X 1l X 1 ) 2, a similar assertion holds for the second sample, and further the two samples are independent. Now you need to recall that in this case ˆT has t-distribution with n 1 + n 2 2 degrees of freedom. What was wished to show. 7. Problem Let X Binom(θ, n). Then for sufficiently large n X θ N(0, θ(1 θ)/n). This yields (according to our previous work) that As a result, P θ ( X θ z α/2 (θ(1 θ)/n) 1/2 ) = 1 α. E α = z α/2 [θ(1 θ)]/n 1/2 is the Error that guarantees the (1 α) confidence. Then the sample size needed to guarantee an error E is n z 2 α/2 θ(1 θ)/e2. Now we note that max θ (0,1) θ(1 θ) = 1/4, and this yields a conservative lower bound n z 2 α/2 /(4E2 ). Remark: Typically the variance θ(1 θ) will be estimated by X(1 X). 8. Problem Here we have two independent Binomial RVs X 1 Binom(θ 1, n 1 ) and X 2 Binom(θ 2, n 2 ). Both θ 1 and θ 2 are unknown, and we are interested only in estimation of the parameter θ = θ 1 θ 2 (the difference between the populations probabilities of success). Denote ˆθ 1 = X 1 /n 1 and ˆθ 2 = X 2 /n 2. Also, denote ˆσ = [ˆθ 1 (1 ˆθ 1 )/n 1 + ˆθ 2 (1 ˆθ 2 )/n 2 ] 1/2. If both samples are large then we can use normal approximation and get ˆθ 1 ˆθ 2 (θ 1 θ 2 ) N(0, 1). ˆσ This implies that the (1 α) confidence maximal Error (half of the confidence interval) is E α = z α/2 /ˆσ = z α/2 [ˆθ 1 (1 ˆθ 1 )/n 1 + ˆθ 2 (1 ˆθ 2 )/n 2 ] 1/2. 4

5 Remark: Please note that if the Error E is given, then the last equality allows us to find the lower bound for corresponding sample sizes. 9. Problem Here n = 120, σ = 10.5, and 1 α =.99. Then (using our formula and the Normal Table in your text) E = z α/2 σ/n 1/2 = z.005 (10.5)/(120) 1/2 = (2.755)(10.5)/(120) 1/ Problem It is given that n = 40, X = 24.05, S = We can assert for θ = E(X) that (assuming n > 30) P θ ( X θ t α/2,n 1 S/n 1/2 ) = 1 α. In your Table IV (page 575) no t-distribution for 39 degrees of freedom can be found because beyond 29 degrees of freedom the t-distribution is practically standard normal. Thus we can use the approximation t α/2,39 = z α/2. Here α/2 =.025, so z.025 = Then E = (1.96)(2.68)/(40) 1/2. 5