Stat 315: HW #6 Fall 018 Due: Wednesday, October 10, 018 Updated: Monday, October 8 for misprints. 1. An airport shuttle route includes two intersections with traffic lights. Let i be the number of lights at which the shuttle must stop during a single trip. Consider a random sample of size two (i.e. two trips) from the following distribution. x 0 1 p(x) 0.3 0.5 0. (a) Compute the sampling distribution of the statistic W 1 +. (i.e. find the pmf of W ) Since 1 and are independent, P ( 1 x 1, x ) P ( 1 x 1 ) P ( x ). For example, P (W 1) P ( 1 1, 0) + P ( 1 0, 1) P ( 1 1) P ( 0) + P ( 1 0) P ( 1) 0.3 0.5 + 0.5 0.3 0.3 w 0 1 3 4 p(w) 0.09 0.3 0.37 0.0 0.04 (b) Compute the mean and variance of W. E[W ] 4 w0 p(w) w 0.09 0 + 0.3 1 + 0.37 + 0. 3 + 4 0.04 1.8 E[W ] E[W ] 4 w0 p(w) w 0.09 0 + 0.3 1 + 0.37 + 0. 3 + 4 0.04 4. V ar(w ) E[W ] (E[W ]) 4. (1.8) 0.98 V ar(w ) 1
. (Devore 5.67) One piece of PVC pipe is to be inserted inside another piece. The length of the first piece is normally distributed with mean value 0 in. and standard deviation.5 in. The length of the second piece is a normal rv with mean and standard deviatio5 in. and.4 in., respectively. The amount of overlap is normally distributed with mean value 1 in. and standard deviation.1 in. Assuming that the lengths and amount of overlap are independent of one another, what is the probability that the total length after insertion is between 34.5 in. and 35 in.? Denote 1 and to be the two pieces of PVC pipe and let 3 be the overlap. The total length is L 1 + 3. E[L] E[ 1 + 3 ] E[ 1 ] + E[ ] E[ 3 ] 0 + 15 1 34 V ar(l) V ar( 1 + 3 ) V ar( 1 )+V ar( )+( 1) V ar( 3 ) V ar( 1 )+V ar( )+ V ar( 3 ) (since 1,, 3 are independent) V ar(l) 0.5 + 0.4 + 0.1 0.4 σ L 0.4 0.6481 Hence, since L 1 + 3, a sum of normal random variables, then L is also normally distributed with L N (µ 34, σ 1) P (34.5 L 35) P ( 34.5 34 L 34 35 34 0.6481 0.6481 0.6481 1.54) P (Z 0.77) 0.938 0.7794 0.1588 P (0.77 Z 1.54) P (Z
3. (Devore 5.70) Consider a random sample of size n from a continuous distribution having median 0 so that the probability of any one observation being positive is.5. Disregarding the signs of the observations, rank them from smallest to largest in absolute value, and let W the sum of the ranks of the observations having positive signs. For example, if the observations are -.3, +.7, +.1, and -.5, then the ranks of positive observations are and 3, so W 5. In Chapter 15, W will be called Wilcoxons signed-rank statistic. W can be represented as follows: W 1 Y 1 + Y + 3 Y 3 +... + n Y n, where the Y i s are independent Bernoulli rv s, each with p.5 (Yi 1 corresponds to the observation with rank i being positive). (a) Determine E[Y i ] and then E[W ] using the equation for W. Hint: The first n positive integers sum to n(n+1). Since Y i Bernoulli(p), E[Y i ] p and V ar(y i ) p(1 p). E[Y i ] 0.5 Since W 1 Y 1 + Y + 3 Y 3 +... + n Y n, E[W ] E[1 Y 1 + Y + 3 Y 3 +... + n Y n ] i E[Y i ] i 0.5 0.5 i 0.5 i1 n(n + 1) 4 i1 E[W ] i1 n(n + 1) (b) Determine V ar(y i ) and then V ar(w ). Hint: The sum of the squares of the first n positive integers can be expressed as n(n+1)(n+1). 6 Hint: If 1,,..., n are IID (i.e., a random sample), then E[ n i1 i] n i1 E[ i] and V ar( n i1 i) n i1 V ar( i). Since Y i is Bernoulli(p), V ar(y i ) p(1 p) 0.5 0.5 0.5 V ar(y i ) V ar(w ) V ar(1 Y 1 + Y + 3 Y 3 +... + n Y n ) 1 V ar(y 1 ) + V ar(y ) +... + n V ar(y n ) i V ar(y i ) 0.5 i i1 n(n + 1)(n + 1) 4 i1 V ar(w ) 3
4. A manufacturer has interest in gap size consistency for the piston rings it produces. Model a randomly selected piston rings gap as a random variable with mean 0.36 mm and standard deviation.04 mm. For each random sample, write the sampling distribution of the sample mean statistic and compute the standard error. Note: E[ ] µ, V ar( ) σ n Original Population n E [ ] SE( ) (a) N (µ 0.36, σ 0.04 ) 16 0.36 16 0.01 (b) N (µ 0.36, σ 0.04 ) 64 0.36 64 0.005 (c) unknown(µ 0.36, σ 0.04 ) 16 0.36 16 0.01 (d) unknown(µ 0.36, σ 0.04 ) 64 0.36 64 0.005 Note: For (c), since the sample size is less than 30, we are not assured that the distribution will be approximately normal. For (d), since the sample size is greater than 30, we can assume the distribution will be approximately normally distributed. 5. For the situation described in part (a) of #4, compute the probability the sample mean is between 0.34 and 0.38 mm. P (0.34 < < 0.38) P ( 0.34 0.36 < x µ 0.01 σ/ < 0.38 0.36 ) P ( < Z < ) 0.9545 n 0.01 6. (Devore 5.53) Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.. (a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 9 pins is at least 51? P ( 51) P ( µ σ/ 51 50./ ) P (Z.5) 1 P (Z <.5) 0.006 9 (b) Without assuming population normality, what is the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51? Since the sample size is greater than 40, by the CLT, the distribution of the sample mean is approximately normal even though the population may not be normally distributed. P ( 51) P ( µ σ/ 51 50./ ) P (Z 5.7) 0 40 4
7. (Devore 6.11) Of randomly selected male smokers, 1 smoked filter cigarettes, whereas of n randomly selected female smokers, smoked filter cigarettes. Let p 1 and p denote the probabilities that a randomly selected male and female, respectively, smoke filter cigarettes. Note: 1 and are binomial random variables, so E[] np and V ar() np(1 p). (a) Show that ( 1 / ) ( /n ) is an unbiased estimator for p 1 p. Hint: E[ i ] n i p i for i 1,. [ ] E 1 n 1 E[ 1 ] 1 n E[ ] p 1 n p n p 1 p (b) What is the standard error of the estimator in part (a)? ( ) ( V ar 1 n V ar 1 )+( 1) V ar n (p )(1 p ) n SE p 1 + p n ( ) 1 n V ar ( 1 n ) ( ) n 1 V ar( n 1 )+ 1 1 n (p 1 )(1 p 1 ) + (p )(1 p ) n V ar( ) (p 1 )(1 p 1 ) + n 1 (c) How would you use the observed values x 1 and x to estimate the standard error of your estimator? You can estimate the standard error by using your sample proportions, ˆp 1 x 1 and ˆp x n. (d) If n 00, x 1 17, and x 176, use the estimator of part (a) to obtain an estimate of p1 p. x 1 x n 17 176 00 00 0.635 0.88 0.45 (e) Use the result of part (c) and the data of part (d) to estimate the standard error of the estimator. ( ) SE 1 ˆp n 1 (1 ˆp ) ˆp (1 ˆp ) 0.635 n + 0.88 0.041 00 00 5