Solution to HW 2, Problem 1

Transcription

1 Gia Elise Barboza A MSU, Department of Statistics STT-825 Sample Surveys Solution to HW 2, Problem 1 Consider text problem #12 (Chapter 2), but only do Buckeye and Gilbert. Use 95% confidence. Buckeye: n 0 = (1.96)2 ( ) (.04) 2 = (1) = = (2) Gilbert: = = (3) The Finite Population Correction makes a difference for Buckeye. b.) Suppose we believe the proportion in Buckeye is at least.65. Plan the sample size for Buckeye to achieve a margin of error of 4 percentage points. n 0 = (1.96)2 ( ) (.04) 2 = (4) = = (5) 1

2 Solution to HW 2, Problem 2 A storage facility contains 1500 shelves (equal sizes), and the problem is to estimate the mean value of items per shelf and the total value of the items in storage. Use this pilot sample to get values of the parameters for planning the sample size. popu<-c(31,40,53,58,60,67,74,77,82,93) mean(popu) [1] 63.5 var(popu) [1] sd(popu) [1] se_mean<-sqrt(((1-(10/1500))*((19.09^2)/10))) se_mean [1] a.) Give the sampling units and define y i. The sampling unit is the shelf and y i is the total value of items per shelf. b.) What sample size is needed to estimate the mean value per shelf to within $3.00 of the actual mean value with 95% confidence? n 0 = (1.96)2 (19.6) 2 (3.00) 2 = (6) = = (7) c.) What sample size is needed to estimate the mean value per shelf to within 4% of the actual mean value with 95% confidence? CV = =.301 (8) n 0 = (1.96)2 (.301) 2 (.04) 2 = (9) 2

3 = = (10) d.) What sample size is needed to estimate the total value of items in storage to within 4500 of the actual total value with 95% confidence? t t D y µ y D = e e = = 3. (11) n 0 = (1.96)2 (19.09) 2 (3.00) 2 = (12) = = (13) e.) What sample size is needed to estimate the total value of items in storage to within 4% of the actual total value with 95% confidence? D =.04 n 0 = (1.96)2 (.301) 2 (.04) 2 = (14) = = (15) Solution to HW 2, Problem 3 Consider the storage facility in #5, but now suppose we want to estimate the number of items in storage that are more than 6 months old. Data from the pilot sample was 12, 18, 07, 08, 15, 18, 21, 13, 08, 18 items on 10 sampled shelves were older than 6 months. a) Give the sampling units and define y i. 3

4 The sampling unit is the shelf and y i is the total number of items that are more than 6 months old. b) Use the sample of size 10 to compute a 95% confidence interval for the number of items in storage that are older than 6 months. popu_6mo<-c(12,18,07,08,15,18,21,13,08,18) > op <- par( bg="light gray", fg="midnight blue") > hist(popu_6mo,main=("histogram of Shelves > 6 Months Old")) > mean(popu_6mo) [1] 13.8 > var(popu_6mo) [1] > sd(popu_6mo) [1] > se_mean<-sqrt(((1-(10/1500))*(( ^2)/10))) > se_mean [1] Histogram of Shelves > 6 Months Old Frequency popu_6mo Figure 1: Histogram of Shelves Greater than 6 Months Old 4

5 se[y] = (1 n )S2 n = 1.57 (16) CI mean = [y Z α/2 SE(y), y + Z α/2 SE(y)] = (17) [ (1.57), (1.57)] = [10.72, 16.88] CI T OT = 1500 CI mean = [16080, ] (18) c) Comment on the validity of your confidence level in (b). The confidence interval relies on the normal approximation but in order for us to be certain in the validity of the interval, n should be large. Although the concept of large is unclear, 10 observations is certainly not large and therefore the 95% confidence interval is probably invalid. Solution to HW 2, Problem 4 A company has freight bills issued during a 10 day period and ranging in value from $2.20 to $ In (a) - (e) below, give the value of S 2 for planning the sample size. Do not find the sample size. a) The bill amounts are approximately uniformly distributed. S 2 = (b a)2 12 = ( )2 12 = b) The bill amounts are approximately normally distributed. S = (b a) 6 = ( ) 6 = S 2 = c) The bill amounts are approximately J-shaped distributed. S 2 = (b a)2 18 = ( )2 18 =

6 d) The bill amounts are approximately unimodal, symmetrically distributed. S 2 = (b a)2 24 = ( )2 24 = e) o distributional shape for the bill amounts is assumed. S 2 (b a)2 4 = ( )2 4 =

7 Solution to HW 2, Problem 5 For problem #7, a manager used information from a similar company to find S = $5.10 to use for planning the sample size. The manager recommended a sample size of 400 to acheieve the accuracy of ±.50 with 95% confidence. The boss noted that the sample size of 400 was larger than the total number of bills in the population. How did the manager get 400? What did the manager do wrong? What the manager did was only half way correct. He applied the right formula to get: n 0 = (1.96)2 (5.10) 2 (.5) 2 = (19) What he failed to do was apply the finite population correction. Had he done so, he would have come to a different conclusion: = (20) As it was, his samples size is based on a sampling plan that is done with replacement. Solution to HW 2, Problem 6 Do text problem 3a (not 3b), Chapter 4. Use stratum sizes 5713, 1272,1288, 5072 which are computed by Area/.039. Also, answer the following questions: h h n h y h t h = h y h S 2 h 2 h (1 n h h ) s2 h n h (5713) 2 ( (1292) 2 ( (1288) 2 ( (5072) 2 ( ) = ) = ) = ) = t h = V ar(ystr ) = SE[y str ] = = b. What are the sampling units? 7

8 Solution to HW 2, Problem 2, Solution to HW 2, Problem 2 8 Tows. c. What is the sampling weight for a tow in stratum #4? W hj = ( h n h ) = ( 1 π hj ) = = d. Compute a 95% confidence interval for the total yield. df = n H = 18 4 = 14; t 0.025,14 = CI = y str ± t α 2 SE(y str) = CI = ± 2.145( ) = [ , 23360] e. Comment on use of the t-distribution in part (d). n is too small for this confidence interval to be valid.

9 Solution to HW 2, Problem 7 We have a population of 500 accounts, stratified by balance of account: a. i. Give a proportional allocation of a sample of n = 100 accounts. h Stratum # of Accounts n = = = ii. Give a sampling weight for an observation in stratum # = 5 b. i. Give a optimal allocation of a sample of n = 100 accounts. h S h n h = ( PL ) n l=1 ls l h i h S h h S h PL l=1 ls l n i 1 150(2.07)= (.2322)= (2.75)= (.6171)= (4.03)= (.1507)=15.07 L l=1 ls l = 1337 ii. Give a sampling weight for an observation in stratum #1. 1 n 1 = = 6.45 c. For a simple random sample of size 100, give the sampling weight for an observation from Stratum #1. = 500 = 5 n 100 9

10 Do problem 10, chapter 4. Solution to HW 2, Problem 8 h y h t h s 2 h h n h t str = H h=1 = 102(3.14) + 310(2.11) + 217(1.23) + 178(.45) = H SE[ t str ] = V ar[ t str ] = h=1 (1 n h h )h 2 s 2 h V ar[ t str ] = [( )2 ]+[( SE[ t str ] = ( ) = n h )2 ]+[( )2 ]+[(1 178)2 ] = b. We need to compare with the previous problem from Chapter 2, therefore we need the estimated total and standard error of this estimate from the SRS. They are: t = y = 807(1.78) = SE[ t] = SE[y] = 807(.367) = ow we can see that both the estimated total and the standard errors are slightly lower than for simple random sampling. c. h n h h p h = = = = =

11 p str = H h h=1 p h = (.143) V ar( p str ) = H h=1 (1 n h h h )2 bp h (1 bp h ) n h 1 (.526) h = 1 ( )2 (.143(1.143) ) = h = 2 ( )2 (.526(1.526) ) = h = 3 ( )2 (.692(1.692) ) = h = 4 ( )2 (.727(1.727) ) = V ar( p str ) =..004 V ar( p str ) = (.692) + (.727) = d. From before, we had p =.56 and SE[ p] = Again, the standard error is slightly smaller than it is in the case of SRS. This means that the estimators are slightly more precise when we use stratified random sampling. 11

12 Solution to HW 2, Problem 9 Refer to text problem 15, Chapter 4. We did part (a) in class, do part (b). o, it is not possible to avoid all selection and measurement bias. In terms of selection bias, it is quite likely that some of the otter dens will be overlooked in the 110- m-wide strip along the coast. Also, any otters that are not found in the 110-m-wide strip will not make it into the sampling frame, and that is another potential source of selection bias. Measurement bias could easily occur as well. For example, if some of the otter dens are counted more than once, if the people doing the counting are poorly trained or if some mistakes are made while counting the dens. BOXPLOTS OF HOLT BY HABITAT HOLTS Group 1 Group 2 Group 3 Group 4 HABITAT Figure 2: Boxplots of Otter (Holt) Data By Habitat 12