CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES

## Fixed effects: reading ~ 1 ## Value Std.Error DF t-value p-value

## Standardized Within-Group Residuals:

## Number of Observations: 200

## Number of Groups: 10

# extract the variance components VarCorr(readmixed)

## factor(schoolid) = pdLogChol(1)

## Variance StdDev

## (Intercept) 19.12775 4.373528

## Residual 58.52342 7.650060

(a) The mean reading score is 30.604. The with-replacement SE is 1.5533.

(b) The without-replacement SE is 1.4612, just a little bit smaller.

5.19 From the output in the previous exercise, the estimated proportion is 0.47829 with SE 0.0657 and 95% CI [0.330, 0.627]. The CI is quite wide!

5.20 From the R code for Exercise 5.18, we have µ ˆ = 31 565 with SE 1.485.

Here is analogous code for the MIXED procedure in SAS software: proc mixed data=schools; class schoolid; model reading = / solution cl; random schoolid; run;

5.21 With the different costs, the last two columns of Table 5.8 change. The table now becomes:

Now the relative net precision is highest when one stem is sampled per site.

5.22 (a) Here is the sample ANOVA table from the GLM procedure of SAS software:

We estimate R2 by

This value is greater than 0, indicating that there is a clustering effect.

(b) Using (5.36), with c1 = 15c2 and R2 = 0.5, and using the approximation M (N 1) ≈ NM 1, we have

Taking m = 4, we would have n = 300/4 = 75.

5.23 Answers will vary.

5.24 Students using the .csv file must change the missing value code from -9 before doing calculations with the data.

(a) Figure SM5.1 shows a histogram for the population values.

The population has mean 25.782, standard deviation 11.3730526, and median 26. These are calculated ignoring the small amount of missing data.

(b, c) Answers will vary, depending on which hour is chosen for the systematic sample. Table SM5.4 gives the summary statistics for each of the 24 possible systematic samples.

CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES

CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES

5.27 (a) From (5.27), we have that equation above (5.9) states that of

(b) The first equality follows directly from (5.13). Because SSTO = SSB + SSW,

(d) The coefficient of S2R2 in (c) is

5.28

(b) In the following, let y

CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES

CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES

The notation O(1/n) denotes a term that tends to 0 as n → ∞.

(b) Follows from the last line of (a).

(c)

5.30 The cost constraint implies that

Setting the derivative equal to zero and solving for into (5.34), we have:

Using Exercise 5.27,

5.31 Using (5.28) and (5.24),

Similarly,

5.32 This exercise does not rely on methods developed in this chapter (other than for a general knowledge of systematic sampling), but represents the type of problem a sampling practitioner might encounter. (A good sampling practitioner must be versatile.)

(a) For all three cases, P (detect the contaminant) = P (distance between container and nearest grid point is less than R). We can calculate the probability by using simple geometry and trigonometry.

Case 1: R < D. See Figure SM5.2(a).