Suppose the local Childhood Lead Poisoning Prevention Council in a metropolitan

Question

Suppose the local Childhood Lead Poisoning Prevention Council in a metropolitan area in western Tennessee undertakes the responsibility of determining the proportion of homes in a certain development of 120 homes with unsafe lead levels. Because of the great expense involved in performing spectrometric testing of interior walls, ceilings, floors, baseboards, cabinets, and other obvious lead hazards such as crib bars, as well as of exterior sidings, porches and porch rails, it was decided to select a sample of homes under study. A good up-to-date frame exists for sampling purposes. This frame is a street listing containing the address and owner of each home for each of the streets in the target area. It was decided to select a one-in-three (1 -in-3) sample of homes. Let us assume that the only houses with serious lead hazard problems are the 26^th, 27^th, 28^th, and 29^th on the list. a. Suppose the random number 2 was chosen to start the sequence. Estimate the proportion of homes with lead hazards from the sample. b. Obtain a 95% confidence interval for the proportion of homes with lead hazards. What assumptions did you make? c. What is the true variance of the distribution of the estimated proportion of homes with lead hazards? How does this compare with the variance estimated in part (b)? d. Suppose that a simple random sample of 40 homes had been selected instead. What is the variance of the distribution of the estimated proportion of lead hazardous homes in this case? How does this value compare with the variance from a 1-in-3 systematic sample?

Explanation / Answer

(a)

Since the list starts with 2, the sample consists of the following numbered houses:

2,5,8,11,14,17,20,23,26,29,32,....

So only 2 out of the 4 affected houses are on the systematic list generated.

So, proportion of homes with lead hazards accoring to this systematic sample = 2/40 = 0.05

(b)

S.d of this systematic sample = (40*0.05*0.95)0.5 = 1.37

So, 95% CI is:

0.05 - (1.96*(1.37/400.5)) < u < 0.05 + (1.96*(1.37/400.5))

0.05-0.424 < u < 0.05+0.424

-0.374 < u < 0.474

Here, u denotes the population proportion.

(c)

True variance = 120*(4/120)*(1 - 4/120) = 3.86

Variance in (b) = 1.87

(d)

In this case we can model this as a Bernoulli's process with parameter p = 4/120 = 0.03

So, variance = n*p*(1-p) = 40*0.03*0.97 = 1.164

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