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Biostatistics

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148 CHAPTER 5 SOME IMPORTANT SAMPLING DISTRIBUTIONS<br />

difference between population means, the probability of obtaining a difference between<br />

sample means as large as or larger than 13 is .0375.<br />

Sampling from Normal Populations The procedure we have just followed<br />

is valid even when the sample sizes, n 1 and n 2 , are different and when the population<br />

variances, s 2 1 and s2 2 have different values. The theoretical results on which this procedure<br />

is based may be summarized as follows.<br />

Given two normally distributed populations with means m 1 and m 2 and variances s 2 1<br />

and s 2 2 , respectively, the sampling distribution of the difference, x 1 x 2 , between the<br />

means of independent samples of size n 1 and n 2 drawn from these populations is<br />

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi<br />

normally distributed with mean m 1 m 2 and variance s 2 1 =n <br />

1 þ s<br />

2<br />

2 =n 2 .<br />

Sampling from Nonnormal Populations Many times a researcher is<br />

faced with one or the other of the following problems: the necessity of (1) sampling from<br />

nonnormally distributed populations, or (2) sampling from populations whose functional<br />

forms are not known. A solution to these problems is to take large samples, since when the<br />

sample sizes are large the central limit theorem applies and the distribution of the<br />

difference between two sample means is at least approximately normally distributed<br />

with a mean equal to m 1 m 2 and a variance of s 2 1 =n <br />

1<br />

þ s<br />

2<br />

2 =n 2 . To find probabilities<br />

associated with specific values of the statistic, then, our procedure would be the same as<br />

that given when sampling is from normally distributed populations.<br />

EXAMPLE 5.4.2<br />

Suppose it has been established that for a certain type of client the average length of a home<br />

visit by a public health nurse is 45 minutes with a standard deviation of 15 minutes, and that<br />

for a second type of client the average home visit is 30 minutes long with a standard<br />

deviation of 20 minutes. If a nurse randomly visits 35 clients from the first and 40 from the<br />

second population, what is the probability that the average length of home visit will differ<br />

between the two groups by 20 or more minutes?<br />

Solution:<br />

No mention is made of the functional form of the two populations, so let us<br />

assume that this characteristic is unknown, or that the populations are not<br />

normally distributed. Since the sample sizes are large (greater than 30) in<br />

both cases, we draw on the results of the central limit theorem to answer the<br />

question posed. We know that the difference between sample means is at<br />

least approximately normally distributed with the following mean and<br />

variance:<br />

m x1 x 2<br />

¼ m 1 m 2 ¼ 45 30 ¼ 15<br />

sx 2 1 x 2<br />

¼ s2 1<br />

þ s2 2<br />

¼ ð15Þ2<br />

n 1 n 2 35 þ ð20Þ2<br />

40 ¼ 16:4286

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