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m. For a general proof of this, see Problem 5.6.
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2 (d) The variance sX of the sampling distribution of means is obtained by subtracting the mean 6 from each number in (1), squaring the result, adding all 25 numbers obtained, and dividing by 25. The final result is
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2 sX
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5.40 so that sX
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This illustrates the fact that for finite populations involving sampling with replacement (or infinite 2 s2 >n since the right-hand side is 10.8>2 5.40, agreeing with the above value. For a populations), sX general proof of this, see Problem 5.7.
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5.2. Solve Problem 5.1 in case sampling is without replacement.
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As in (a) and (b) of Problem 5.1, m
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10.8, s
(c) There are 5C2 10 samples of size two which can be drawn without replacement (this means that we draw one number and then another number different from the first) from the population, namely, (2, 3), (2, 6), (2, 8), (2, 11), (3, 6), (3, 8), (3, 11), (6, 8), (6, 11), (8, 11).
The selection (2, 3), for example, is considered the same as (3, 2).
CHAPTER 5 Sampling Theory
The corresponding sample means are 2.5, 4.0, 5.0, 6.5, 4.5, 5.5, 7.0, 7.0, 8.5, 9.5
and the mean of sampling distribution of means is mX 2.5 4.0 5.0 m. 6.5 4.5 10 5.5 7.0 7.0 8.5 9.5 6.0
illustrating the fact that mX
(d) The variance of the sampling distribution of means is
2 sX
(2.5
6.0)2
(4.0
6.0)2
(5.0 10
6.0)2
(9.5
6.0)2
10.8 5 s2 N n n a N 1 b, since the right side equals 2 a 5 For a general proof of this result, see Problem 5.47.
2 This illustrates sX
and sX
2 b 1
4.05, as obtained above.
5.3. Assume that the heights of 3000 male students at a university are normally distributed with mean 68.0 inches and standard deviation 3.0 inches. If 80 samples consisting of 25 students each are obtained, what would be the mean and standard deviation of the resulting sample of means if sampling were done (a) with replacement, (b) without replacement
The numbers of samples of size 25 that could be obtained theoretically from a group of 3000 students with and without replacement are (3000)25 and 3000C25, which are much larger than 80. Hence, we do not get a true sampling distribution of means but only an experimental sampling distribution. Nevertheless, since the number of samples is large, there should be close agreement between the two sampling distributions. Hence, the mean and standard deviation of the 80 sample means would be close to those of the theoretical distribution. Therefore, we have (a) mX m 68.0 inches and sX s !n 3 !25 n 1 0.6 inches
68.0 inches and
s N !n A N
3 3000 !25 A 3000
25 1
which is only very slightly less than 0.6 inches and can for all practical purposes be considered the same as in sampling with replacement. Thus we would expect the experimental sampling distribution of means to be approximately normally distributed with mean 68.0 inches and standard deviation 0.6 inches.
5.4. In how many samples of Problem 5.3 would you expect to find the mean (a) between 66.8 and 68.3 inches, (b) less than 66.4 inches
# The mean X of a sample in standard units is here given by Z (a) 66.8 in standard units 68.3 in standard units (66.8 (68.3 68.0)>0.6 68.0)>0.6 2.0 0.5 # X mX sX # X 68.0 . 0.6
Proportion of samples with means between 66.8 and 68.3 inches (area under normal curve between z (area between z (area between z 0.4772 0.1915 2 and z 0 and z 0.6687 (80) (0.6687) or 53 (Fig. 5-2). 0) 0.5) 2.0 and z 0.5)
Then the expected number of samples
CHAPTER 5 Sampling Theory
Fig. 5-2
(b) 66.4 in standard units
(66.4
68.0)>0.6
Proportion of samples with means less than 66.4 inches (area under normal curve to left of z (area to left of z (area between z 0.5 0.4962 0) 2.67 and z 0) 0.0038 0.304 or zero (Fig. 5-3). 2.67)
Then the expected number of samples
(80)(0.0038)
Fig. 5-3
5.5. Five hundred ball bearings have a mean weight of 5.02 oz and a standard deviation of 0.30 oz. Find the probability that a random sample of 100 ball bearings chosen from this group will have a combined weight, (a) between 496 and 500 oz, (b) more than 510 oz.
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