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17 . A3 2s2 1 This illustrates the general result for independent samples, sU1 U2 U and (e). The proof of the general result follows from Theorem 3-7, page 78.
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s2 2, as is seen from (d) U
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5.13. The electric light bulbs of manufacturer A have a mean lifetime of 1400 hours with a standard deviation of 200 hours, while those of manufacturer B have a mean lifetime of 1200 hours with a standard deviation of 100 hours. If random samples of 125 bulbs of each brand are tested, what is the probability that the brand A bulbs will have a mean lifetime that is at least (a) 160 hours, (b) 250 hours more than the brand B bulbs
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CHAPTER 5 Sampling Theory
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# # Let XA and XB denote the mean lifetimes of samples A and B, respectively. Then mXA and
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mXA sXA
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1400 s2 B nB
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200 hours (200)2 125 20 hours
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(100)2 B 125
The standardized variable for the difference in means that Z # (XA # XB) sXA (mXA
XB XB)
# (XA
# XB) 20
and is very nearly normally distributed. (a) The difference 160 hours in standard units Required probability (160 200)>20 2. 2)
(area under normal curve to right of z 0.5000 0.4772 (250 0.9772 2.50.
(b) The difference 250 hours in standard units Required probability
200)>20
(area under normal curve to right of z 0.5000 0.4938 0.0062
2.50)
5.14. Ball bearings of a given brand weigh 0.50 oz with a standard deviation of 0.02 oz. What is the probability that two lots, of 1000 ball bearings each, will differ in weight by more than 2 oz
# # Let X1 and X2 denote the mean weights of ball bearings in the two lots. Then mX1 sX1
X2 X2
mX1 s2 2 n2
0.50 (0.02)2 1000
0 0.000895
s2 1 B n1
(0.02)2 B 1000
# # (X1 X2) 0 The standardized variable for the difference in means is Z and is very nearly normally 0.000895 distributed. A difference of 2 oz in the lots is equivalent to a difference of 2>1000 0.002 oz in the means. This can # # # # occur either if X1 X2 0.002 or X1 X2 0.002, i.e., Z Then P(Z 2.23 or Z 2.23) P(Z 2.23) P(Z 2.23) 2(0.5000 0.4871) 0.0258 0.002 0 0.000895 2.23 or Z 0.002 0 0.000895 2.23
5.15. A and B play a game of heads and tails, each tossing 50 coins. A will win the game if he tosses 5 or more heads than B, otherwise B wins. Determine the odds against A winning any particular game.
Let PA and PB denote the proportion of heads obtained by A and B. If we assume the coins are all fair, the probability p of heads is 1. Then 2 mPA and sPA
mPA s2 B P
0 pq nB 2 A1 B A1 B 2 2 B 50 0.10
2s2 A P
pq A nA
The standardized variable for the difference in proportions is Z (PA PB 0)>0.10. On a continuous-variable basis, 5 or more heads means 4.5 or more heads, so that the difference in proportions should be 4.5>50 0.09 or more, i.e., Z is greater than or equal to (0.09 0)>0.10 0.9
CHAPTER 5 Sampling Theory
(or Z 0.9). The probability of this is the area under the normal curve to the right of Z 0.9, which is 0.5000 0.3159 0.1841. Therefore, the odds against A winning are (1 0.1841) : 0.1841 0.8159 : 0.1841, or 4.43 to 1.
5.16. Two distances are measured as 27.3 inches and 15.6 inches, with standard deviations (standard errors) of 0.16 inches and 0.08 inches, respectively. Determine the mean and standard deviation of (a) the sum, (b) the difference of the distances.
If the distances are denoted by D1 and D2, then (a) mD1
27.3 s2 2 D 27.3 s2 2 D
15.6 2(0.16)2 15.6 2(0.16)2
42.9 inches (0.08)2 11.7 inches (0.08)2 0.18 inches 0.18 inches
sD1 (b) mD1
2s2 1 D mD1 mD2
2s2 1 D
5.17. A certain type of electric light bulb has a mean lifetime of 1500 hours and a standard deviation of 150 hours. Three bulbs are connected so that when one burns out, another will go on. Assuming the lifetimes are normally distributed, what is the probability that lighting will take place for (a) at least 5000 hours, (b) at most 4200 hours
Denote the lifetimes as L1, L2, and L3. Then mL1 sL1
L2 L3 L2 L3
mL1 2s2 1 L
mL2 s2 2 L (5000
mL3 s2 3 L
4500 hours
23(150)2 1.92.
260 hours
(a) 5000 hours in standard units
4500)>260
Required probability
(area under normal curve to right of z 0.5000 0.4726 0.0274 1.15.
1.92)
(b) 4200 hours in standard units
(4200
4500)>260
Required probability
(area under normal curve to left of z 0.5000 0.3749 0.1251
1.15)
Sampling distribution of variances 5.18. With reference to Problem 5.1, find (a) the mean of the sampling distribution of variances, (b) the standard deviation of the sampling distribution of variances, i.e., the standard error of variances.
(a) The sample variances corresponding to each of the 25 samples in Problem 5.1 are 0 0.25 4.00 9.00 20.25 0.25 0 2.25 6.25 16.00 4.00 2.25 0 1.00 6.25 9.00 6.25 1.00 0 2.25 20.25 16.00 6.25 2.25 0
The mean of sampling distribution of variances is mS 2 sum of all variances in the table above 25 1)(s2)>n, since for n 135 25 2 and s2 5.40 10.8 [see Problem 5.1(b)], the
This illustrates the fact that mS2 (n right-hand side is 1 (10.8) 5.4. 2
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