¼ ¼ 6:0. sum of all sample means in ð8þ 25

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1 1. Samling Distribution of means. A oulation consists of the five numbers 2, 3, 6, 8, and 11. Consider all ossible samles of size 2 that can be drawn with relacement from this oulation. Find the mean of the oulation, the standard deviation of the oulation, (c) the mean of the samling distribution of means, and (d) the standard deviation of the samling distribution of means (i.e., the standard error of means). 2 þ 3 þ 6 þ 8 þ :0 2 ð2 6Þ2 þð3 6Þ 2 þð6 6Þ 2 þð8 6Þ 2 þð11 6Þ 2 and 3:29. (c) 16 þ 9 þ 0 þ 4 þ 2 10:8 There are ðþ 2 samles of size 2 that can be drawn with relacement (since any one of the five numbers on the first draw can be associated with any one of the five numbers on the second draw). These are The corresonding samle means are (2, 2) (2, 3) (2, 6) (2, 8) (2, 11) (3, 2) (3, 3) (3, 6) (3, 8) (3, 11) (6, 2) (6, 3) (6, 6) (6, 8) (6, 11) (8, 2) (8, 3) (8, 6) (8, 8) (8, 11) (11, 2) (11, 3) (11, 6) (11, 8) (11, 11) (8) and the mean of samling distribution of means is X illustrating the fact that X. sum of all samle means in ð8þ :0 (d ) The variance 2 X of the samling distribution of means is obtained by subtracting the mean 6 from each number in (8), squaring the result, adding all 2 numbers thus obtained, and dividing by 2. The final result is 2 X 13=2 :40, and thus X ffiffiffiffiffiffiffiffiffi :40 2:32. This illustrates the fact that for finite oulations involving samling with relacement (or infinite oulations), 2 X 2 = since the right-hand side is 10:8=2 :40, agreeing with the above value. 2. Assume that the heights of 3000 male students at a university are normally distributed with mean 68.0 inches (in) and standard deviation 3.0 in. If 80 samles consisting of 2 students each are obtained, what would be the exected mean and standard deviation of the resulting samling distribution of means if the samling were done with relacement and without relacement?

2 The numbers of samles of size 2 that could be obtained theoretically from a grou of 3000 students with and without relacement are ð3000þ 2 and ð Þ, which are much larger than 80. Hence we do not get a true samling distribution of means, but only an exerimental samling distribution. evertheless, since the number of samles is large, there should be close agreement between the two samling distributions. Hence the exected mean and standard deviation would be close to those of the theoretical distribution. Thus we have: X 68:0 in and X ffiffiffiffi 3 ffiffiffiffiffi in 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi X 68:0 in and X ffiffiffiffi 3 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi which is only very slightly less than 0.6 in and can therefore, for all ractical uroses, be considered the same as in samling with relacement. Thus we would exect the exerimental samling distribution of means to be aroximately normally distributed with mean 68.0 in and standard deviation 0.6 in. 3. For roblem (2), how many samles would you exect to find the mean between 66.8 and 68.3? Less than 66.4 in? The mean X of a samle in standard units is here given by z X X X 66:8 in standard units 68:3 in standard units X 68:0 66:8 68:0 68:3 68:0 2:0 0: Fig. 8-2 Areas under the standard normal curve. Standard normal curve showing the area between z 2 and z 0.; Standard normal curve showing the area to the left of z 2:67.

3 As shown in Fig. 8-2, Proortion of samles with means between 66:8 and 68:3 in ðarea under normal curve between z 2:0 and z 0:Þ ðarea between z 2 and z 0Þþðarea between z 0 and z 0:Þ 0:4772 þ 0: Thus the exected number of samles is ð80þð687þ 3:496, or 3. 66:4 in standard units As shown in Fig. 8.2, 66:4 68:0 2:67 Proortion of samles with means less than 66:4 in ðarea under normal curve to left of z 2:67Þ ðarea to left of z 0Þ ðarea between z 2:67 and z 0Þ 0: 0:4962 0:0038 Thus the exected number of samles is ð80þð0:0038þ 0:304; or zero. 4. Find the robability that in 120 tosses of a fair coin less than 40% or more than 60% will be heads and /8 or more will be heads. P 1 2 0:0 sffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffi q ð 1 2 P Þð1 2 Þ 0: % in standard units 60% in standard units 0:40 0:0 0: :0 0:046 2:19 2:19 Required robability ðarea to the left of 2:19 lus area to the right of 2.19) ð2ð0:0143þ 0:0286Þ Although this result is accurate to two significant figures, it does not agree exactly since we have not used the fact that the roortion is actually a discrete variable. To account for this, we subtract 1=2 1=2ð120Þ from 0.40 and add 1=2 1=2ð120Þ to 0.60; thus, since 1=240 0:00417, the required roortions in standard units are 0:40 0: :0 0:046 2:28 and 0 þ 0: :0 0:046 2:28 so that agreement with the first method is obtained. ote that ð0:40 0:00417Þ and ð0 þ 0:00417Þ corresond to the roortions 47:=120 and 72:=120 in the first method.

4 ð20 0:00417Þ in standard units 20 0: :0 0:046 2:6 Required robability ðarea under normal curve to right of z 2:6Þ ðarea to right of z 0Þ ðarea between z 0 and z 2:6Þ 0: 0:4960 0:0040 Estimation Theory. In a samle of five measurements, the diameter of a shere was recorded by a scientist as 6.33, 6.37, 6.36, 6.32, and 6.37 centimeters (cm). Determine unbiased and efficient estimates of the true mean and the true variance. The unbiased and efficient estimate of the true mean (i.e., the oulations mean) is P X X 6:33 þ 6:37 þ 6:36 þ 6:32 þ 6:37 6:3 cm The unbiased and efficient estimate of the true variance (i.e., the oulation variance) is ^s 2 P ðx ^XÞ 2 1 s2 1 ð6:33 6:3Þ2 þð6:37 6:3Þ 2 þð6:36 6:3Þ 2 þð6:32 6:3Þ 2 þð6:37 6:3Þ 2 1 0:000 cm 2 ote that although ^s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:000 0:023 cm is an estimate of the true standard deviation, this estimate is neither unbiased nor efficient. 6. In measuring reaction time, a sychologist estimates that the standard deviation is 0.0 seconds (s). How large a samle of measurements must he take in order to be 9% and 99% confident that the error of his estimate will not exceed 0.01 s? The 9% confidence limits are X 1:96= ffiffiffiffi ffiffiffiffi, the error of the estimate being 1:96= ffiffiffiffi. Taking ffiffiffiffi s 0:0 s, we see that this error will be equal to 0.01 s if (1.96)(0.0)/ 0:01; that is, ð1:96þð0:0þ=0:01 9:8, or 96:04. Thus we can be 9% confident that the error of the estimate will be less than 0.01 s if is 97 or larger.

5 Another method ð1:96þð0:0þ 0:01 if ð1:96þð0:0þ 1 0:01 or ð1:96þð0:0þ 9:8 0:01 Then 96:04, or 97. The 99% confidence limits are X 2:8= ffiffiffiffi. Then (2.8)(0.0)/ 0:01, or 166:4. Thus we can be 99% confident that the error of the estimate will be less than 0.01 s only if is 167 or larger. 7. A random samle of 0 mathematics grades out of a total of 200 showed a mean of 7 and a standard deviation of 10. What are the 9% confidence limits for estimates of the mean of the 200 grades? With what degree of confidence could we say that the mean of all 200 grades is 7 +/- 1? Since the oulation size is not very large comared with the samle size, we must adjust for it. Then the 9% confidence limits are X 1:96 X X 1:96 ffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 7 1:96 10 ffiffiffiffiffi 0 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi : The confidence limits can be reresented by X z c X X z c ffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 7 z 1 c 10 ffiffiffiffiffi 0 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi :23z c Since this must equal 7 1, we have 1:23z c 1, or z c 0:81. The area under the normal curve from z 0 to z 0:81 is ; hence the required degree of confidence is 2ð0:2910Þ 0:82, or 8.2%. 8. The voltages of 0 batteries of the same tye have a mean of 18.2 volts (V) and a standard deviation of 0. V. Find the robable error of the mean and the 0% confidence limits. Probable error of the mean 74 X 74 ffiffiffiffi 74 ^s ffiffiffiffi s 74 ffiffiffiffiffiffiffiffi 74 0: ffiffiffiffiffi 0:048 V 1 49 ote that if the standard deviation of 0: V is comuted as ^s, the robable error is 74ð0:= ffiffiffiffiffi 0 Þ0:048 also, so that either estimate can be used if is large enough. The 0% confidence limits are 18 0:048 V.

6 Hyothesis Testing 9. Find the robability of getting between 40 and 60 heads inclusive in 100 tosses of a fair coin. To test the hyothesis that a coin is fair, adot the following decision rule: Accet the hyothesis if the number of heads in a single samle of 100 tosses is between 40 and 60 inclusive. Reject the hyothesis otherwise. Find the robability of rejecting the hyothesis when it is actually correct. (c) Grah the decision rule and the result of art (d) What conclusions would you draw if the samle of 100 tosses yielded 3 heads? And if it yielded 60 heads? (e) Could you be wrong in your conclusions about art (d)? Exlain. According to the binomial distribution, the required robability is þ þþ Since 100ð 1 2 Þ and q 100ð1 2Þ are both greater than, the normal aroximation to the binomial distribution can be used in evaluating this sum. The mean and standard deviation of the number of heads in 100 tosses are given by 100ð 1 ffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2Þ0 and q ð100þð 1 2 Þð1 2 Þ On a continuous scale, between 40 and 60 heads inclusive is the same as between 39. and 60. heads. We thus have 39: in standard units 39: 0 2:10 60: in standard units 60: 0 Required robability area under normal curve between z 2:10 and z 2:10 2ðarea between z 0 and z 2:10Þ 2ð0:4821Þ 0:9642 2:10 The robability of not getting between 40 and 60 heads inclusive if the coin is fair is = Thus the robability of rejecting the hyothesis when it is correct is (c) The following figure shows the robability distribution of heads in 100 tosses of a fair coin. If a single samle of 100 tosses yields a z score between and 2.10, we accet the hyothesis; others, we reject the hyothesis and decide that the coin is not fair. The error made in rejecting the hyothesis when it should be acceted is the Tye 1 error of the decision rule; and the robability of making this error, equal to from art, is reresented by the total shaded area of the figure. If a single samle of 100 tosses yields a number of heads whose z score (or z statistic) lies in the shaded regions, we would say that

7 this z score differed significantly from what would be exected if the hyothesis were true. For this reason, the total shaded area (i.e., the robability of a Tye I error) is called the significance level of the decision rule and equals in this case. Thus we seak of rejecting the hyothesis at the (or 3.8%) significance level. Rejection region Accetance region Rejection region z = 2.10 z = 2.10 (d) According to the decision rule, we would have to accet the hyothesis that the coin is fair in both cases. One might argue that if only one more head had been obtained, we would have rejected the hyothesis. This is what one must face when any shar line of division is used in making decisions. (e) Yes. We could accet the hyothesis when it actually should be rejected as would be the case, for examle, when the robability of heads is really 0.7 instead of 0.. The error made in acceting the hyothesis when it should be rejected is the Tye II error of the decision.