Practice Exam 01 SOLVED

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University of Alabama *

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Electrical Engineering

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Apr 3, 2024

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xlsx

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a. n EV SE 16 66 1.375 36 66 0.917 b. c. We cannot assume that the sampling distribution of the sample mean is normally distributed. 0 d. We can assume that the sampling distribution of the sample mean is normally distributed and the probability 0.485439 a. n ev se 16 66 1.375 36 66 0.916667 a. n ev se 16 66 1.375 36 66 0.916667 d. 0.485439 No, only the sample mean with n = 36 will have a normal distribution.
that the sample mean falls between 66 and 68 is
a. n ev se 16 66 1.375 36 66 0.917 d 0.4854 a. n ev se 16 66 1.375 36 66 0.917 d. 0.4854 a. n eV se 16 66 1.375 36 66 0.9166666667 d. 0.48543852
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a. EV -9.5 SE 0.282843 b. 0.03855 c. 0.9229 a. ev -9.5 se 0.282843 b. 0.03855 c. 0.9229 ev -9.5 se 0.282843 0.03855 c. 0.9229 b. 0.03855
a. ev -9.5 se 0.2828 b. 0.0385 c. 0.9229 a. ev -9.5 se 0.28284271 b. 0.03854994 c. 0.9229
a. n EV SE 20 0.12 0.072664 50 0.12 0.045957 b. No, only the sample proportion with n = 50 will have a normal distribution. c. 0 d. 0.1682892 a. n ev se 20 0.12 0.0727 50 0.12 0.0460 d. 0.1683 a. n ev se 20 0.12 0.072664 50 0.12 0.045957 d. 0.168289 n ev se 20 0.12 0.072664 50 0.12 0.045957 0.168289 When n = 20 the approximation condition is not satisfied because np = 20(0.12) = 2.4 < 5. When n = 50 the app For the sample with n = 20, we cannot assume that P is approximately normally distributed.
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proximation conditions are satisfied: np = 50(0.12) = 6 > 5 and n(1 − p) = 50(1 − 0.12) = 44 > 5.
a. n ev se 20 0.12 0.07266361 50 0.12 0.0459565 d. 0.16828924
a. 0.692896 b. 0.84331 a. 0.692896 b. 0.8433 a. 0.692896 0.84331 As the sample number increases, the probability of more than 20% also increases, due to the lower z value an The difference between the two probabilities is from the sample size. The standard error is reduced with large
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nd decreased standard error er sample sizes, which brings the sample proportion closer to the population proportion. Therefore, when n = 2
200 there is a greater probability that the sample proportion will be closer to the population proportion of 23%
%, so there is a greater probability that the sample proportion will be more than 20%.
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n 80 x 6.4 S 1.8 alpha 0.05 alpha/2 0.025 1-alpha/2 0.975 UB 6.794435 LB 6.005565 Yes, we can conclude with 95% confidence that the mean sleep in this Midwestern town is not 7 hours becaus n 80 x 6.4 s 1.8 alpha 0.05 aplha/2 0.025 1-aplha/2 0.975 UB 6.79 LB 6.0056 n 80 x 6.4 s 1.8 alpha 0.05 alpha/2 0.025 1-aplha/2 0.975 lb 6.005565 up 6.794435
se the value 7 does not fall within the confidence interval.
Debt x 25470.6 24040 alpha 0.05 19153 alpha/2 0.025 26762 1-alpha/2 0.975 31923 31533 LB 23921.11 34207 UB 27020.09 14623 24370 31016 20107 22090 17089 16306 20653 21673 14951 22701 23521 26215 23714 23906 23690 32254 25115 24610 22975 35402 29869 37419 22266 33848 17479 28236 30052 35136 25118 22922 28917 23634 29329
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x alpha alpha/2 1-alpha/2 lb ub
25470.6 0.05 0.025 0.975 23921.11 27020.09
n 36 n 36 x 158000 x 158000 s 36000 s 36000 alpha 0.05 alpha 0.05 alpha/2 0.025 alpha/2 0.025 1-alpha/2 0.975 1-alpha/2 0.975 t 2.030108 t 2.030108 MOE 12180.65 moe 12180.65 LB 145819.4 lb 145819.4 UB 170180.6 up 170180.6
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n 36 n 36 x 158000 x 158,000 s 36000 s 36,000 alpha 0.05 alpha 0.05 alpha/2 0.025 alpha/2 0.025 1-alpha/2 0.975 1-alpha/2 0.975 t 2.030108 t 2.030108 moe 12180.65 moe lb 145819.4 ub 170180.6 t 2.030108 moe 12180.65 t 2.030108 moe 12180.65
Stock Price x 75.66667 stock price 71 s 3.559026 71 73 n 6 73 76 alpha 0.1 76 78 alpha/2 0.05 78 81 1-alpha/2 0.95 81 75 75 t 2.015048 MOE 2.927797 UB 78.59446 LB 72.73887
x 75.66667 stock price x 75.66667 s 3.559026 71 s 3.559026 n 6 73 alpha 0.1 alpha 0.1 76 alpha/2 0.05 alpha/2 0.05 78 1-alpha/2 0.95 1-alpha/2 0.95 81 n 6 75 t 2.015048 t 2.015048 moe 2.927797 moe 2.927797 lb 72.73887 lb 72.73887 ub 78.59446 UB 78.59446
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n 80 n 80 p 0.625 prob of success p 0.625 prob of success 1-p 0.375 prob. Of failure 1-p 0.375 prob of failure s 0.054127 s 0.054127 alpha 0.05 alpha 0.05 alpha/2 0.025 alpha/2 0.025 1-alpha/2 0.975 1-alpha/2 0.975 LB 0.518914 Lb 0.518914 UB 0.731086 up 0.731086 LB 0.268914 LB 0.268914 UB 0.481086 UB 0.481086 n 80 p 0.625 prob of success 1-p 0.375 prob of failure alpha 0.05 alpha/2 0.025 1-alpha/2 0.975 s 0.054127 lb 0.518914 ub 0.731086 lb 0.268914 up 0.481086
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RIGHT TAIL Pop mean 12.6 n 25 x 13.4 S 3.2 a-1 z test stat 1.25 p-value 0.10565 p-value ≥ 0.10 a-2 a-3 We cannot conclude that the population mean is greater than 12.6 b-1 n 100 z test stat 2.5 p-value 0.00621 p-value < 0.01 b-2 b-3 We conclude that the population mean is greater than 12.6. Do not reject H0 since the p-value is greater than α Reject H0 since the p-value is smaller than α.
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TWO-TAIL TEST pop mean 120 S 46 x 132 108 n 50 se 6.505382 a1 test stat 1.844626 sig. level 0.05 a2 p-value 0.065092 DO NOT REJECT Ho bec. P-value is higher than sig. level. a3 We cannot conclude that the population mean differs from 120. b1 test stat -1.844626 sig. level 0.1 b2 p-value 0.065092 REJECT Ho bec. P-value is lower than sig. level. b3 We conclude that the population mean differs from 120.
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LEFT TAIL 95 pop mean 100 99 x 92.33333 85 n 6 80 s 7.890923 98 t test stat -2.379876 97 p-value 0.03159 Conclusion Interpret We cannot conclude that the population mean is less than 100 Use T.DIST function to find p-value because there is NO population standard deviation known to us. We would use 1-T.DIST function if we had a positive t test statistic (e.g. +2.38). 0.025 ≤ p-value < 0.05 Do not reject H0 since the p-value is greater than the significance level
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TWO-TAIL TEST pop mean 1.2 s 0.06 x 1.22 n 36 se 0.01 b1 test stat 2 b2 p-value 0.053308 sig. level 0.05 c1 c2 We canNOT conclude that the machine is working improperly. Do not reject Ho since the p-value is greater than significance level.
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Firm PE_Ratio a. Ha: Mu is not = to 15 1 14 Ho: Mu is = to 15 2 24 3 13 b1. 1.065593 4 49 5 15 b2. 0.2954 6 12 7 11 c. 8 7 p-value is higher than 0.05 significance level. 9 12 10 20 11 13 12 10 13 15 14 8 15 19 16 11 17 15 18 17 19 8 20 19 21 17 22 17 23 11 24 21 25 20 26 18 27 13 28 44 29 14 30 16 Do not reject H0; there is insufficient evidence to state that the m
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mean P/E ratio differs from 15
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Lottery Expenditure LEFT TAIL 790 pop mean 860.7 594 n 100 899 x 841.94 1105 s 217.1525 1090 t test stat -0.863909 1197 p-value 0.194863 p-value > 0.10 413 803 1069 633 712 512 481 654 695 426 736 769 877 777 785 776 1119 833 813 747 1244 1023 1325 719 1182 528 958 1030 1234 833 745 985 774 1002 561 681 546 777 Do not reject H0; there is insufficient evidence to state that the average Massachusetts
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844 856 785 1289 502 703 334 1140 594 719 1002 943 1025 969 576 627 989 915 662 802 876 962 878 668 1227 947 864 1016 1022 723 665 1072 610 538 992 978 1291 1139 1111 873 850 941 845 639 495
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1016 939 974 893 645 1098 788 682 686 764 759
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resident spent less than $860.70 on the lottery in 2010
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