Final Exam Review (Math 1342)

Final Exam Review (Math 1342) 1) 5.5 5.7 5.8 5.9 6.1 6.1 6.3 6.4 6.5 6.6 6.7 6.7 6.7 6.9 7.0 7.0 7.0 7.1 7.2 7.2 7.4 7.5 7.7 7.7 7.8 8.0 8.1 8.1 8.3 8.7 Min = 5.5 Q 1 = 25th percentile = middle of first half of data set = 8th value = 6.4 Q 2 = Median = 50th percentile = middle of entire data set = (15th value + 16th value)/2 = 7.0 Q 3 = 75th percentile = middle of second half of data set = 23rd value = 7.7 Max = 8.7 5.5 6.4 7.0 7.7 8.7 2) 25 35 43 44 47 48 54 55 56 57 59 62 63 65 66 68 69 69 71 71 73 73 74 76 77 77 78 79 80 81 81 82 83 85 89 92 93 94 97 98 Min = 25 Q 1 = 25th percentile = middle of first half of data set = (10th value + 11th value)/2 = 58 Q 2 = Median = 50th percentile = middle of entire data set = (20th value + 21st value)/2 = 72 Q 3 = 75th percentile = middle of second half of data set = (30th value + 31st value)/2 = 81 Max = 98 25 58 72 81 98

3) 100 150 x - 900-1100 For x 900, its z - score = = - 1.33 150 Go to Table II, look for z = -1.33; corresponding area is 0.0918 Area = 0.0918-1.33 0 Thus, percentage of trainees earning less than $900 month 9.18% 4) 32.3 1.2 x - 32-32.3 For x 32, its z - score = = - 0.25 1.2 Go to Table II, look for z = -0.25; corresponding area is 0.4013 Area = 0.4013-0.25 0 Thus, the probability that the volume of soda in a randomly selected bottle will be less than 32 oz is 0.4013

5) 200 50 x - 200-200 For x 200, its z - score = = 0 50 x - 275-200 For x 275, its z - score = = 1.5 50 Go to Table II, look for z = 0; corresponding area is 0.5 Go to Table II, look for z = 1.5; corresponding area is 0.9332 Area = 0.9332-0.5 = 0.4332 0 1.5 Thus, the probability of a rating that is between 200 and 275 is 0.4332 6) 200 50 x - 170-200 For x 170, its z - score = = -0.6 50 x - 220-200 For x 220, its z - score = = 0.4 50 Go to Table II, look for z = -0.6; corresponding area is 0.2743 Go to Table II, look for z = 0.4; corresponding area is 0.6554 Area = 0.6554-0.2743 = 0.3811-0.6 0 0.4 Thus, probability of a rating that is between 170 and 220 is 0.3811

7) 268 15 x - 300-268 For x 300, its z - score = = 2.13 15 Go to Table II, look for z = 2.13; corresponding area is 0.9834 Area = 1-0.9834 = 0.0166 0 2.13 Thus, the probability that a pregnancy lasts at least 300 days is 0.0166. 8) 5.67 0.70 x - 5.48-5.67 For x 5.48, its z - score = = -2.714 0.07 x - 5.82-5.67 For x 5.82, its z - score = = 2.142 0.07 Go to Table II, look for z = -2.71; corresponding area is 0.0034 Go to Table II, look for z = 2.14; corresponding area is 0.9838 Area = 0.0034 Area = 1-0.9838 = 0.0162-2.714 0 2.142 Thus, percentage of legal quarters will be rejected is 0.0034 + 0.0162 = 0.0196

9) Since is known, we use the formula x z/ 2, x z/ 2 n x 4.153 0.7 n15 n 0.0456 =1-95.44% 0.0456 / 2 0.0228 2 Use Table II and look inside table for area of 0.0228; corresponding z-score is -2. Area = 0.228 Area = 0.228-2 0 2 Hence /2 2 z x z/ 2, x z/ 2 n n 0.7 0.7 4.153 2, 4.153 2 = 3.79, 4.51 15 15

10) Since is known, we use the formula x z/ 2, x z/ 2 n x 526 29 n106 0.10 =1-90% 0.10 / 2 0.05 2 n Use Table II and look inside table for area of 0.0228; corresponding z-score is -1.64. Area = 0.05 Area = 0.05-1.64 0 1.64 Hence z /2 1.64 x z/ 2, x z/ 2 n n 29 29 526 1.64, 526 +1.64 = 521.4, 530.6 106 106

11) Since is known, we use the formula x z/ 2, x z/ 2 n x 20 3.9 n131 0.05 =1-95% 0.05 / 2 0.025 2 n Use Table II and look inside table for area of 0.025; corresponding z-score is -1.96. Area = 0.025 Area = 0.025-1.96 0 1.96 Hence z /2 1.96 x z/ 2, x z/ 2 n n 3.9 3.9 20 1.96, 20 +1.96 = 19.3, 20.7 131 131

12) s s Since is unknown, we use the formula x t/ 2, x t/ 2 n n x 243 s 16.2 n12 df = 12-1 = 11 0.05 =1-95% 0.05 / 2 0.025 2 Go to Table IV; look under t with df = 11; corresponding t-score is 2.201. Thus, 2.201 t /2 0.025 Area = 0.025 Area = 0.025-2.201 0 2.201 x z/ 2, x z/ 2 n n 16.2 16.2 243 2.201, 243 +2.201 = 232.7, 253.3 12 12

13) s s Since is unknown, we use the formula x t/ 2, x t/ 2 n n x 83 s 14.1 n 30 df = 30-1 = 29 0.01 =1-99% 0.01 / 2 0.005 2 Go to Table IV; look under t with df = 29; corresponding t-score is 2.756. Thus, 2.756 t /2 0.005 Area = 0.025 Area = 0.025-2.756 0 2.756 x z/ 2, x z/ 2 n n 14.1 14.1 83 2.756, 83 2.756 = 75.91, 90.09 30 30

14) s s Since is unknown, we use the formula x t/ 2, x t/ 2 n n x 76.2 s 21.4 n 27 df = 27-1 = 26 0.05 =1-95% 0.05 / 2 0.025 2 Go to Table IV; look under t with df = 26; corresponding t-score is 2.055. Thus, 2.055 t /2 0.025 Area = 0.025 Area = 0.025-2.055 0 2.055 x z/ 2, x z/ 2 n n 21.4 21.4 76.2 2.055, 76.2 +2.055 = 67.7, 84.66 27 27 15) See Answer Sheet 16) See Answer Sheet 17) See Answer Sheet

18) a) H : 35 H : 35 (Two-tailed test) o a b) = 0.01; since this is a two-tailed test, we need to divide by 2; hence /2 = 0.005 x - c) Since is unknown, we use the formula t to compute test statistic. s n x - 41-35 t 7.252 s 3.7 n 20 d) For two-tailed test: Go to Table IV; look for t 0.005 with df = n - 1 = 20-1 = 19. Critical values are is ±2.861 Area = 0.005 (Rejection Region) Area = 0.005 (Rejection Region) -2.861 0 2.861 7.52 e) Since Test Statistic value of 7.52 is inside of rejection region, we reject H o : 35 f) At the 1% level of significance, there is sufficient evidence to conclude that the mean score for sober women differs from 35.0, the mean score for men.

19) a) H : 160 H : 160 (Right-tailed test) o a b) = 0.05; since this is a right-tailed test, we do not need to divide by 2; x - c) Since is unknown, we use the formula t to compute test statistic. s n x - 183-160 t 9.583 s 12 n 25 d) For right-tailed test: Go to Table IV; look for t 0.05 with df = n - 1 = 25-1 = 24. Critical value is 1.711 Area = 0.05 (Rejection Region) 0 1.711 9.583 e) Since Test Statistic value of 9.583 is inside of rejection region, we reject H o : 160 f) At the 5% level of significance, there is sufficient evidence to conclude that the mean score for students from this university is greater than 160.

20) a) H : 18.7 H : 18.7 (Two-tailed test) o a b) = 0.05; since this is a two-tailed test, we need to divide by 2; hence /2 = 0.025 x - c) Since is unknown, we use the formula t to compute test statistic. s n x - 20.7-18.7 t 0.86 s 7.7 n 11 d) For two-tailed test: Go to Table IV; look for t 0.025 with df = n - 1 = 11-1 = 10. Thus, critical values are is ±2.228 Area = 0.025 (Rejection Region) Area = 0.025 (Rejection Region) -2.228 0 0.86 2.228 e) Since Test Statistic value of 0.86 is inside of non-rejection region, we do not reject H o : 18.7 f) At the 5% level of significance, there is not sufficient evidence to conclude that the mean amount of time served by convicted burglars in her hometown is different from 18.7 months.

21) We use the formula - 1 1 n n x1 x2 t / 2 S p 0.05 =1-95% 0.05 / 2 0.025 2 1 2 Go to Table IV; look for t 0.025 with df = n 1 n 2 114 17 1 30. Hence, critical values are is ±2.042 Thus, t /2 = 2.042 Area = 0.025 Area = 0.025-2.042 0 2.042 1 1 n 2 2 1 s1 n2 s2 S p = = 4.055477352 n n 2 1 2 1 1 x1 - x2 t / 2 S p = (-8.99, -3.01) n n 1 2

22) We use the formula - 1 1 n n x1 x2 t / 2 S p 0.005 =1-99% 0.01 / 2 0.005 2 1 2 Go to Table IV; look for t 0.005 with df = n 1 n 2 111 9 1 19. Hence, critical values are is ±2.861 Thus, t /2 = 2.861 Area = 0.005 Area = 0.005-2.861 0 2.861 S 2 2 n 1 s n 1 s 103.6 83.1 2 2 1 1 2 2 p = = = 3.386902879 n1n2 2 18 1 1 x1 - x2 t / 2 S p = (-1.38, 7.38) n n 1 2

23) From the frequency table, we know that there are 40 data values (5+9+10+8+8). Since we do not know the exact values of these 40 data values, we have to approximate them. We approximate them by using the midpoint value of each class. For the first class of 50-60, the midpoint value is 55. We know that there are 5 data values between 50 and 60. Since we don't know exactly what they are, these five values will be approximated by 55, 55,55,55,55 For the second class of 60-70,, the midpoint value is 65. We know that there are 9 data values between 60 and 70. Since we don't know exactly what they are, these nine values will be approximated by 65, 65,65,65,65, 65, 65,65,65 For the third class of 70-80,, the midpoint value is 75. We know that there are 10 data values between 70 and 80. Since we don't know exactly what they are, these ten values will be approximated by 75, 75,75,75,75,75, 75,75,75,75 For the fourth class of 80-90, the midpoint value is 85. We know that there are 8 data values between 80 and 90. Since we don't know exactly what they are, these eight values will be approximated by 85, 85,85,85,85,85, 85,85 For the fifth of class of 90-100, the midpoint value is 95. We know that there are 8 data values between 90 and 100. Since we don't know exactly what they are, these eight values will be approximated by 95, 95,95,95,95,95, 95,95 Thus, the 40 data values (approximated) are: 55, 55,55,55,55, 65, 65,65,65,65, 65, 65,65,65 75, 75,75,75,75,75, 75,75,75,75, 85, 85,85,85,85,85, 85,85, 95, 95,95,95,95,95, 95,95 The sample mean = x = 76.25. 2 xi x s= n 1 = 40-1 55-76.25 55-76.25 55-76.25 95-76.25 2 2 2 2 = 13.2 For more information, see page 119 (in textbook) for more information. Standard deviation for this sample is 13.2.

24) Waiting time(minutes) Number of customer Approximated data values 0-4 14 2,2,2,2,2, 2,2,2,2,2, 2,2,2,2 4-8 11 6,6,6,6, 6,6,6,6,6,6,6 8-12 7 10,10,10,10,10,10,10 12-16 16 14,14,14,14, 14,14,14,14, 14,14,14,14, 14,14,14,14 16-20 0 20-24 2 22,22 The data values (approximated) are: 2,2,2,2,2, 2,2,2,2,2, 2,2,2,2, 6,6,6,6, 6,6,6,6,6,6,6, 10,10,10,10,10,10,10, 14,14,14,14, 14,14,14,14, 14,14,14,14, 14,14,14,14, 22,22 Standard deviation for this sample is 5.6.

25) a) H : H : (Two-tailed test) o 1 2 a 1 2 b) = 0.01; since this is a two-tailed test, we need to divide by 2; /2 = 0.005 d c) We use the formula t to compute test statistic. Sd n Before After diff = Before - after 70 74-4 73 82-9 76 74 2 80 83-3 77 89-12 d mean of differences = -5.2; S d standard dev. of differences = = x i x n 1 2 2 2 2 2 2-4 - (-5.2) -9 - (-5.2) 2 - (-5.2) -3 - (-5.2) -12 - (-5.2) = 5.449770637 5-1 d 5.2 t = = -2.134 S d 5.449 n 5 d) For two-tailed test: Go to Table IV; look for t 0.005 with df = n 1 5-1 = 4. Critical values are ±4.604.

Area = 0.005 (Rejection Region) Area = 0.005 (Rejection Region) -2.228-2.134 0 2.228 e) Since Test Statistic value of -2.134 is inside of non-rejection region, we do not reject H o : 1 2 f) At the 1% significance level, the data do not provide sufficient evidence to conclude that the mean score before tutoring differs from the mean score after tutoring. 26) a) H : H : > (Right-tailed test) o 1 2 a 1 2 b) = 0.05; since this is a right-tailed test, we do not need to divide by 2; d c) We use the formula t to compute test statistic. Sd n Before After diff = Before - after 118.4 119-0.6 111.8 110.5 1.3 108.8 106.4 2.4 115.3 116.1-0.8 112.8 111 1.8 113.8 113.9-0.1 114.9 111.3 3.6 110.5 106.6 3.9 d mean of differences = 1.4375; S d S standard dev. of differences d = = x x i n 1 2 2 2 2 2 2 2 2 = 1.8259 2 1.4375 - (-0.6) 1.4375 - (1.3) 1.4375 - (2.4) 1.4375 - (-0.8) 1.4375 - (1.8) 1.4375 - (-0.1) 1.4375 - (3.6) 1.4375 - ( 3.9) d 1.4375 t = = 2.227 Sd 1.8259 n 8 d) For two-tailed test: Go to Table IV; look for t 0.05 with df = n 1 8-1 = 7. Critical value is 1.895. 8-1

Area = 0.05 (Rejection Region) 0 1.895 2.227 e) Since Test Statistic value of 2.227 is inside of rejection region, we reject H o : 1 2 f) At the 5% significance level, the data provide sufficient evidence to conclude that the training helps to improve times for the 800 meters. 27) = 1-90% = 0.10; and /2 = 0.05 We use the formula d S d t/2 n d mean of differences = 0.7; S standard dev. of differences = 1.12 d Go to Table IV; look for t 0.05 with df = n 1 5-1 = 4. Critical values are ±2.13. Thus t /2 = 2.13 Area = 0.05 Area = 0.05-2.13 0 2.13 S d 1.12 d t/2 0.7 2.13 = (-0.37, 1.7668) n 5 28) = 1-99% = 0.01; and /2 = 0.005 We use the formula d S d t/2 n d mean of differences = 1.22; S standard dev. of differences = 1.826 d

Go to Table IV; look for t 0.05 with df = n 1 9-1 = 8. Critical values are ±3.355. Thus t /2 = 3.355 Area = 0.005 Area = 0.005-3.355 0 3.355 S d 1.826 d t/2 1.22 3.355 = (-0.82, 3.26) n 9 29) pq ˆˆ We use the formula pˆ z /2 n n 770 pˆ 0.067 qˆ 1 0.067 0.933 1.95 0.05 / 2 0.025 Use Table II and look inside table for area of 0.025; corresponding z-score is -1.96 Hence z /2 1.96 Area = 0.025 Area = 0.025-1.96 0 1.96

pq ˆˆ pˆ z /2 = n 0.0670.933 0.067 1.96 0.0493, 0.0846 770 30) We use the formula n 713 pˆ 0.22 qˆ 1 0.22 0.78 1.90 0.10 / 2 0.05 pq ˆˆ pˆ z /2 n Use Table II and look inside table for area of 0.05; corresponding z-score is -1.645 Hence z /2 1.645 Area = 0.05 Area = 0.05-1.645 0 1.645

pq ˆˆ pˆ z /2 = n 0.220.78 0.22 1.645 0.1945, 0.2455 713 31) pq ˆˆ We use the formula pˆ z /2 n n 300 112 pˆ 0.373 300 qˆ 1 0.373 0.627 1.98 0.02 / 2 0.01 Use Table II and look inside table for area of 0.01; corresponding z-score is -2.32 Hence z /2 2.32 Area = 0.01 Area = 0.01-2.32 0 2.32

pq ˆˆ pˆ z /2 = n 0.3730.627 0.373 2.32 0.3082, 0.4377 300 32) n 85 pˆ 5.9% 0.059 qˆ 1 0.059 0.941 0.01 H : p0.03 H : p0.03 (right-tailed test) o a pˆ p We use the formula z to compute test statistic. p(1 p) / n pˆ p 0.059 0.03 0.029 z 1.567 p(1 p) / n 0.03(1 0.03) /85 0.0185 For right-tailed test: Go to Table II; look inside Table II for area of 0.99 (Note: 1 -.01 = 0.99); corresponding z-score is 2.32

Area = 0.01 (Rejection Region) 0 1.567 2.32 Since Test Statistic is inside of non-rejection region, we do not reject At the 1% significance level, the data do not provide sufficient evidence to conclude that the percentage of defects exceeds 3%. H o : p 0.03 33) n 160 133 pˆ 0.83125 160 0.05 H : p0.88 H : p0.88 (left-tailed test) o a pˆ p We use the formula z to compute test statistic. p(1 p) / n pˆ p 0.83125 0.88 0.04875 z 1.90 p(1 p) / n 0.88(1 0.88) /160 0.0256 For right-tailed test: Go to Table II; look inside Table II for area of 0.05; corresponding z-score is -1.64

Area = 0.05 (Rejection Region) -1.90-1.64 Since Test Statistic is inside of rejection region, we reject H o : p 0.88 At the 5% level of significance, the data provide sufficient evidence to conclude that the proportion of times that luggage is returned within 24 hours is less than 0.88 34) n 1,068 pˆ 0.48 0.05 H : p0.05 H : p0.05 (left-tailed test) o a pˆ p We use the formula z to compute test statistic. p(1 p) / n pˆ p 0.48 0.5 0.02 z p(1 p) / n 0.05(1 0.05) /1068 0.0153 1.307 For right-tailed test: Go to Table II; look inside Table II for area of 0.05; corresponding z-score is -1.64

Area = 0.05 (Rejection Region) -1.64-1.307 Since Test Statistic is inside of non-rejection region, we do not reject At the 5% level of significance, the data do not provide sufficient evidence to conclude that the percentage of voters who prefer the Democrat is less than 50%. H o : p 0.05 35) pˆ pˆ 1 1 1 2 2 2 z /2 21/ 43 0.4883 qˆ 1 0.4483 0.5117 34 / 58 0.5862 qˆ 1 0.5862 0.4138 n 43 n 58 1 90% 10% 0.10 / 2 0.05 1.645

Area = 0.05 Area = 0.05-1.645 0 1.645 Confidence Interval: ˆ ˆ ˆ ˆ ˆ ˆ p p z p 1 p / n p 1 p / n 1 2 / 2 1 1 1 1 1 2 ( 2.62, 0.066) 36) 35)

pˆ pˆ 1 1 1 2 2 2 z /2 37 / 76 0.4868 qˆ 1 0.4868 0.5132 43/ 77 0.5584 qˆ 1 0.5584 0.4416 n 76 n 77 1 95% 5% 0.05 / 2 0.025 1.96 Area = 0.05 Area = 0.05-1.96 0 1.96 Confidence Interval: ˆ ˆ ˆ ˆ ˆ ˆ p p z p 1 p / n p 1 p / n 1 2 / 2 1 1 1 1 1 2 ( 0.229, 0.086)