Final Exam Review (Math 1342)

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1 Final Exam Review (Math 1342) 1) 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 = ) 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 =

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

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

4 7) x For x 300, its z - score = = Go to Table II, look for z = 2.13; corresponding area is Area = = Thus, the probability that a pregnancy lasts at least 300 days is ) x For x 5.48, its z - score = = x For x 5.82, its z - score = = Go to Table II, look for z = -2.71; corresponding area is Go to Table II, look for z = 2.14; corresponding area is Area = Area = = Thus, percentage of legal quarters will be rejected is =

5 9) Since is known, we use the formula x z/ 2, x z/ 2 n x n15 n = % / Use Table II and look inside table for area of ; corresponding z-score is -2. Area = Area = Hence /2 2 z x z/ 2, x z/ 2 n n , = 3.79,

6 10) Since is known, we use the formula x z/ 2, x z/ 2 n x n =1-90% 0.10 / n Use Table II and look inside table for area of ; corresponding z-score is Area = 0.05 Area = Hence z / x z/ 2, x z/ 2 n n , = 521.4,

7 11) Since is known, we use the formula x z/ 2, x z/ 2 n x n =1-95% 0.05 / n Use Table II and look inside table for area of 0.025; corresponding z-score is Area = Area = Hence z / x z/ 2, x z/ 2 n n , = 19.3,

8 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 = =1-95% 0.05 / Go to Table IV; look under t with df = 11; corresponding t-score is Thus, t / Area = Area = x z/ 2, x z/ 2 n n , = 232.7,

9 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 = =1-99% 0.01 / Go to Table IV; look under t with df = 29; corresponding t-score is Thus, t / Area = Area = x z/ 2, x z/ 2 n n , = 75.91,

10 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 = =1-95% 0.05 / Go to Table IV; look under t with df = 26; corresponding t-score is Thus, t / Area = Area = x z/ 2, x z/ 2 n n , = 67.7, ) See Answer Sheet 16) See Answer Sheet 17) See Answer Sheet

11 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 = x - c) Since is unknown, we use the formula t to compute test statistic. s n x t s 3.7 n 20 d) For two-tailed test: Go to Table IV; look for t with df = n - 1 = 20-1 = 19. Critical values are is ±2.861 Area = (Rejection Region) Area = (Rejection Region) 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.

12 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 t 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 Area = 0.05 (Rejection Region) e) Since Test Statistic value of 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.

13 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 = x - c) Since is unknown, we use the formula t to compute test statistic. s n x t 0.86 s 7.7 n 11 d) For two-tailed test: Go to Table IV; look for t with df = n - 1 = 11-1 = 10. Thus, critical values are is ±2.228 Area = (Rejection Region) Area = (Rejection Region) 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.

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

15 22) We use the formula n n x1 x2 t / 2 S p =1-99% 0.01 / Go to Table IV; look for t with df = n 1 n Hence, critical values are is ±2.861 Thus, t /2 = Area = Area = S 2 2 n 1 s n 1 s p = = = n1n x1 - x2 t / 2 S p = (-1.38, 7.38) n n 1 2

16 23) From the frequency table, we know that there are 40 data values ( ). 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 , 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 = xi x s= n 1 = = 13.2 For more information, see page 119 (in textbook) for more information. Standard deviation for this sample is 13.2.

17 24) Waiting time(minutes) Number of customer Approximated data values ,2,2,2,2, 2,2,2,2,2, 2,2,2, ,6,6,6, 6,6,6,6,6,6, ,10,10,10,10,10, ,14,14,14, 14,14,14,14, 14,14,14,14, 14,14,14, ,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.

18 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 = d c) We use the formula t to compute test statistic. Sd n Before After diff = Before - after d mean of differences = -5.2; S d standard dev. of differences = = x i x n (-5.2) -9 - (-5.2) 2 - (-5.2) -3 - (-5.2) (-5.2) = d 5.2 t = = S d n 5 d) For two-tailed test: Go to Table IV; look for t with df = n = 4. Critical values are ±4.604.

19 Area = (Rejection Region) Area = (Rejection Region) e) Since Test Statistic value of 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 d mean of differences = ; S d S standard dev. of differences d = = x x i n = (-0.6) (1.3) (2.4) (-0.8) (1.8) (-0.1) (3.6) ( 3.9) d t = = Sd n 8 d) For two-tailed test: Go to Table IV; look for t 0.05 with df = n = 7. Critical value is

20 Area = 0.05 (Rejection Region) e) Since Test Statistic value of 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 = 4. Critical values are ±2.13. Thus t /2 = 2.13 Area = 0.05 Area = S d 1.12 d t/ = (-0.37, ) n 5 28) = 1-99% = 0.01; and /2 = We use the formula d S d t/2 n d mean of differences = 1.22; S standard dev. of differences = d

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

22 pq ˆˆ pˆ z /2 = n , ) We use the formula n 713 pˆ 0.22 qˆ / pq ˆˆ pˆ z /2 n Use Table II and look inside table for area of 0.05; corresponding z-score is Hence z / Area = 0.05 Area =

23 pq ˆˆ pˆ z /2 = n , ) pq ˆˆ We use the formula pˆ z /2 n n pˆ qˆ / Use Table II and look inside table for area of 0.01; corresponding z-score is Hence z / Area = 0.01 Area =

24 pq ˆˆ pˆ z /2 = n , ) n 85 pˆ 5.9% qˆ 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 z p(1 p) / n 0.03(1 0.03) / For right-tailed test: Go to Table II; look inside Table II for area of 0.99 (Note: = 0.99); corresponding z-score is 2.32

25 Area = 0.01 (Rejection Region) 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 ) n pˆ 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 z 1.90 p(1 p) / n 0.88(1 0.88) / For right-tailed test: Go to Table II; look inside Table II for area of 0.05; corresponding z-score is -1.64

26 Area = 0.05 (Rejection Region) 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 ) n 1,068 pˆ 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 z p(1 p) / n 0.05(1 0.05) / For right-tailed test: Go to Table II; look inside Table II for area of 0.05; corresponding z-score is -1.64

27 Area = 0.05 (Rejection Region) 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 ) pˆ pˆ z /2 21/ qˆ / qˆ n 43 n % 10% 0.10 /

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

29 pˆ pˆ z /2 37 / qˆ / qˆ n 76 n % 5% 0.05 / Area = 0.05 Area = Confidence Interval: ˆ ˆ ˆ ˆ ˆ ˆ p p z p 1 p / n p 1 p / n 1 2 / ( 0.229, 0.086)