MATH 1340 Practice Final Exam 1. A. Nominal Level Nominal measures classify data into names or labels for certain characteristics. For example, rocks can be generally classified as igneous, sedimentary, and metamorphic. Each classification is mutually exclusive and does not require any order or ranking. 2. B. statistic A parameter is a single measure about a population while a statistic is a single measure about a sample. 3. E. pareto chart A pareto chart is a series of bars whose heights, arranged in descending order of height from left to right, reflect the frequency or impact of problems. The categories represented by the tall bars on the left are relatively more significant than those on the right. Pareto chart is used to separate the “vital few” from the “trivial many”. 4. D. standard deviation Measure of central tendency like mean, median, mode, percentile, quartile, and interquartile, is a single value that describes a set of data by identifying the central position within the set of data. Standard deviation, on the other hand, is an example of measure of variability, or the measure of difference among variables (interval and ratio variables) and the mean within a certain data set. 5. B. The mean of the distribution will be increased by seven points, but the standard deviation will remain unchanged. For a given data se {63, 72, 67, 69, 65}, the mean is 67.2 and a standard deviation of 3.49. If each number of the given data set is incremented by 5 points, the new data set will be {68, 77, 72, 74, 70}. The new mean is 72.2 (increased by 5 points) but the standard deviation remains 3.49. This simply means that an increment of a uniform value to each number on a given data set will increase the mean that is equal to the value added to each number on the data set. On the other hand, the standard deviation remains because the variation or differences between each value and the mean does not change. 6. E.

= 19 and s = 13 Using TI – 84 Plus: Encode (4, 10, 21, 22, 38) into L1. Then press STAT, go to CALC then select 1-Var Stats 7. E. mean and midrange In the presence of extreme value(s) in a given data set, both the mean and the midrange are most influenced. The mean is the average of all scores in a data set, and midrange is the difference between the highest score and the lowest score. So, an extreme value can highly affect the value of the mean and midrange. Example: (4, 10, 21, 22, 38, 40) : = 22. 5, midrange = 36, median = 21.5 (4, 10, 21, 22, 38, 1000) : = 182.5, midrange = 996, median = 21.5 8. D. 3/14 12 black, 10 brown, 6 white; total number of socks = 18 P(one white sock) = 6/28 = 3/14

9. C. 0.2401 P(a person will get a cold) = 0.7 4 people randomly selected P(all 4 will get a cold) = P(one will get a cold) x P(one will get a cold) x P(one will get a cold) x P(one will get a cold) P(all 4 will get a cold) = P(one will get a cold)4 = (0.7)4 = 0.2401 10. A. σ = 15.08 n = 1000 trials; p = 0.65; q = 1 – 0.65 = 0.35 σ = √(n ∙ p ∙ q) σ = √(1000 ∙ 0.65 ∙ 0.35) σ = 15.08 11. C. P(at most 2) = 0.0398 40% have brown eyes n = 14 people randomly selected P(at most 2) = binomcdf (14, 0.40, 2) P(at most 2) = 0.0398 12. B. False, because ∑ P (x) ≠ 1 ∑ P (x) = 0.1 + 0.2 + 0.3 + 0.3 + 0.2 = 1.1 ≠ 1 Probability can only take a value within 0 ≤ P (x) ≤ 1, where 0 means impossible to happen and 1 means almost certain to happen 13. C. 0.9834 Standard normal distribution, μ = 0 and σ = 1 and the area that is greater than z = -2.13 is the area to the right of z = -2.13 Area = normalcdf (- 2.13, E99, 0, 1) Area = 0.9834 14. A. 14.6 students < μ < 17.4 students Given: n = 61 = 16 s=4 Find: 99% confidence interval Solution: since, σ is unknown, use TInterval: STAT, TESTS, TInterval, Stats, = 16, s = 4, n =61, C-Level = .99 Interval estimate = (14.638, 17.362) 15. C. r = 0.05 r value can be:

-1 ≤ r ≤ 1;

where -1 denotes perfect negative correlation and 1 denotes perfect positive correlation 0 means no correlation The nearer the value of r to 0 means the weaker the correlation between two variables

16. D. 36 Given: ten regions (number of columns) vs. five income levels (number of rows) To determine the degrees of freedom for χ² (chi square) test of independence: df = (c – 1)(r – 1) df = (10 – 1)(5 – 1) df = (9)(4) = 36 17. E. 7.2 Day of the week: No. of Accidents: Expected:

Sun 11 10

Mon 9 10

Tue 6 10

Wed 5 10

Thu 12 10

Fri 13 10

Sat 14 10

Total 70

Expected = ∑observed/number of category (k) Using TI – 84; encode the observed and expected values on L1 and L2 respectively. df = k – 1; df = 7 – 1 = 6 STAT; TESTS; χ²GOF-Test; Observed: L1; Expected: L2; df = 6 χ² = 7.2 (test statistic) 18. C. Test statistic: G = 5, Critical Values 3 and 10; Fail to Reject H0 HHH

M

HHHH

MMM

H

(1st Run)

(2nd Run)

(3rd Run)

(4th Run)

(5th Run)

n1 = 8 – shots made;

n2 = 4 – missed shots G = 5 – number of runs

Since n1 ≤ 20 and n2 ≤ 20 and α = 0.05, the test statistic is G = 5 (number of runs) and we refer to Table A-10 to find the critical values of 3 and 10. Because G = 5 is neither less than or equal to the critical value of 3, nor is it greater than or equal to the critical value of 10, we do not reject randomness (fail to reject H0). There is not sufficient evidence to reject randomness. 19. ans. D. The process is not in statistical control, because there is a pattern, trend, or cycle that is obviously not random. n = 5 students for each sample UCL = 15 (upper control limit)

Ṝ = 10 (mean of the sample means) LCL = 5 (lower control limit)

16 14 12 10 8 6 4 2 0

UCL Mean LCL

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15

There is an obvious upward trend that corresponds to values that are increasing over time.

20. C. colors Qualitative or categorical data consist of names and labels that are numbers representing counts or measurements. a. heights – interval data (can be measured) b. weights – interval data (can be measured) d. test scores – interval (counts) 21. D. systematic In systematic sampling, selection of sample starts at a point and then continues on selecting every kth element in the population. 22. B. convenience sampling In convenience sampling, it is the researcher’s prerogative to simply use results or consider samples that are easy to get. 23. C. Cross-sectional In a cross-sectional study, data are observed, measured, and collected at one point in time. Since in the given situation, it specifies the current employment of its citizens this month. 24. E. 43.8 Class 1 – 10 11 – 20 21 – 30 31 – 40 41 – 50 51 – 60

Frequency(L2) 3 0 4 8 12 20

∑f = 47

XM(L1) 5.5 15.5 25.5 35.5 45.5 55.5

XM ∙ f C 16.5 0 102 284 546 1110

∑( XM ∙ fC) = 2058.5

= ∑( XM ∙ fC) / ∑f = 2058.5 / 47 = 43.8 or Using TI-84; encode lower class limits on L1 and the upper class limits on L2. Set L3 as the class midpoint (L1+L2)/2. Encode the frequencies on L4. Then, use STAT, CALC, 1-Var Stats, List: L3, Frequency: L4 Ans: = 43.79787234 25. C. 50% 1st Quartile

Median

P25

P50

3rd Quartile

P75

The percent of values in a data set that lie between the first (lower) quartile and the third (upper) quartile is 50% Also known as Interquartile Range (Q3 – Q1).

26. C. 256 standard deviation (s) = 16 To obtain the variance (σ); take the square of the standard deviation s2 = (16)2 = 256 27. C. 98 Range = Highest score (HS) – Lowest score (LS) Range = 73; LS = 25 HS = Range + LS = 73 + 25 = 98

28. D. 89.48 Given: z = 2.1; x = 92; s = 1.2 Find: (mean) z = (x - ) / s = x – zs = 92 – (2.1 x 1.2) = 89.48

29. B. 21% P39 – P18 = 39th percentile – 18th percentile = 39% - 18 % = 21% 30. C. 26.5 Using TI-84; encode the data on L1 L1: 3, 5, 8, 9, 10, 10, 11, 12, 13, 20, 24, 26, 27, 31, 31, 49 STAT, CALC, 1-Var Stats, List: L1, Frequency: __ (scroll down to find Q3 = 26.5)

31. E. 10 and 31 Mode is the value that occurs with the greatest frequency appeared in given data set. For the given stem and leaf, 10 and 31 appeared twice which is the greatest frequency within the data set.

32. D. 1.5 Since the standard deviation is the measure of deviation of each value from the mean, the farther (higher) the value from the mean, the more “unusual” it would be. a. 3.5 – 2.8 = 0.7 b. 2.8 – 2.7 = 0.1 c. 2.8 – 1.7 = 1.1 d. 2.8 – 1.5 = 1.3 (greatest deviation from the mean) 33. A. x = 50 pounds = 40 lbs s = 8 lbs z = 1.25 z = (x - ) / s x = + zs = 40 + (1.25)(8) = 50 34. C. 0.314 = 0.295 lbs s = 0.029 lbs 75th percentile (P75) = 3rd quartile (Q3) – corresponds an area of 0.75 under the normal distribution curve Using TI-84: 2nd Function; DISTR; invNorm; area = 0.75; μ = 0.295; σ = 0.029 x = 0.314 35. A. P(A) = 0.042 Probability values can take a value between 0 and 1. A small probability corresponds to an event that unusual. 36. D. (4/52) (3/51) (2/50) There are four aces in a standard deck of 52 cards. In a three-card hand, the probability of the first card to be a ace is 4/52. Since we assumed that the first card is an ace, the remaining cards would be 51 with 3 aces. So, the probability of getting an ace for the second card is 3/51. Accordingly, the probability of getting another ace for the third card is 2/50. 37. A. 19/27 8 – ABC 7 – NBC 12 – CBS P(watch NBC or CBS) = (# of occurrence A + # of occurrence B) / total # of different events = (NBC + CBS) / (ABC + NBC + CBS) = (7 + 12) / (8 + 7 + 12) = 19/27

38. D. 1/6 1

2

1

4

5

6

4

2 3

3

4 4

4

10

5

10

6

10

P(4) = 3/36 P(10) = 3/36 P(4 or 10) = P(4) + P(10) = 3/36 + 3/36 = 6/36 = 1/6 39. A. 1/26 52 cards, half (26) are red cards P(red king) = 1/26

40. C. 1/84,750,400 P(jackpot) = 1/ (50C5)(40C1) = 1/ (2,118,760)(40) = 1/84,750,400 Using TI-84:

1 / (50, MATH, nCr, 5)(40, MATH, nCr, 1)

41. C. 0.3736 P(A) = 0.13, P(B) = 0.28 P(A or B) = P(A) + P(B) – P(A and B)

both are not mutually exclusive (independent)

P(A and B) = P(A) * P(B) Then, P(A or B) = P(A) + P(B) – [ P(A) * P(B) ] = 0.13 + 0.28 – (0.13 * 0.28) = 0.3736

42. C. mutually exclusive Events A and B are disjoint (or mutually exclusive) if they cannot occur at the same time.

43. C. 32 Gender can be male (M) and female (F); 2 possible result n=5 25 = (2)(2)(2)(2)(2) = 32

44. A. All of the CD’s are good. The phrase “at least one” is equivalent to “one or more”. The complement of getting at least one item of a particular type is that you get NO items of that type.

45. C. 1.5 μ = ∑x * P(x) x

P(x)

x* P(x)

-2

0.1

-0.2

0

0.2

0

1

0.1

0.1

2

0.2

0.4

3

0.4

1.2

= (-2)(0.1) + (0)(0.2) + (1)(0.1) + (2)(0.2) + (3)(0.4) = -0.2 + 0 + 0.1 + 0.4 + 1.2 = 1.5 Using TI-84:

STAT, CALC, 2-Var Stats ∑xy = 1.5 = μ

46. C. 2.65 μ = 1.5 x

P(x)

x2

x2 * P(x)

-2

0.1

4

0.4

0

0.2

0

0

1

0.1

1

0.1

2

0.2

4

0.8

3

0.4

9

3.6

σ2 = (∑x2 * P(x)) – μ2 = 4.9 – 1.52 = 4.9 – 2.25 = 2.65 47. D. The probability of success always equals the probability of failure. A Binomial Probability Distribution results from a procedure that meets all the following requirements: 1. The procedure has a fixed number of trials. 2. The trials must be independent. 3. Each trial must have all outcomes classified into two categories. 4. The probability of a success remains the same in all trials. 48. D. 0.9236 Less than z = 1.43 means the area to the left of z = 1.43 Using TI-84: 2nd Function, DISTR, normalcdf lower: -E99 upper: 1.43 μ=0 σ=1 area = 0.9236 49. C. 0.2417 μ = 10 and σ = 2 P(lies between seven and nine) = normalcdf(7,9,10,2) = 0.2417

50. C. 0.0409, yes

p= 0.74 and n= 60 np =44.4 and n(1− p)=15.6 μ = np = 44.4, σ = √(np(1− p)) = √(60*.74*.26) ≈3.398. Using the continuous correction factor we see that we need the area under the normal curve with these parameters and to the left of 38.5. The z-score of 38.5 is z = (x – μ) / s = 38.5 - 44.4) / 3.398 = - 1.74 The area under the standard normal curve to the left of z = -1.74 is .0409. (see Table A-2)

51. D. 87.35% Percentage of the scores that lie between z = -1.31 and z = 1.86; μ = 0 and σ =1. P(-1.31 < z < 1.86) = normalcdf (-1.31, 1.86, 0, 1) = 0.8734593179 Percentage = P(-1.31 < z < 1.86) x 100 = 0.8734593179 x 100 = 87.35%

52. D. 1/2 n = 64,

= 16 and s = 4

σ = σ/√n = 4/√64 = 4/8 = 1/2 53. A. 0.9962 μ = 100; σ = 15; n = 4;

x = 120

μ =μ Central Limit σ = σ/√n Theorem σ = 15 / √4 = 7.5 z = (x - μ ) / σ or normalcdf (-E99, 120, 100, 7.5) = 0.9961695749 = (120 – 100) / 7.5 = 2.67 Table A-2: area corresponds to z = 2.67 is 0.9962

54. D. 2.575 From Table A-2, the common critical value for 0.99 (99%) confidence level is z = 2.575 55. A. wider than The greater the confidence level for a data means the wider the confidence interval for a population mean for the same data.

56. C. 53 and 67 @ 99% confidence interval is (52 < μ < 68); the Range is 16 The possible 95% confidence limit is 53 and 67, because the range is 14, and is lower than the range for 99%. Since the confidence interval decreases from 99% to 95%, the interval range should become narrower.

57. A. 3458 books E = 2 months, σ = 60 months and CI = 95% n = [ (zα/2 * σ) / E ]2 = [ (invNorm (1 – 0.025) * 60) / 2 ]2 ~ 3458 58. B. False The width of a confidence interval estimate for the population mean will become smaller if α (alpha) is increased (e.g. α decreased from 0.01 to 0.05 means the confidence level will decrease from 99% to 95%). Thus the interval becomes narrower.

59. B. 49 E = 1.4 lbs; σ = 5 and α = 0.05 n = [ (zα/2 * σ) / E ]2 = [ (invNorm (1 – 0.025) * 5) / 1.4 ]2 ~ 49

60. A. 78.20 < μ < 89.20 σ = 7.4; n = 12;

= 83.7

CI @ 99% = ZInterval (7.4, 83.7, 12) = 78.20 < μ < 89.20

61. C. 20 The degree of freedom for t-distribution is df = n – 1 df = 21 – 1 = 20

62. A. rejected If the test statistic falls in the critical region (rejection region), the null hypothesis is REJECTED.

63. B. t-statistic Although n is less than 30, the population standard deviation σ is unknown, we use ttest.

64. B. H1 : p < 0.20 The claim is “less than 20% of all college freshmen can read at the 12th grade reading level. The claim generally becomes the alternative hypothesis except for the statement of equality were we assign the claim as the null hypothesis(e.g: test the claim that 20% of all college freshmen). H0 : p ≥ 0.20 H1 : p < 0.20

65. B. Reject the null hypothesis α = 0.05 but do not reject the null hypothesis at α = 0.01. n = 51; ̅ = 31.8;

= 0.75

H0 : μ ≥ 32 oz H1 : μ < 32 oz Since σ is unknown, we use -test: Using TI-84: STAT; TESTS; -Test; Stats: μ0 = 32,

= 31.8,

= 0.75, n = 51,

<0

p – value = 0.0313; @ α = 0.05 (Reject H0) and @ α = 0.01 (Fail to reject H0) 66. C. The sample data supports the claim that the Mill Valley Brewery is cheating. Using TI-84: STAT; TESTS; -Test; Stats: μ0 = 32, = 0.75, = 31.8, n = 51, <0 p – value = 0.0313 < 0.05 (Reject H0) There is a sufficient evidence to support the claim. 67. D. 0.0035 n = 703; p = 0.1593; α = 0.05 H0 : p ≥ 0.20 H1 : p < 0.20 x = (703)(0.1593) = 112 Using TI-84: 1-PropZTest: p0 = 0.2; x = 112; n = 703; prop:
68. C. H1 : μ1 – μ2 > 0 Since the claim states that the mean braking distance of four-cylinder (μ1) is greater than the mean braking distances of six-cylinder cars (μ2), their difference must be greater than zero (μ1 – μ2 > 0). 69. E. 9.5 < μd < 27.6 Since σ is unknown, we use t-test. Using TI-84: STAT; TESTS; TInterval: Data; List: L1-L2; Freq: 1; C-Level: 0.99 Interval: (9.5247, 27.642) or 9.5 < μd < 27.6

70. C. low scores on x are associated with high scores on y A correlation coefficient of r = -0.94 denotes a negative correlation. Thus, an increase in value of one variable will decrease the value of the other variable, or vice versa.

71. A. positive The value of the correlation coefficient can be between -1 and 1. A negative r denotes a negative correlation, and a positive r denotes a positive correlation.

72. B. the amount of variation in the y-variable that is explained by the regression line The coefficient of determination r2 is a measure of the amount of variation in the yvariable that is explained by the regression line.

73. B. 36.3 The best predicted value is the mean ӯ = 36.3

74. A. 5.991 Column (smokers and non-smokers) x Row (0 -1 visit, 2 – 3 visits, 4 or more) 2 x 3 table;

df = (c-1)(r-1) = 2;

α = 0.05

Using Table A-4 (Chi-Square Distribution); critical value is X2 is 5.991

75. D. Critical Values rs = ± .505; There does not appear to be a correlation between the variables. rs = 0.412;

n = 27 < 30

z = 0.01 (significance level, alpha)

Using Table A-9: Critical value of Spearman’s Rank Correlation Coefficient rs = ± 0.505. If rs is between the negative and positive critical values, fail to reject the null hypothesis (no correlation).

76. A. Test Statistic: rs = - 0.285, Critical Values: rs = ± 0.648, There does not appear to be a correlation.

Comfort

Price

d

d2

5

1

4

16

8

7

1

1

9

3

6

36

10

5

5

25

4

4

0

0

3

2

1

1

2

10

8

64

1

9

8

64

7

6

1

1

6

8

2

4

∑d2 = 212

n = 10

H0 : ρs = 0 (no correlation) H1 : ρs ≠ 0 rs = 1 – {6∑d2 / [n(n2 – 1)} = 1 – [6*212 / (10 * 99)] = 1 – 1.28484

Critical Values: n = 10 and z = 0.05 (Table A-9) rs = ± 0.648 Since rs is between the negative and positive critical values, fail to reject H0.

= - 0.285

77. E. Test Statistic: x = 5; Critical value: 3; Fail to reject no difference Positive signs = 10 Negative signs = 5 (this is x since it occur less) Ties = 3 (we disregard ties) H0 : No difference (equal) H1 : Different (not equal) Using Table A-7 (Critical Values for the Sign Test) α = 0.05 and n = 15 (total number of + and – signs only); two-tail test Critical Value: 3

Since 10 > 3, Fail to reject H0

78. D. Not random The fundamental principle of the runs test is: Reject randomness if the number of runs is very low or very high. 79. A. Ho: The Math scores are equal to the verbal scores H1 : The Math scores are greater than the verbal scores The claim is programmers do better on the mathematics portion of the test than in verbal portion. The claim must be the alternative hypothesis (except when it says equal). Here it says math is greater than verbal scores.

80. B. Test statistic: x = 0; Critical Value x = 1; Reject H0 + sign = 9 - sign = 0 (less occur, x = 0) Using Table A-7: α = 0.05; x = 0; n = 9; one-tail Critical Value: x = 1 Since 0 < 1, Reject H0 81. B A process is statistically stable (or within statistical control) if it has only natural variation, with no patterns, cycles, or unusual points. A – has upward and downward trend C – has cyclical pattern D – has exceptional high value

MATH 1340 Practice Final Exam 1. A. Nominal Level ...

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