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STA 2023: FINALS HANDS-ON -- A Practice Guide with FEW solutions for ‘A’ Better Grade

Mudunuru, Venkateswara Rao Spring 2016

Venkat VnV

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© Venkateswara Rao Mudunuru Page | 1

1. Identify the class width, class midpoints, and class boundaries. Complete the

chart including the relative frequencies and cumulative frequencies.

Solution:

Class Width = 3

Notes:

Class width: = (𝑈𝐶𝐿 − 𝐿𝐶𝐿) + 𝑇𝑜𝑙𝑒𝑟𝑎𝑛𝑐𝑒 = (13 − 11) + 1 = 3.

Where LCL = Lower Class Limit and UCL = Upper Class Limit

Tolerance = Lower Class Limit of second class – Upper Class limit of first class

𝑇𝑜𝑙𝑒𝑟𝑎𝑛𝑐𝑒 = 14 − 13 = 1

Class Midpoint = 𝐿𝐶𝐿 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑐𝑙𝑎𝑠𝑠+𝑈𝐶𝐿 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑐𝑙𝑎𝑠𝑠

2=

11+13

2= 12

Class Boundaries:

Lower Class Boundary (LCB) of each class = LCL of each class – (tolerance/2)

LCB of first class = 11 − (1/2) = 10.5

Upper Class Boundary (UCB) of each class = UCL of each class + (tolerance/2)

UCB of first class = 11 + (1/2) = 11.5

Relative Frequency % = (𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 𝐸𝑎𝑐ℎ 𝐶𝑙𝑎𝑠𝑠

𝑡𝑜𝑡𝑎𝑙 𝑜𝑓 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠) × 100;

Relative Frequency for the First Class: = (10

45) × 100 = 22.22%

Cumulative Frequency = Sum of (frequency of present class + frequency of previous class) and so on

Class

Limits Frequency

Class

Midpoints

Class Boundaries Relative

Frequency (%)

Cumulative

frequency Lower Upper

11-13

10

14-16 15

17-19 8

20-22 12

Class

Limits

Frequency Class

Midpoints

Class

Boundaries

Relative

Frequency

(%)

Cumulative

frequency

Lower Upper

11-13

10 12 10.5 13.5 22.22 10

14-16 15

15 13.5 16.5 33.33 25

17-19 8

18 16.5 19.5 17.78 33

20-22 12

21 19.5 22.5 26.67 45

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2. Identify the class width, class midpoints, and class boundaries. Complete the chart including the relative frequencies and cumulative frequencies.

3. Identify the class width, class midpoints, and class boundaries. Complete the

chart including the relative frequencies and cumulative frequencies.

4. One may define an outlier as any observation outside the interval (Q1 - (1.5 x

IQR), Q3 + (1.5 x IQR)), where IQR is the interquartile range. A random sample of 16 airline carry-on luggage bags gave the following weights.

0, 4, 5, 6, 7, 7, 8, 10, 10, 11, 11, 12, 21, 14, 27, 30

a. Give the five number summary of the data and use it to draw a rough box-and-whisker plot for the same.

b. Identify the outliers (if any) based on the above criterion.

Solution:

a. Five Number Summary: [First write the given data in ascending order]

𝑀𝑖𝑛𝑖𝑚𝑢𝑚 = 0; 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 = 30; 𝑄1 =6 + 7

2= 6.5; 𝑄2 =

10 + 10

2= 10; 𝑄3 =

12 + 14

2= 13

b. IQR = Q3 – Q1 = 13 – 6.5 = 6.5 The interval is (Q1 - (1.5 x IQR), Q3 + (1.5 x IQR)) = (6.5 – (1.5 x 6.5), 13 + (1.5 x 6.5)) = (-3.25, 22.75) So clearly 27 and 30 are the outliers in the given data.

Notes:

For more information, refer to slides 46 – 50 of Chapter-3

Class

Limits Frequency

Class

Midpoints


Frequency (%)

Cumulative


1.1-1.3

15

1.4-1.6 12

1.7-1.9 8

2.0-2.2 10

Cost in $ Sales

Frequency

Class

Midpoints


Frequency (%)

Cumulative


10-26 2 27-43 15

44-60 48

61-77 23

0 6.5 10 13 30

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5. One may define an outlier as any observation outside the interval (Q1 - (1.5 x IQR), Q3 + (1.5 x IQR)), where IQR is the interquartile range. A random sample of 10 values are given below

1, 5, 8, 9, 8, 11, 12, 16, 14, 13



6. One may define an outlier as any observation outside the interval (Q1 - (1.5 x IQR), Q3 + (1.5 x IQR)), where IQR is the interquartile range. A random sample of 10 values are given below.

19, 11, 7, 24, 13, 15, 10, 3, 10, 20



7. Use the table of listed data below to find the sample mean, the sum of squares and sample standard deviation of the data - label each clearly.

Count

1 2 3 2 4 7 3 6 3

TOTALS


20 3 60 -34.615 1198.198 3594.595 50 7 350 -4.615 21.298 149.086 80 1 80 25.385 644.398 644.398 110 2 220 55.385 3067.498 6134.996

TOTALS 13 710 10523.075 Solution:

Sample Mean: �̅� = ∑ 𝑥𝑓

∑ 𝑓=

710

13= 54.615

Sum of Squares: SS = ∑(𝑥 − �̅�)2 × 𝑓 = 10523.075

Sample Standard Deviation: s = √∑(𝑥−�̅�)2×𝑓

𝑛−1= √

10523.075

13−1= 29.613

Notes:

Remember that 𝑛 = ∑ 𝑓

For more information on these problems, refer to slides 16 and 30 of Chapter-3.

)(i ix if ii fx xxi 2xxi ii fxx

2

ix if ii fx xxi 2xxi ii fxx

2

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5 8 10 10 15 12 20 7

TOTALS

10. In a survey of 100 individuals who use one of two brands of detergent, 43 used brand A, 37 used brand B, and 20 used both brands A and B. a. What proportion (probability) used either brand A or brand B?

b. What proportion used neither of these two brands?

11. A student takes two core courses, statistics and biology. In any given semester

the probability of failing statistics course is 3% and the probability of failing biology is 5%. In a given semester, if courses are independent: a. What is the probability that Student will fail both the exams?

b. What is the probability that Student will fail one or the other (but not both)

courses?

Solution:

a. 𝑃(𝑆′ ∩ 𝐵′) = 𝑃(𝑆′) × 𝑃(𝐵′) = 0.03 × 0.05 = 0.0015;

b. 𝑃((𝑆′ ∩ 𝐵)𝑜𝑟(𝑆 ∩ 𝐵′)) = 𝑃((𝑆′ ∩ 𝐵) ∪ (𝑆 ∩ 𝐵′))

= 𝑃(𝑆′) × 𝑃(𝐵) + 𝑃(𝑆) × 𝑃(𝐵′) = (0.03 × 0.95) + (0.97 × 0.05) = 0.077;

Notes:

For more information on these problems, refer to Exam-2 Review Packet

12. A box contains 10 defective and 18 non-defective bulbs. Two are drawn out together. One of them is tested and found to be non-defective bulb. a. What is the probability that the other one is also non-defective?

b. What is the probability that the other one is a defective bulb?

13. An urn contains 4 red, 3 orange, and 7 yellow balls.

a. What is the conditional probability that the second ball is red given the first ball was yellow without replacement?

b. What is the probability that two yellow balls are drawn in succession when

chosen without replacement?

c. What is the probability that two yellow balls are drawn in succession when

chosen with replacement?

ix if ii fx xxi 2xxi ii fxx

2

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14. A box contains 16 colored mugs: 4 red, 6 yellow and 6 green. a. A mug is selected at random. What is the probability that the mug chosen is

either red or green? Is this an independent event or dependent event or mutually exclusive event?

b. A mug is selected at random. What is the probability that the mug chosen is either red or yellow?

c. Two mugs are drawn in succession with replacement. What is the probability that both mugs chosen are yellow? Is this an independent event or dependent event or mutually exclusive event?

d. Two mugs are drawn in succession without replacement. What is the probability that both mugs chosen are yellow? Is this an independent event or dependent event or mutually exclusive event?

Solution:

a. 𝑃(𝑅𝑒𝑑 ∪ 𝐺𝑟𝑒𝑒𝑛) = 𝑃(𝑅𝑒𝑑) + 𝑃(𝐺𝑟𝑒𝑒𝑛) =4

16+

6

16=

10

16= 0.625; 𝑀𝑢𝑡𝑢𝑎𝑙𝑙𝑦𝐸𝑥𝑐𝑙𝑢𝑠𝑖𝑣𝑒

b. 𝑃(𝑅𝑒𝑑 ∪ 𝑌𝑒𝑙𝑙𝑜𝑤) = 𝑃(𝑅𝑒𝑑) + 𝑃(𝑌𝑒𝑙𝑙𝑜𝑤) =4

16+

6

16=

10

16= 0.625

c. 𝑃(𝑌1 ∩ 𝑌2) = 𝑃(𝑌1) × 𝑃(𝑌2) =6

16×

6

16= 0.141; 𝐼𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡

d. 𝑃(𝑌1 ∩ 𝑌2) = 𝑃(𝑌1) × 𝑃(𝑌2|𝑌1) =6

16×

5

15= 0.125; 𝐷𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡

Notes:

For more information on these problems, refer to Exam-2 Review Packet.

15. A card is drawn from a deck of 52 cards. a. What is the probability that the card is diamond or a face card? b. What is the probability that the card is not a spade card? c. What is the probability that the card is either a 6 or heart card?

16. Given the experiment: a fair die is rolled; if a multiple of three appears uppermost, then the die is rolled again, otherwise a fair coin is tossed. a. Draw the probabilistic tree diagram of this experiment with probabilities

mentioned on branches. b. What is the sample space of this experiment? c. What is the associated probability distribution of outcomes in this distribution?

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1/6

0.5

0.5

0.5

17. Given the experiment: A fair die is rolled; if a prime number appears uppermost, then a fair coin is tossed, otherwise an unfair coin with P (tail) = 0.55 is tossed. a. Draw a probabilistic tree diagram for the following experiment. b. What is the sample space of this experiment? c. What is the associated probability distribution in this experiment?

Solution:

a.

b. 𝑆 = {1𝐻, 1𝑇, 2𝐻, 2𝑇, 3𝐻, 3𝑇, 4𝐻, 4𝑇, 5𝐻, 5𝑇, 6𝐻, 6𝑇}

c. 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛:

𝑃(2𝐻) = 𝑃(2𝑇) = 𝑃(4𝐻) = 𝑃(4𝑇) = 𝑃(6𝐻) = 𝑃(6𝑇) =1

6× 0.5 = 0.083

𝑃(1𝐻) = 𝑃(3𝐻) = 𝑃(5𝐻) =1

6× 0.45 = 0.075

𝑃(2𝑇) = 𝑃(3𝑇) = 𝑃(5𝑇) =1

6× 0.55 = 0.092

Notes:

For more details on these problems, refer to Exam-2 Review Packet.

18. Given the experiment: A fair die is rolled; if an even number appears uppermost, then a fair coin is tossed, otherwise a four sided die is rolled. a. Draw a probabilistic tree diagram for the following experiment. b. What is the sample space of this experiment? c. What is the associated probability distribution in this experiment?

Start

1H

T

2H

T

3H

T

4H

T

5H

T

6H

T

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19. Consider a small population consisting of 150 students enrolled in STA4442 course. All 150 students are cross-classified by two classification variables; Gender & Eye color.

Male Female Total Brown 5 20 25 Blue 25 25 50

Others 30 45 75 Total 60 90 150

a. Find the probability that a student selected at random is a male or has blue eye color.

b. Find the probability that a student selected at random has blue eye color given that the student is a female.

20. From the following table of students with Math anxiety, find the probabilities: Math Test

Anxiety Pass Fail Total No 20 30 50 Yes 40 60 100

Total 60 90 150

a. Find the probability that the student selected at random passed Math given that the student has Math anxiety?

b. Find the probability that the student selected at random has no anxiety? c. Find the probability that the student selected at random failed Math test and

has no anxiety?

21. The number of girls and boys enrolled for different majors in a community college are listed in the table below

a. What is the probability that the student is a girl given that she is a biology major? b. What is the probability that the student is biology major given it’s a girl? c. If a student is chosen at random, what is the probability that the student is a boy?

Solution:

a. 𝑃(𝐺𝑖𝑟𝑙|𝐵𝑖𝑜) =𝑃(𝐺𝑖𝑟𝑙∩𝐵𝑖𝑜)

𝑃(𝐵𝑖𝑜)=

150710⁄

270710⁄

= 0.5555

b. 𝑃(𝐵𝑖𝑜|𝐺𝑖𝑟𝑙) =𝑃(𝐺𝑖𝑟𝑙∩𝐵𝑖𝑜)

𝑃(𝐺𝑖𝑟𝑙)=

150710⁄

360710⁄

= 0.417

c. 𝑃(𝐵𝑜𝑦) =𝑛(𝑏𝑜𝑦𝑠)

𝑛(𝑆)=

350

710= 0.493

Notes: For more information on these problems, refer to Exam-2 Review Packet and

Slides 33 – 35 of Chapter-4.

Major Girls Boys Total Biology 150 120 270

Chemistry 100 100 200 Stats 110 130 240 Total 360 350 710

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22. A company purchases large shipments of lemons and uses this acceptance sampling plan: randomly select and test 200 lemons, then accept the whole batch if there is less than two found to be rotten bitter; that is, at most one lemon is bad. If a particular shipment actually has a 3% rate of defects:

a. Using the binomial probability distribution, find the probability that this whole shipment is accepted?

b. Show that a normal approximation to this binomial distribution is appropriate.

c. Using normal approximation both with and without continuity correction, find the probability that this whole shipment is accepted?

23. In a survey on campus it is reported that about 75% of students graduate in time. Suppose 90 students are considered and let r represent the number of students who will graduate in time. Use the normal approximation to the binomial with correction for continuity to find the probability that more than 64 students graduate in time.

24. In a survey on campus it is reported that about 75% of students graduate in time. Suppose 90 students are considered and let r represent the number of students who will graduate in time. Use the normal approximation to the binomial without correction for continuity to find the probability that at least 65 students graduate in time.

25. In economics, of all new products put on the market, 60% failed and are taken off the market within 2 years. Suppose a store introduces 80 new products. Use the normal approximation for the binomial distribution what is the probability that within 2 years fewer than 52 will fail with correction for continuity and without correction for continuity?

Solution:

𝐺𝑖𝑣𝑒𝑛 𝑛 = 80, 𝑝 = 0.60, 𝑞 = 1 – 𝑝 = 0.40

Assumptions: 𝑛𝑝 ≥ 5 & 𝑛𝑞 ≥ 5 ⇒ 𝑛𝑝 = (80)(0.60) = 48 > 5 & 𝑛𝑞 = (80) (0.40) = 32 > 5

⟹ 𝐴𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛𝑠 𝑎𝑟𝑒 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑. 𝐻𝑒𝑛𝑐𝑒 𝑤𝑒 𝑐𝑎𝑛 𝑝𝑟𝑜𝑐𝑒𝑒𝑑 𝑢𝑠𝑖𝑛𝑔 𝑁𝑜𝑟𝑚𝑎𝑙 𝑎𝑝𝑝𝑟𝑜𝑥𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑡𝑜 𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙.

𝑁𝑜𝑤, 𝜇 = 𝑛𝑝 = (80)(0.60) = 48; 𝜎 = √𝑛𝑝𝑞 = √80 (0.60)(0.40) = 4.38

With correction: 𝑃( 𝑥 < 52 ) = 𝑃(𝑥 ≤ 51) = 𝑃(𝑥 ≤ 51.5) = 𝑃 (𝑥−𝜇

𝜎 ≤

51.5−48

4.38)

= 𝑃 (𝑍 ≤ 0.80) = 𝟎. 𝟕𝟖𝟖𝟏

Without correction: 𝑃( 𝑥 < 52 ) = 𝑃(𝑥 ≤ 51) = 𝑃 (𝑥−𝜇

𝜎 ≤

51−48

4.38)

= 𝑃( 𝑍 ≤ 0.68) = 𝟎. 𝟕𝟓𝟏𝟕

Notes:

For more information on these problems, refer to slides 33 and 38 – 40 of Chapter-6 and Exam-3 Review Solutions Packet. © Ve

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26. A group consists of five (5) men and five (5) women; assuming four people are selected at random to form a committee, what is the probability that the committee consists of exactly two women?

Solution:

𝑃(2𝑀 ∩ 2𝑊) =𝑛(2𝑀 ∩ 2𝑊)

𝑛(4 𝑚𝑒𝑚𝑏𝑒𝑟 𝑐𝑜𝑚𝑚𝑖𝑡𝑡𝑒𝑒)=

5𝐶2 × 5𝐶2

10𝐶4

= 0.4762

Note:

For more information on these problems, refer to slides 52 - 54 of Chapter-5 and Exam-2 Review Packet.

27. Given seven women and eight men, what is the probability of selecting one woman and two men to form a three-member committee?

28. Given seven women and eight men, what is the probability of selecting one man and two women to form a three-member committee?

29. Suppose that a random variable 𝑋 is normally distributed with mean 55 and standard deviation 9.

a. Find 𝑃(44 < 𝑥 < 67)

b. Find 𝒂 such that 𝑃(𝑥 < 𝑎) = 0.95 c. When a random sample of size 100 is selected, what is the probability that

the sample mean is greater than 56.2?

30. Scores on exam-2 for statistics are normally distributed with mean 75 and standard deviation 12.

a. Find 𝑃(55 < 𝑋 < 67)

b. Find 𝒂 such that 𝑃(𝑥 < 𝑎) =0.8888

c. If 16 students are considered, what is the probability that the average score of these students is between 55 and 67 points?

Solution:

a. 𝑃(55 < 𝑥 < 67) = 𝑃 ( 55−75

12 ≤

𝑥−𝜇

𝜎 ≤

67−75

12)

= 𝑃(−1.67 ≤ 𝑧 ≤ −0.67 ) = 𝑃( 𝑧 ≤ −0.67) − 𝑃(𝑧 ≤ −1.67)

= [1 − 𝑃(𝑧 < 0.67)] − [1 − 𝑃(𝑧 < 1.67)] = [1 − 0.7486] − [1 − 0.9525]

𝑃(55 < 𝑥 < 67) = 0.2514 − 0.0475 = 0.2039

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b. 𝑃(𝑥 < 𝑎) = 0.8888;

⟹ 𝑃 (𝑧 <𝑎 − 75

12) = 0.8888

𝐴𝑛𝑑 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑧 − 𝑡𝑎𝑏𝑙𝑒 𝑤𝑒 ℎ𝑎𝑣𝑒

𝑃(𝑧 < 1.22) = 0.8888

𝑒𝑞𝑢𝑎𝑡𝑖𝑛𝑔 ⟹𝑎 − 75

12= 1.22 ⟹ 𝑎 = 75 + 12(1.22) = 89.64

c. 𝑃(55 < �̅� < 67)

= 𝑃 ( 55 − 75

12√16

⁄ ≤

�̅� − 𝜇𝜎

√𝑛⁄ ≤

67 − 75

12√16

⁄)

= 𝑃(−6.67 ≤ 𝑧 ≤ −2.67 ) = 𝑃( 𝑧 ≤ −2.67) − 𝑃(𝑧 ≤ −6.67)

= [1 − 𝑃(𝑧 < 2.67)] − [1 − 𝑃(𝑧 < 6.67)] = [1 − 0.9962] − [1 − 1] = 0.0038 − 0 = 0.0038

Note:

For more information on these problems, refer to slides 30; 41 – 43; and 50 - 52 of Chapter-6 and Exam-3 Review Packet.

31. Given that the POPULATION MEAN is 90 and the STANDARDDEVIATION is 8 on

a continuous NORMALLY DISTRIBUTED scale. Find the following:

a. 𝑃( 𝑥 < 94)

b. 𝑃(𝑥 > 86)

c. 𝑃(82 < 𝑥 < 94)

d. If 25 students are considered, what is the probability that the average

score of these students is between 80 and 90 points 𝑃(80 < �̅� < 90)

e. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑎 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑃(𝑥 < 𝑎) = 0.9656

f. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑎 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑃(𝑥 > 𝑎) = 0.9418

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32. In a medical trial, the researcher wishes to estimate the true mean height of a

man age 18-25 to within 5 inches with a 95% degree of confidence. How many

subjects need to be included in the study given the standard deviation is 8

inches?

Solution:

Given Maximal Margin of error, E = 5; and σ =8. Also given c = 95% = 0.95 and

z0.95 = 1.96

𝑛 = (𝑧𝑐𝜎

𝐸)

2

= (1.96 × 8

5)

2

= 9.834 ≈ 10

33. In a poll to determine the percent of actors that are extroverts, how many actors

need to be surveyed to ensure with a 99% degree of confidence that the

estimated proportion is within 3% of the true proportion given approximately 75%

of all actors are extroverts.

Solution:

Since preliminary estimate of proportion p is given as p = 0.75.

Given p = 0.75, c = 99% = 0.99, and z0.99 = 2.58 and E = 0.03

𝑛 = 𝑝 × (1 − 𝑝) × (𝑧𝑐

𝐸)

2

= 0.75 × 0.25 × (2.58

0.03)

2

= 1386.75 ≃ 1387

34. In a poll to determine the percent of individuals who voted for proposition 8, how

many individuals need to be surveyed to ensure with a 90% degree of confidence

that the estimated proportion is within 3% of the true proportion given no

preliminary proportion is known.

Solution:

Since NO preliminary estimate of proportion p is given we use p = 0.5 in the

formula for estimation of sample size, n.

Given c = 90% = 0.90, and z0.95 = 1.645 and E = 0.03

𝑛 =1

4(

𝑧𝑐

𝐸)

2

=1

4(

1.645

0.03)

2

= 751.6 ~ 752 © Ven M

udun

uru




35. A random sample of medical files used to estimate the proportion of all people

who have blood type B. If you have no preliminary estimate of p, how many

medical files should you include in a random sample in order to be 95% sure that

the point estimate �̂� within a distance of 0.05 from p; that is error is 5%?

Solution:

Since NO preliminary estimate of proportion p is given we use p = 0.5 in the

formula for estimation of sample size, n.

Given c = 95% = 0.95, and z0.95 = 1.96 and E = 0.05

𝑛 =1

4(

𝑧𝑐

𝐸)

2

=1

4(

1.96

0.05)

2

= 384.16 ≃ 385

36. Given a preliminary estimate of proportion as 40%, how many business phone

calls should you include in a random sample to be 90% sure that the point

estimate �̂� will be within a distance of 3% from true proportion.

37. Total plasma volume is important in determining the required component in blood

replacement therapy for a person undergoing surgery. Plasma volume is

influenced by the overall health and physical activity of an individual. Previous

studies indicate that the standard deviation of the distribution of blood plasma

was 7.50ml/kg. A physician wants to estimate the population mean plasma

volume for male firefighters within 2.50ml/kg of the population mean. How many

male firefighters should be included in the sample to be 95% confident that the

sample mean plasma volume is within the given margin of error?

38. A random sample of 49 text books purchased at a local bookstore showed an

average price of $122 with a standard deviation of $15. Let 𝝁 be the true mean

cost of a text book sold by this store. Construct a confidence interval with a 90%

degree of confidence. Clearly label the point estimate, the critical value, the

margin of error, the confidence interval and the interpretation.

© Ven M

udun

uru




39. A random sample of 9 sleeping bags purchased at a local store showed an

average price of $84 with a sample standard deviation of $13. Let 𝝁 be the true

mean cost of a sleeping bag sold by this store. Construct a confidence interval

with a 90% degree of confidence. Clearly label the point estimate, the critical

value, the margin of error, the confidence interval and the interpretation.

Solution:

Given n = 9; sample mean, �̅� = 84; sample standard deviation, 𝑠 = 13, ⇒ 𝜎

-unknown AND n = 9 < 30 ⇒ t-distribution with v = (n-1) degrees of freedom.

v = 9 – 1 = 8; Given c = 90% = 0.90, therefore t8, 0.90 = 1.8595

The Point Estimate of Mean µ is �̅� = 84

The 90% Confidence Interval is (�̅� − 𝐸, �̅� + 𝐸), 𝑤ℎ𝑒𝑟𝑒 𝐸 = 𝑡𝑣.𝑐𝑠

√𝑛

𝐸 = 𝑡𝑣,𝑐

𝑠

√𝑛⟹ 𝐸 = 1.8595 × (

13

√9) = 8.058

90% CI for mean µ is, (�̅� − 𝐸, �̅� + 𝐸) = (84 − 8.058, 84 + 8.058) = (75.942, 92.058)

I am 90% confident that the true mean cost of sleeping bags is between 75.942 and 92.058

40. A random sample of 200 books purchased at a local bookstore showed that 72 of

the books were murder mysteries. Let 𝑝 be the true proportion of books sold by

this store that is murder mystery. Construct a confidence interval with a 90%

degree of confidence. Clearly label the point estimate, the critical value, the

margin of error, the confidence interval and the interpretation.

41. In a random sample of 181 students, it was found that 101 were biology majors.

Compute and interpret a 99% confidence interval for the total number of biology

majors in the campus.

Solution:

Given n = 181, x = 101 implying that �̂� =𝑥

𝑛=

101

181= 0.558, which is the point

estimate of population (true) proportion.

Assumptions: 𝑛𝑝 = 181 × 0.558 = 101 > 5 𝑎𝑛𝑑 𝑛𝑞 = 181 × 0.442 = 80 > 5

Given c = 99% = 0.99, and z0.99 = 2.58 © Ven M

udun

uru




The 99% Confidence Interval is (�̂� − 𝐸, �̂� + 𝐸), 𝑤ℎ𝑒𝑟𝑒 𝐸 = 𝑧𝑐√�̂��̂�

𝑛

99% CI for p is (�̂� − 𝐸, �̂� + 𝐸)

= (0.558 − 2.58 × √0.558 × 0.442

181, 0.558 + 2.58 × √

0.558 × 0.442

181)

= (0.558 − 0.095, 0.558 + 0.095) = (0.463, 0.653)

I am 99% confident that the true proportion of biology majors in campus is

between 46% (or 0.46) and 65% (or 0.65).

42. Sixty six wind speed readings gave an average speed of 76 m/sec with a

standard deviation is 20.5 m/sec. Find and interpret a 90% confidence interval for

the parameter of interest.

Solution:

Given n = 66; sample mean, �̅� = 76; sample standard deviation, 𝑠 = 20.5, ⇒ 𝜎

-unknown BUT n = 66 > 30 ⇒ z-distribution. Given c = 0.90, and z0.90 = 1.645

The Point Estimate of Mean µ is �̅� = 76

The 90% Confidence Interval is (�̅� − 𝐸, �̅� + 𝐸), 𝑤ℎ𝑒𝑟𝑒 𝐸 = 𝑧𝑐𝜎

√𝑛

𝐸 = 1.645 × (20.5

√66) = 4.151

90% CI for mean µ is, (�̅� − 𝐸, �̅� + 𝐸) = (76 − 4.151, 76 + 4.151) = (71.849, 80.151)

I am 90% confident that the true mean wind speed is between 71.849 m/sec and 80.151 m/sec.

43. Given a random sample of twelve measurements recorded a sample mean 42 with

a standard deviation 6. Using 90% confidence, what is the maximal margin of

error for this problem? Find an interval estimate for the true mean? Interpret the

result.

44. Given a random sample of fifty measurements recorded a sample mean 52 with a

standard deviation 6. Using 90% confidence, what is the maximal margin of error

for this problem? Find an interval estimate for the true mean? Interpret the result.

Note:

For more information on these problems, refer to slides 30; 41 – 43; and 50 - 52 of Chapter-6 and Exam-3 Review Packet. © Ve

n Mud

unuru




45. A Packaging Company produces boxes out of cardboard and has a specified

weight of 30 oz. It is known that the weight of a box is normally distributed with

standard deviation1.3 oz. A random sample of 36 boxes yielded a sample mean

of 30.5 oz. At 5% level of significance, test the claim that the mean weight of a

box is 30 oz or is there significant evidence that the mean weight is not equal to

30 oz.

Solution:

a. Details given in the problem:

Given n = 36, �̅� = 30.5, 𝛼 = 0.05, 𝜎 = 1.3, 𝜇 = 30.

b. State the Null hypothesis, Alternate hypothesis and what tail test is this:

𝐻0: 𝜇 = 30 𝑣𝑠 𝐻1: 𝜇 ≠ 30; This is a two tail test.

c. Find the Critical Statistic(Table value) and Illustrate: zα/2 = ±1.96

d. State and Compute test Statistic:

𝑧 =�̅� − 𝜇

𝜎

√𝑛

=30.5 − 30

1.3

√36

= 2.31

e. Compute p-value:

𝑝 = 2𝑃(𝑍 > |𝑧|) = 2𝑃(𝑍 > |2.31|) = 2[1 − 𝑃(𝑍 < 2.31)] = 2[1 − 0.9896] = 0.0208

f. Conclusion using critical method:

At 5% level of significance, we have sufficient evidence to reject null hypothesis.

g. Conclusion using p-value method:

Here p-value = 0.0208 < 0.05 = 𝛼 , so we reject the null hypothesis.

-1.96 1.96

© Ven M

udun

uru




46. A rope manufacture claims that their 10 gauge rope can hold up to 350 pounds.

Let 𝜇 be the mean of the weight supported by this type of rope. To ensure the

safety of those individuals using the rope, quality management selected 13 ropes.

The sample mean of 341.3 and the sample standard deviation of 9.9 pounds. At

1% level of significance, test the claim that the mean weight supported by the

rope is 350 pounds or is there significant evidence that the mean weight

supported by the rope is less than 350 pounds.

Solution:


Given n = 13, �̅� = 341.3, 𝛼 = 0.01, 𝑠 = 9.9, 𝜇 = 350

Here 𝜎 is unknown, AND n = 13 < 30, so we will proceed using t-distribution

with (n-1) = 13 – 1 = 12 degrees of freedom.


𝐻0: 𝜇 = 350 𝑣𝑠 𝐻1: 𝜇 < 350; This is a left tail test.

c. Find the Critical Statistic(Table value) and Illustrate: tα,v = -2.681


𝑡 =�̅� − 𝜇

𝑠

√𝑛

=341.3 − 350

9.9

√13

= −3.17

e. Compute p-value:

0.0005 < p-value < 0.005

f. Conclusion using critical method: At 1% level of significance, we do have sufficient evidence to reject null hypothesis.


Here p-value < 0.01 = 𝛼 , we reject the null hypothesis.

-2.681

© Ven M

udun

uru




47. A random sample of 207 individuals who voted in the last election showed that

122 voted for the given amendment. Let 𝑝 be the true proportion of population of

voters that voted for the given amendment. Test the claim at the 1% level of

significance, that this is two-thirds majority.

48. As per records the claim of average weight of a female lion is 95.0 lbs. To check

this claim, a random sample of 42 wild mountain lions (18 months or older) were

captured, weighed, tagged and released; the weights recorded in pounds yielded

a sample mean of 90.0 lbs. with a standard deviation of weights is 30.7. At a 5%

level of significance to does this data indicate that the true mean is less than the

claim.

Solution:


Given n = 42, �̅� = 90, 𝛼 = 0.05, 𝑠 = 30.7, 𝜇 = 95.

Here 𝜎 is unknown, BUT n = 42 > 30, so we will proceed using z-distribution.


𝐻0: 𝜇 = 95 𝑣𝑠 𝐻1: 𝜇 < 95; This is a left tail test.

c. Find the Critical Statistic(Table value) and Illustrate: zα = −1.645


𝑧 =�̅� − 𝜇

𝜎

√𝑛

=90 − 95

30.7

√42

= −1.06

e. Compute p-value:

𝑝 = 𝑃(𝑍 < 𝑧) = 𝑃(𝑍 < −1.06) = 1 − 𝑃(𝑍 < 1.06) = 1 − 0.8554 = 0.1446

f. Conclusion using critical method:

At 5% level of significance, we do not have sufficient evidence to reject null hypothesis.


Here p-value = 0.1446 > 0.05 = 𝛼 , so we fail to reject the null hypothesis.

-1.645

© Ven M

udun

uru




49. As per records the claim of average weight of a female lion is 95.0 lbs. To check

this claim, a random sample of 12 wild mountain lions (18 months or older) were

captured, weighed, tagged and released; the weights recorded in pounds yielded

a sample mean of 90.0 lbs. with a standard deviation of weights is 30.7. At a 5%

level of significance to does this data indicate that the true mean is less than the

claim.

50. To test at the 5% level of significance, whether the true proportion of diabetics in

the state of Florida between the ages of 18 and 44 years is 15%. It is shown that

out of 1,260 diabetics in the state of Florida in 2007, 146 were in the outlined

age group. Test the claim that true proportion of diabetics between the ages of

18 and 44 years is actually less than15%.

Solution:


Given n = 1260, x = 146, 𝛼 = 0.05, p = 15% = 0.15; �̂� =𝑥

𝑛=

146

1260= 0.116

b. Assumptions:

Here both the categories have at least a count of 5 in each of them.

c. State the Null hypothesis, Alternate hypothesis and what tail test is this:

𝐻0: 𝑝 = 0.15 𝑣𝑠 𝐻1: 𝑝 < 0.15; This is a left tail test.

d. Find the Critical Statistic(Table value) and Illustrate: zα = −1.645

e. State and Compute test Statistic:

𝑧 =�̂� − 𝑝

√𝑝𝑞𝑛

=0.116 − 0.15

√0.15 × 0.851260

= −3.38

f. Compute p-value:

𝑝 = 𝑃(𝑍 < 𝑧) = 𝑃(𝑍 < −3.38) = [1 − 𝑃(𝑍 < 3.38)] = [1 − 0.9996] = 0.0004 g. Conclusion using critical method:


h. Conclusion using p-value method:

Here p-value = 0.0004 < 0.05 = 𝛼 , we reject the null hypothesis.

-1.645

© Ven M

udun

uru




51. Studies report that 50% of college students are biology majors. In a random

sample of 519 students, it was found that 285 were biology majors. At 5% level of

significance, do the data indicate that the results are different to the reports

published?

Solution:


Given n = 519, x = 285, 𝛼 = 0.05, p = 50% = 0.5; �̂� =𝑥

𝑛=

285

519= 0.549

b. Assumptions:

Here both the categories have at least a count of 5 in each of them. We

proceed using z-distribution

c. State the Null hypothesis, Alternate hypothesis and what tail test is this:

𝐻0: 𝑝 = 0.5 𝑣𝑠 𝐻1: 𝑝 ≠ 0.50; This is a two tail test.

d. Find the Critical Statistic(Table value) and Illustrate: zα/2 = ±1.96

e. State and Compute test Statistic:

𝑧 =�̂� − 𝑝

√𝑝𝑞𝑛

=0.549 − 0.5

√0.5 × 0.5519

= 2.23

f. Compute p-value:

𝑝 = 2𝑃(𝑍 > |𝑧|) = 2𝑃(𝑍 > |2.23|) = 2[1 − 𝑃(𝑍 < 2.23)] = 2[1 − 0.9871] = 0.0258

g. Conclusion using critical method:


h. Conclusion using p-value method:

Here p-value = 0.0258 < 0.05 = 𝛼 , we reject the null hypothesis.

-1.96 1.96

© Ven M

udun

uru




Handy Formulas: [Will be given on Final Exam]

𝑪𝒐𝒏𝒇𝒊𝒅𝒆𝒏𝒄𝒆 𝑰𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝑻𝒆𝒔𝒕 𝑺𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄

𝒇𝒐𝒓 𝝁: 𝑨𝒔𝒔𝒖𝒎𝒊𝒏𝒈 𝑵𝒐𝒓𝒎𝒂𝒍𝒊𝒕𝒚 𝝈 𝒖𝒏𝑲𝒏𝒐𝒘𝒏 𝑨𝑵𝑫 𝒏 < 𝟑𝟎:

(�̅� − 𝐸, �̅� + 𝐸), 𝑤ℎ𝑒𝑟𝑒 𝐸

= 𝑡𝑣,𝑐

𝑠

√𝑛

𝑡 =�̅� − 𝜇

𝑠

√𝑛

𝒇𝒐𝒓 𝝁: 𝝈 𝑲𝒏𝒐𝒘𝒏 𝑶𝑹 𝒏 ≥ 𝟑𝟎: (�̅� − 𝐸, �̅� + 𝐸), 𝑤ℎ𝑒𝑟𝑒 𝐸

= 𝑧𝑐

𝜎

√𝑛

𝑧 =�̅� − 𝜇

𝜎

√𝑛

𝒇𝒐𝒓 𝒑𝒓𝒐𝒑𝒐𝒓𝒕𝒊𝒐𝒏𝒔, 𝒑:

(�̂� − 𝐸, �̂� + 𝐸), 𝑤ℎ𝑒𝑟𝑒 𝐸

= 𝑧𝑐√�̂��̂�

𝑛

𝑧 = �̂� − 𝑝

√𝑝𝑞𝑛

𝑺𝒂𝒎𝒑𝒍𝒆 𝒔𝒊𝒛𝒆 𝒇𝒐𝒓 𝒑𝒓𝒐𝒑𝒐𝒓𝒕𝒊𝒐𝒏𝒔: 𝑛 = 𝑝(1 − 𝑝) (𝑧𝑐

𝐸)

2

𝑺𝒂𝒎𝒑𝒍𝒆 𝒔𝒊𝒛𝒆 𝒇𝒐𝒓 𝒎𝒆𝒂𝒏𝒔: 𝑛 = (𝑧𝑐𝜎

𝐸)

2

𝒛 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒔𝒄𝒐𝒓𝒆 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒇𝒐𝒓 𝒓𝒂𝒘 𝒅𝒂𝒕𝒂: 𝑧 =𝑥 − 𝜇

𝜎

𝒛 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒔𝒄𝒐𝒓𝒆 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒇𝒐𝒓 𝒅𝒊𝒔𝒕𝒓𝒊𝒃𝒖𝒕𝒊𝒐𝒏 𝒐𝒇 𝒎𝒆𝒂𝒏 𝒔𝒄𝒐𝒓𝒆𝒔:

𝑧 =�̅� − 𝜇𝜎

√𝑛⁄

© Ven M

udun

uru




Standard Normal TABLE P(Z < z)

z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09

0.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359

0.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5753

0.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141

0.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517

0.4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879

0.5 .6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224

0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549

0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852

0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133

0.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389

1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621

1.1 .8643 .8665 .8686 .8708 .8729 .8749 .8770 .8790 .8810 .8830

1.2 .8849 .8869 .8888 .8907 .8925 .8944 .8962 .8980 .8997 .9015

1.3 .9032 .9049 .9066 .9082 .9099 .9115 .9131 .9147 .9162 .9177

1.4 .9192 .9207 .9222 .9236 .9251 .9265 .9279 .9292 .9306 .9319

1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441

1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545

1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633

1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706

1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767

2.0 .9772 .9778 .9783 .9788 .9793 .9798 .9803 .9808 .9812 .9817

2.1 .9821 .9826 .9830 .9834 .9838 .9842 .9846 .9850 .9854 .9857

2.2 .9861 .9864 .9868 .9871 .9875 .9878 .9881 .9884 .9887 .9890

2.3 .9893 .9896 .9898 .9901 .9904 .9906 .9909 .9911 .9913 .9916

2.4 .9918 .9920 .9922 .9925 .9927 .9929 .9931 .9932 .9934 .9936

2.5 .9938 .9940 .9941 .9943 .9945 .9946 .9948 .9949 .9951 .9952

2.6 .9953 .9955 .9956 .9957 .9959 .9960 .9961 .9962 .9963 .9964

2.7 .9965 .9966 .9967 .9968 .9969 .9970 .9971 .9972 .9973 .9974

2.8 .9974 .9975 .9976 .9977 .9977 .9978 .9979 .9979 .9980 .9981

2.9 .9981 .9982 .9982 .9983 .9984 .9984 .9985 .9985 .9986 .9986

3.0 .9987 .9987 .9987 .9988 .9988 .9989 .9989 .9989 .9990 .9990

3.1 .9990 .9991 .9991 .9991 .9992 .9992 .9992 .9992 .9993 .9993

3.2 .9993 .9993 .9994 .9994 .9994 .9994 .9994 .9995 .9995 .9995

3.3 .9995 .9995 .9995 .9996 .9996 .9996 .9996 .9996 .9996 .9997

3.4 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9997 .9998

For z values greater than 3.49, use 1.000 to approximate the area.

(b) Confidence Interval, Critical Values zc

Level of Confidence c Critical Value zc

0.90, or 90% 1.645

0.95, or 95% 1.96

0.99, or 99% 2.58

(c) Hypothesis Testing, Critical Values z0

Level of Significance = 0.05 = 0.01 =0.1

Critical value for a left-tailed test 1.645 2.33 -1.28

Critical value for a right-tailed test 1.645 2.33 1.28

Critical value for a two-tailed test 1.96 2.58 ±1.645 © Ve

n Mud

unuru




One-tail

area 0.250 0.125 0.100 0.075 0.050 0.025 0.010 0.005 0.0005

Two-tail

area 0.500 0.250 0.20 0.150 0.100 0.050 0.020 0.010 0.0010

d.f./c 0.500 0.750 0.800 0.850 0.900 0.950 0.980 0.990 0.999

1 1.000 2.414 3.078 4.165 6.314 12.706 31.821 63.657 636.619

2 0.816 1.604 1.886 2.282 2.920 4.303 6.965 9.925 31.599

3 0.765 1.423 1.638 1.924 2.353 3.182 4.541 5.841 12.924

4 0.741 1.344 1.533 1.778 2.132 2.776 3.747 4.604 8.610

5 0.727 1.301 1.476 1.699 2.015 2.571 3.365 4.032 6.869

6 0.718 1.273 1.440 1.650 1.943 2.447 3.143 3.707 5.959

7 0.711 1.254 1.415 1.617 1.895 2.365 2.998 3.499 5.408

8 0.706 1.240 1.397 1.592 1.860 2.306 2.896 3.355 5.041

9 0.703 1.230 1.383 1.574 1.833 2.262 2.821 3.250 4.781

10 0.700 1.221 1.372 1.559 1.812 2.228 2.764 3.169 4.587

11 0.697 1.214 1.363 1.548 1.796 2.201 2.718 3.106 4.437

12 0.695 1.209 1.356 1.538 1.782 2.179 2.681 3.055 4.318

13 0.694 1.204 1.350 1.530 1.771 2.160 2.650 3.012 4.221

14 0.692 1.200 1.345 1.523 1.761 2.145 2.624 2.977 4.140

15 0.691 1.197 1.341 1.517 1.753 2.131 2.602 2.947 4.073

16 0.690 1.194 1.337 1.512 1.746 2.120 2.583 2.921 4.015

17 0.689 1.191 1.333 1.508 1.740 2.110 2.567 2.898 3.965

18 0.688 1.189 1.330 1.504 1.734 2.101 2.552 2.878 3.922

19 0.688 1.187 1.328 1.500 1.729 2.093 2.539 2.861 3.883

20 0.687 1.185 1.325 1.497 1.725 2.086 2.528 2.845 3.850

21 0.686 1.183 1.323 1.494 1.721 2.080 2.518 2.831 3.819

22 0.686 1.182 1.321 1.492 1.717 2.074 2.508 2.819 3.792

23 0.685 1.180 1.319 1.489 1.714 2.069 2.500 2.807 3.768

24 0.685 1.179 1.318 1.487 1.711 2.064 2.492 2.797 3.745

25 0.684 1.198 1.316 1.485 1.708 2.060 2.485 2.787 3.725

26 0.684 1.177 1.315 1.483 1.706 2.056 2.479 2.779 3.707

27 0.684 1.176 1.314 1.482 1.703 2.052 2.473 2.771 3.690

28 0.683 1.175 1.313 1.480 1.701 2.048 2.467 2.763 3.674

29 0.683 1.174 1.311 1.479 1.699 2.045 2.462 2.756 3.659

30 0.683 1.173 1.310 1.477 1.697 2.042 2.457 2.750 3.646

35 0.682 1.170 1.306 1.472 1.690 2.030 2.438 2.724 3.591

40 0.681 1.167 1.303 1.468 1.684 2.021 2.423 2.704 3.551

45 0.680 1.165 1.301 1.465 1.679 2.014 2.412 2.690 3.520

50 0.679 1.164 1.299 1.462 1.676 2.009 2.403 2.678 3.496

60 0.679 1.162 1.296 1.458 1.671 2.000 2.390 2.660 3.460

70 0.678 1.160 1.294 1.456 1.667 1.994 2.381 2.648 3.435

80 0.678 1.159 1.292 1.453 1.664 1.990 2.374 2.639 3.416

100 0.677 1.157 1.290 1.451 1.660 1.984 2.364 2.626 3.390

500 0.675 1.152 1.283 1.442 1.648 1.965 2.334 2.586 3.310

1000 0.675 1.151 1.282 1.441 1.646 2.962 2.330 2.581 3.300

0.674 1.150 1.282 1.440 1.645 1.960 2.326 2.576 3.291

Student t-distribution table

© Ven M

udun

uru



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