Biostats Homework 1 Answers PDF

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Cobbs: Bio350 Biostatistics Spring 2019 Homework 1 Questions & Answers Name(Print last, first): 1.

ANSWERS

Consider the two data sets A and B. Sample A is ungrouped data and sample B is grouped data. (3 pts.) a.

Fill out the table below. Here Sample C is the natural log transform of the data in sample A and Sample D is the square root transform of Sample B. (1 pt.) A. For each data sets compute statistics below Sample A Sample B Sample C Sample D Mean 1.572 2.016 0.2596 1.239831 Median

1.7

2

Standard deviation

0.4242833

1.657858

0.1774937

0.1884932

25%tile

1.325

1

0.18

1.1509356

75%tile

1.875

3

0.28

1.3692139

interquartile range

1.875-1.325 = 0.55

2

0.1

1.36921-1.1509 3= 0.2182783

b. Sample A (1/4 pt.)

0.2

1.30384

Make a histogram for each of the data sets A, Band C and draw, paste or tape them in the boxes below. (1/4 pt.)

Cobbs: Bio350 Biostatistics Spring 2019 Homework 1 Questions & Answers Sample B (1/4 pt.)

Sample C

c .

(1/4 pt.). Skewed (yes or no)

If skewed describe how it is skewed

Sample A

yes

skewed to left

Sample B

yes

highly skewed to right

Sample C

yes

very highly skewed to right

Cobbs: Bio350 Biostatistics Spring 2019 Homework 1 Questions & Answers d.

Make a boxplot for each of data sets A - B and draw them in the boxes below. (1/4 pt.)

Box plot for sample A

e . Name of transform histogram

(1/4 pt.) 1/y

Box plot for sample B

Box plot for sample C

Cobbs: Bio350 Biostatistics Spring 2019 Homework 1 Questions & Answers 2.

(1 pt.)

a

b

c

d

Cobbs: Bio350 Biostatistics Spring 2019 Homework 1 Questions & Answers 3.

(1 pt.)

a

Use Mendels law of segregation Pr(D/d) = 1/2

b

D = dwarf allele, c = normal allele Unconditional probabilities are 1/4 D/D, 1/2 D/d, 1/4 d/d so conditional probability is Pr(D/d| not D/D) = (1/2)/(1/2 + 1/4) = (2/4)/(3/4) = 0.6667

c

Pr(D/D | boy lives to 10 yrs) = Pr(D/D and lives to 10 yrs)/Pr(lives to 10 yrs) = 0/(3/4) = 0

4.

(1 pt.)

Use Bayes Rule:

5.

, Using A= smoke and B= allergy

(2 pts.)

a

b

c

d

e

Pr(GCGCGC or CGCGCG) = Pr(GCGCGC)+Pr(CGCGCG) =

Cobbs: Bio350 Biostatistics Spring 2019 Homework 1 Questions & Answers 6.

(2 pts.)

a

1/5 =0.2

b

3/5 = 0.6

c

1/(5P2) = 1/(5!/(5-2)!) = 1/(5!/3!) = 1/(5C4) = 1/20 = 0.05 or (1/5)(1/4)

d

(1/5)(3/4) = 3/20 = 0.15

e

1/(5P3) = 1/(5!/(5-3)!) = 1/(5!/2!) = 1/(5C4C3) = 1/60 = 0.016667 or (1/5)(1/4)(1/3)

f

(3P3)/(5P3) = (3!/(3-3)!) /(5!/(5-3)!) = (3!/(0)!) /(5!/(2)!) = (3C2C1) /(5C4C3) = 6/60 = 0.1

g

1/(5C2) = 1/[5!/{2!(5-2)!}] = 1/[(5C4C3C2C1)/{(2C1)(3C2C1)}] = 1/10 = 0.1

h

Pr(A->notAorB)+Pr(notAorB->A) = (1/5)(3/4)+(3/5)(1/4) = 3/20 + 3/20 = 6/20=3/10 = 0.3

i

1/(5C3) = 1/[5!/{3!(5-3)!}] = 1/[(5C4C3C2C1)/{(3C2C1)(2C1)}] = 1/10 = 0.1

j

(3C3) /(5C3) = [3!/{3!(3-3)!]/[5!/{3!(5-3)!}] = 1/[(5C4C3C2C1)/{(3C2C1)(2C1)}] = 1/10 = 0.1

7.

(2 pts.)

a

Pr(A/A) = 320/800 = 0.4

b

Pr(B/b) =195/800 = 39/160 = 0.24375

c

Pr(A/A | not b/b) = (150+145)/(375+195) = 295/570 = 59/114 = 0.5175438

d

Pr(b/b | a/a) =105/190 = 21/38 = 0.552632

e

Pr(b/b | not a/a) =(25+100)/(320+290) = 125/610 = 25/122 = 0.204918

f

Pr(a/a and b/b) = 105/800 = 21/160 = 0.13125

g

Pr(A/a and B/b | not (a/a or b/b) ) 40/(150+145+150+40) = 40/485 = 8/97 = 0.082474

Cobbs: Bio350 Biostatistics Spring 2019 Homework 1 Questions & Answers 8. a

(2 pts.) i

ii

b

i

ii

Cobbs: Bio350 Biostatistics Spring 2019 Homework 1 Questions & Answers 9.

(1 pt.)

a

b

c

10. a

b

(1 pt.)

Cobbs: Bio350 Biostatistics Spring 2019 Homework 1 Questions & Answers Appendix 1. a. b.

For sample A use hand calculator or R For sample B i. Using hand calculator (1) mean

(2)

sd

ii. Using R y...