AMS 310 Review for Winter Final PDF

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AMS 310 Winter Online, Fred Rispoli Review Material for Final Exam and List of Formulas supplied on Final

Resource Materials and Problems:

1. Chapters 5, 6, 7, 8 & 9 all sections. The summary at the end of each chapter is very good 2. Class Notes for each chapter posted on Blackboard 3. Overview videos posted on Blackboard 4. Example videos for each chapter posted on Blackboard 5. Special Topic-YouTube Video: T Tests and the Five Step Method (By me) 6. Homework 2 and 3, Solutions to Homework 2 Posted in Assignments 7. Practice Test for Final 8. Solutions to selected problems in Ahn from the set below. Chapter 5: Problems: # 1, 2, 3, 4, 6, 7, 9 10, 11, 13, 15, 17, 19, 21, 23, 25, 27 Chapter 6: Problems # 1, 2, 3, 5, 6, 9, 11, 12, 13, 15, 19, 21, 27 Chapter 7: Problems: # 1, 3, 4, 5, 6, 7, 8, 9, 10, 11 Chapter 8: Problems: # 1, 3, 5, 7, 9, 11, 13, 17, 21, 23, 27, 29, 31, 33, 35, 37, 39, 41, 45, 47, 49, 51 Chapter 9: Problems: # 3, 5, 7, 8, 9, 11, 13, 15, 17, 18, 20, 21, 23, 24, 25, 27, 29, 31, 33, 37

9. Tables to be familiar with for Test. All given below Standard Normal distribution T Distribution Table Chi-Square Distribution Table F Distribution

Possibly Useful Formulas and Tables Included on Final Exam Binomial(n,p): p(i) =

(ni ) p (1− p) i

n−i

, i = 0, 1,..., n, Mean is np, variance is np(1-p)

From Chapter 5 b

∫ x f 1 ( x ) dx

Given continuous rv’s X and Y, and joint pdf f(x,y), E(X) =

, where a < x < b

a

Conditional Distribution

f 1 ( x 1|x2 )=

f ( x 1 , x 2) f 2 (x 2)

From Chapter 6 For known  and large samples, if X1, … , Xn be a random sample from a normal distribution and S2 is given, then Z =

´X−μ is approximately N(0,1). ❑ √n

For unknown  and small samples, if X1, … , Xn be a random sample from a normal distribution and S2 is given, then T =

´X−μ is approximately tn-1. s √n

If X1, …, Xn be a random sample from N(µ, 2), then

(n−1) s 2 σ

2

~

From Chapter 8

( ) zα σ

Sample size needed to attain maximum error n =

2

E

2

,

χ 2n−1 .

Sample size with  at µ = µ’ for a one sided test: n =

z σ (¿ ¿ α+ z β ) μ0−μ ' , for a two-sided test ¿ ¿ ¿ ¿

z σ (¿ ¿ α/2+ z β ) μ 0−μ ' ¿ ¿ ¿ ¿ Type II error probability

(

'

Փ z α /2 +

μ0−μ

σ / √n

) (

β

(') for a level  test, for H1: µ  µ’

−Փ −z α/2 +

μ0−μ' σ / √n

)

Normal population with known . A 100(1-)% confidence interval:

´x −z α 2

¿ Small samples with unknown . A 100(1-)% confidence interval:

´x −t α 2

, n−1

¿ For hypothesis testing t =

σ σ √ n , ´x + z α2 √n

)

s s √ n , ´x +t α2 ,n −1 √ n )

x´ −μ 0 . s √n

() zα

Sample sizes for proportions, n = p(1-p)

2

2

( ) zα

when estimate of p is known. Use n =

E

1 2 4 E

2

if no

estimate A confidence interval for p: ( ^p−z α 2

^p − p0

√ p0 (1−p 0)/ n

√

√

^p (1− ^p ) p^ ( 1− ^p ) , ^p +z α n n 2

2 2

( n−1)s 2 σ

2

X n

.

(n−1)s 2 Confidence interval for variance: ( χ α2 χ=

) where ^p=

.n−1

(n−1)s , χ2 α 1−

2

2

.n−1

) For hypothesis testing

For testing z =

From Chapter 9 Large independent samples. A 100(1-)% confidence interval:

(

( ´x − ´y ) −z α 2

√

√ )

s21 s22 s 21 s22 + , ( ´x − ´y ) + z α + m n m n 2

x (¿− ´y´ )− Δ For hypothesis testing z =

√

s21 s22 + , m n ¿

.

Small independent samples, when 1 = 2. A 100(1-)% confidence interval:

(

( ´x − ´y ) −t α 2

Where s 2p=

, m+n−2

√

sp

√ )

1 1 1 1 + , ( ´x − ´y ) +t α + sp ,m +n−2 m n m n 2

( m−1 ) s21+(n−1)s22 m+ n−2

x (¿− ´y´ )− Δ

√

For hypothesis testing t =

sp

1 1 + , m n ¿

Small independent samples, when 1 ≠ 2, A 100(1-)% confidence interval:

´x 1− x´ 2 ± t α 2

,ν

√

s21 s22 + m n

where

ν =

(

2 2 s1 s2 + m n

( s12 /m ) ( s22 /n )2 2

m−1 For paired data. A confidence interval for the mean difference

(

sd ´ sd , d+ t α /2, n−1 ,n −1 √ n √n 2

´d−t α

(

^p1−^p2−z α

x+ y ^p= m+ n

n μ 0 (1-sided alternative)

H 0 : μ=μ0 , H 1 : μ < μ 0 (1-sided alternative)

(2-sided alternative) Step 3

z=

´x −μ0 σ/√n

z=

Rejection Region:

¿ z∨≥ z α /2 Z ≥∨z∨¿ p=2 P ¿

Step 5

z= z ≥ zα

Rejection Region:

p=P(Z ≥ z )

´x −μ0 σ / √n z ≤−z α

Rejection Region:

p=P(Z ≤ z)

μ for a normal population with unknown

One Sample Tests about

Step 1

´x −μ0 σ/√n

σ

and small samples

H 0 : μ=μ0 , H 1 : μ ≠ μ 0

H 0 : μ=μ0 , H 1 : μ > μ 0

H 0 : μ=μ0 , H 1 : μ < μ 0

(2-sided alternative)

(1-sided alternative)

(1-sided alternative)

Step 3

t=

x´ −μ 0

t=

s /√ n ¿ t∨≥t α /2, n−1

Rejection Region:

t=

s /√ n t ≥ t α , n−1

Rejection Region:

x´ −μ 0 s/√n

Rejection Region: t ≤−tα , n−1

p=P(T ≥t )

T ≥∨t∨¿ p=2 P ¿

Step 5

x´ −μ 0

p=P(T ≤t )

One sample proportion test Step 1

Step 3

H 0 : p= p0 , H 1 : p ≠ p 0

z=

^p − p0

√ p ( 1− p ) /n 0

0

Rejection Region:

¿ z∨≥ z α /2

H 0 : p= p0 , H 1 : p> p0

z=

^p − p0

√ p ( 1− p ) /n 0

0

Rejection Region:

z ≥ zα

H 0 : p= p0 , H 1 : p< p0

z=

^p − p0

√ p ( 1− p ) /n 0

0

Rejection Region:

z ≤−z α

Step 5

p=P(Z ≥ z )

Z ≥∨z∨¿ p=2 P ¿

One sample test for variance Step 1 H : σ 2= σ 2 , 0

0

H 1 : σ 2 ≠ σ 02

p=P(Z ≤ z)

H 0 : σ 2= σ 02 , H 1 : σ 2>σ 02

H 0 : σ 2= σ 02 , 2

2

H 1: σ < σ0 Step 3

χ 2=

(n−1) s σ20

2

χ 2=

Rejection Region:

(n−1) s σ20

2

χ 2=

Rejection Region:

2 2 χ ≥ χ α /2, n−1 or

2

(n−1) s σ20

2

Rejection Region: 2

2

2

χ ≥ χ α , n−1

χ ≤ χ 1−α , n−1

2 p= P ( X ≥ χ )

2 p= P ( X ≤ χ )

χ 2 ≤ χ 21−α /2, n−1 2 p=2 min { P ( X ≥ χ ) ,

Step 5

2 P ( X ≤ χ )} where 2 X χ n−1

2 X χ n−1

where

Large Independent Two Sample Test for Means Step 1 H 0 : μ 1−μ2=∆0 , H 0 : μ 1−μ2=∆0

H 1 : μ1 −μ 2 ≠ ∆ 0 (2-sided alternative) Step 3

z=

( ´x − ´y )−∆0 

√

2

2

¿ z∨≥ z α /2 Step 5

H 1 : μ1 −μ 2> ∆0

Z ≥∨z∨¿ p=2 P ¿

z=

(2-sided alternative)

( ´x − ´y )−∆0 

√

2

2

s1 s2 + m n

Rejection Region: z ≥ z α

p=P(Z ≥ z )

Small Independent Samples, Equal Variances Step 1 H 0 : μ 1−μ2=∆0 ,

H 1 : μ1 −μ 2 ≠ ∆ 0

2 X χ n−1

H 0 : μ 1−μ2=∆0 , H 1 : μ1 −μ 2< ∆0 (1-sided alternative)

(1-sided alternative)

s1 s2 + m n

Rejection Region:

,

where

H 0 : μ 1−μ2=∆0 ,

H 1 : μ1 −μ 2>∆0 (1-sided alternative)

z=

( ´x − ´y ) −∆0 

√

2

2

s1 s2 + m n

Rejection Region:

z ≤−z α

p=P(Z ≤ z)

H 0 : μ 1−μ2=∆0 , H 1 : μ1 −μ 2< ∆0 (1-sided alternative)

Step 3

t=

(x´ −´y )−∆ 0  sp

√

t=

1 1 + m n

(x´ −´y )−∆ 0  sp

Rejection Region:

√

1 1 + m n

Rejection Region:

t=

(x´ −´y )−∆ 0  sp

√

1 1 + m n

Rejection Region:

|t| ≥ t α ,m+ n−2

t ≥ t α , m+n−2

t ≤−t α , m+n−2

T ≥∨t∨¿ p=2 P ¿

p=P(T ≥t )

p=P(T ≤t )

2

Step 5

Paired Data

Two Sample Proportions Step 1

H 0 : p 1− p 2=0 ,

H 0 : p 1− p 2=0 ,

H 0 : p 1−p 2=0 ,

H 1 : p1− p 2 ≠ 0

H 1 : p1− p 2>0

H 1 : p1− p 2< 0

(1-sided alternative)

(1-sided alternative)

(2-sided alternative) Step 3

z=

√

^p1−^p2 

z=

( m1 + 1n )

^p (1− ^p )

Rejection Region:

|z|≥ z α

√

^p1− ^p2  ^p (1− ^p)

( m1 + 1n )

Rejection Region:

z ≥ zα

z=

√

^p1−^p2 

Rejection Region:

2

Step 5

Z ≥∨z∨¿ p=2 P ¿

Probability Tables

p=P(Z ≥ z )

( m1 + 1n )

^p (1− ^p )

p=P(Z ≤ z)

z ≤−z α

1. The Standard Normal Tables

2. The T distribution

3. The chi-square distribution Table

4. The F Distribution Table...