Final Review

Question 1

Random Sample

Answer 1

Every member of the pop. has an equal chance of being selected.

Question 2

Why are simple random samples preferred?

Answer 2

To prevent bias; so the sample represents the population.

Question 3

Why not voluntary response sample?

Answer 3

May not represent the entire population because only those with a strong opinion respond.

Question 4

Why do organizations do voluntary response samples?

Answer 4

Cheaper, easier, faster.

Question 5

Correlation and causality

Answer 5

Does not show one causes the other, or that one prevents the other.

Question 6

Loaded questoins

Answer 6

The answer is neither one or the other; evokes a passionate or politically incorrect response.

Question 7

Frequency distribution

Answer 7

Title (measurement) , classes (name), frequency

*Make sure sum of fx=# in sample

Question 8

Histogram

Answer 8

Title, frequency, name of classes (can use class widths or midpoints)

Question 9

Mean of frequency distribution

Answer 9

Sum of class midpoints * frequency/sample size
x̄=Σ(f*x)/Σf

Question 10

Standard deviation

Answer 10

s= ✓Σ(x-x̄)^2/(n-1)

Question 11

Variance

Answer 11

s^2

Question 12

Or case

Answer 12

ADD

Question 13

And case

Answer 13

MULTIPLY

Question 14

And case with replacement

Answer 14

Denominator doesn’t change

Question 15

And case without replacement

Answer 15

Denominator changes

Question 16

Combination rule (P of getting x answers correct)

Answer 16

Px= nCrp^xq^n-x

Question 17

Mean of probability distribution

Answer 17

μ=Σ(x*Px)

Question 18

Standard deviation of probability distribution

Answer 18

σ=✓Σ(x^2*Px)-μ^2

Question 19

Binomial dist. (M&Ms) mean

Answer 19

μ=np

Question 20

Binomial dist. (M&Ms) standard deviation

Answer 20

σ=✓npq

Question 21

Binomial dist. (M&Ms) range

Answer 21

μ+/- 2σ (no partial pieces!)

Question 22

Find the indicated birth weight/pregnancy length

Answer 22

find z score (A2) z=(x-μ)/σ -> x=z*σ+μ

Question 23

Critical Value for zα/2

Answer 23

α/2=1-confidence level/2 1-α A2=Zα/2

Question 24

Critical Value for tα/2

Answer 24

df=n-1 1-confidence level=% in 2 tails (A3)=tα/2

Question 25

Confidence level using margin of error

Answer 25

E=zα/2✓(p̂q̂/n)

Question 26

Confidence interval using margin of error

Answer 26

p̂-E<p><p̂+E

Question 27

8 step method

Answer 27

Step 1, Symbolic form
Step 2, If step 1 is false then (opposite step 1)
Step 3 If there is no equal sign, use I as H1.
Step 4 Select a significance level).05 or 0.01.
Step 5 Determine what formula and chart to use
Step 6 Find test stat and (P-value or critical region)
Step 7 Fail to or reject the null
Step 8 Restate with “significant evidence”

Question 28

Find P-value

Answer 28

Find z for p. Look at area in tail(s) compared to α. Make sure area relates to tails.
P-value≤α=no go P-value>α=null fly

Question 29

Traditional method

Answer 29

Find z of p. Find z of α. Compare z’s. If test stat falls in critical region, reject Ho

Question 30

2 samples: P value method

Answer 30

Ho:P1=P2, step 5 p̅= q(bar)=, step 6 z=(p̂1-p̂2)-0/✓p̅qbar/n1+p̅qbar/n2

Question 31

2 samples: Confidence interval method

Answer 31

Use (A3), step 6 tα/2= E=tα/2✓s1^2/n1+s2^2/n2 (x̄1-x̄2)-E<(x̄1-x̄2)+E , step 7 μ1-μ2=0 if CI contains zero fail to reject, step 8 not sufficient to support claim.