AP EXAM Flashcards
(104 cards)
1
Q
describe sistribution of quantitative data
A
- shape
- center
- variability
- outliers
must use comparative words
2
Q
A
3
Q
skewed left
A
mean<median
4
Q
skewed right
A
mean> median
5
Q
low outlier
A
low< Q1-1.5*IQR
6
Q
High outlier
A
High? Q3+1.5*IQR
7
Q
normal distribution emperical rule
A
1 sx: 68%
2sx: 95%
3sx: 99.7%
8
Q
variance
A
standard deviation squared
9
Q
find area of normal distribution on calc
A
menu 652
normcdf
10
Q
find a value given eprcentile for normal distribution on calc
A
menu 653
invnorm
11
Q
get z score for normal distribution
A
menu653 invnorm
mean: 0 sx:1
12
Q
marginal relative frequency
A
% of individual with specific value in 2 way table
13
Q
association?
A
yes, if knowing the value of 1 variable helps predict the other
-segmented bar graphs are same: no association; different bar graphs: association
14
Q
scatterplot: 2 qauntitative variables
describe scatterplot
A
- direction: positive/negatove/none
2.form: linear/non linear - strenght: strong/ moderate/ weak
- unusual features: outsider patter/ clusters
-describe in context of problem using both variables
15
Q
correlation r
A
shows strength and direction of linear association between 2 quantitativ e variables
-between -1 and 1
-unitless/ doesnt change w unit chaneg
16
Q
interpret correlation r
A
About r% of variation in (responce Y) can be explained by the LSRL using (explanitory x)
16
Q
how to calculate correlation r in calc
A
- spreadsheet of points
- menu 614
17
Q
extrapolate
A
use regression line to prediuct w/ x vlaues outside of interval; less reliable
18
Q
residual
ac
A
actual-predicted
“RAP”
19
Q
interpret residual
A
Context had (residual) more./less context than the number predicted on the LSRL
20
Q
residual plot linear?
A
yes if random scatter
-no if U shaped
21
Q
coefficent of determination r^2
A
-proportion of variation respince (Y) that is explained by explanatory (x) in the lSRL
22
Q
high leverage
A
-much larger/smaller x-valies than other points in data set
23
Q
outliers (influential points)
A
-doesnt follow pattern and has large residual
24
observational study
-cant prove causation
-retrospective: examine existing data and prospective: track individuals into future
25
simple random sample
every group of n individuals has an equal chance to be selected as the sample
| using table: ignore spaces and duplicates
26
how make simple random sample with calc
1. assign each indiv. a # 1-N
2. random # generator 542
3. choose indiv that responds with #
| (1,20) : # between 1 and 20
(1,20,3) L 3 #s between 1 and 20
27
stratified random sample
sample selected by choosing an SRS from each statum and combining SRS into one large sample
27
cluster
gorup of indiv. win population thast are located/related near eachother
27
systematic random sampling
-randomly selecting one of the first "k" individuals and choosing every kth individual after
27
heterogenus groups/census
everyone gets asked/sampled
28
convenience sampling
individuals in population that are easy to reach
-likely biased from over estimate/under estimate
29
voluntary responce sample
-individuals chose to be in sample
-could have poeple who are over passioante about topic
30
undercoverage
when members of population are less likely to be chosen
31
nonresponce
individual chosen cant be contacted/refuses to participate
32
responce bias
consistent pattern of innacurate responces
33
confounding
2 variables are associated in a way that their effects on a respionce variable can't be distinguished
34
completely randomized experiemnt
units are randomly assigned treatments
35
purpous of random assignment
-create roughly equivakent groupds
2 can lead to cause and effect
36
principles of experimental design
1.comparison: control group/ compares 2+ tratments
2.random assignments: roughly - groups and cause and effect
3.replication: treat many experimental units, so effects can be distinguished
4.control: keep other varaibles same for all groups
37
statistically significant
-when differnces in responces between groups in as experiment are so big it is unlikley explained by chance variation in random assignment
-outlier/ not on graph= statistically significant
38
interpret statistically significant result
Because a difference of _ or greater never occured in the simulation, the difference is statistically significant. its unlikley to get a difference this large due to chance variation in random assignment
39
block
group of experimental units tgat are similar in some way (in regards to responce to treatment)
40
law of large numbers
if we observe more and more trials of many random process, proporition that speicific outcome occurs approaches its probability
41
how to preform simulation
1. descrieb set up and use a random process to preform one trial. identify what you will record at the end of each trial
2. preform many trials
3. use results of simulation to answer ? of interest
42
sample space
list of al lpossible outcomes
43
mutually exclusive
2 events A and B can never occur together
44
# mutually exclusive
P(A or b)
P(A)+ P(B)
AUB
45
# A+B resulting from same random process
A upsidedown U B
intersection
-where A+B overlap
46
independenet events
if knowing one event's outcome doesn;t change the probablity of other event happening
47
check independence
P(A|B)=P(A|B^c)=P(A)
P(B|A)=P(B|A^c)=P(B)
| if not equal, not independent
48
tree diagram
sample sapce of a random process with multiple stages
48
# independent events
P(A and B)
## Footnote
)P
P(A) * P(B)
49
sum/difference for mean of 2 random variables
mean (x)+ mean (Y)
mean (x)- Mean(y)
| order matters
50
sum/difference of Sx of 2 random variables
square root of Sx (x)^2 + Sx (y)^2
| sx^2= variance
51
binomial setting
prefrom independent trials of the same random rpocess and ocunt sucesss
52
binomial conditions
Binary: possible sucess/failure
I: trials are independent/ meet <10%
N: # of trials is fixed in advance
S: same probability
53
binomial in calculatior
menu 5 A
menu 5 B
54
graph binomial distribution
1. type all values into list
2. highlight and enter command menu 5B
3. add hraph menu 29
55
binomial mean
## Footnote
n
n*p
56
binomial Sx
square root of n*p(1-p)
57
geometic setting
preform independent trials of the same random process and reocrd # of trials it takes to get a success
58
geometric in calculator
menu 5 5 H
menu 5 5 I
59
geometric mean
1/p
59
geometric Sx
sqaure root of (1-p)/p
60
central limit theorem
-sample size gets big enough, the shape becomes more apporximatley normal
61
statistic
-descrimes characteristic of sample parameter (# that describes characteristics of a population)
62
interpret Confidence Interval
we are C% confident the interval from _ to _ caputures the (parameter in conctext)
63
interpret confidence level
Of we wereto select many random sampels of the same size from the same population and construct a confidence interval using each sample, about C% of the intervals would capture the (Parameter in context)
64
conditions fro calculation condifence interval for population proportion (1-prop z interval)
1. random
2. <10 %
3. large counts
65
1 prop Z interval in calc
menu 665
66
conclusion on significance test
PA: Because y=our P val of_ is greater than a of _, we fail to reject Ho. there isn't convincing evidence of (Ha in context)
67
interpret p value
Assuming Ho, there's a {p-value]} probablitlity of getting [sample in context]
68
do 1 prop-z test in calc
menu 675
-always identify z and p value
69
significance test chart
----Ho true ---Ha true
Reject Ho -false (+); Type I; a -correct solu 1-B
fail to reject Ho -correct solu -False (-), Type II, B
-1-a
70
power
chance of correctly rejecting null
71
how to increase power
1. increase sample size
2. increase significance level a
3. null and alternative parameter values are farther apart
4. standard error is smaller
72
conditions for confidence interval for difference between 2 populations
| 2 sample z interval for difference in population
1. random
2. <10%
3. large counts: for both values
73
2 sample z interval for difference in population in calc
menu 666
74
make conclusion with 2 sample z interval for difference in population
if zero is in interval: no!
if zero ISNT in interval: yes!
75
combined sample prortion
Pc: # of sucesses in both samples/ # of individuals in both samples
76
conditions for calculation confudence interval for population mean
1 random
2. <10%
3. N≥30 or approx normal by graph
77
calc for paired t-interval for a mean diff
menu 662
78
calc for t-interval for population mean diff
menu 662
79
find t with calculator
menu 656
80
calc for 1 sample t-test for means
menu 672
81
2 sample t-interval fro mean diff
menu 664
82
2 sample t-test for means
menu 674
83
when to use chi-square test?
-counts or table, or matrix
ho: no difference
ha: there is a differnce
84
chi square hypothesis for goodness of fit
ho: distribution of _ is the same
ha: disttibution of __ isn;t the same
85
chi square test for independence hypothesis
ho: no association
ha: is assocaition
homogeneity: 2 samples
| indeoendence: 1 sample
86
chi square test conditions
1. random sample
2. all expected counts are at least 5
3. <10%
87
calc chi square test for goodness of fit
-make matrix and store
-menu 677
88
chi suare test for homogeneity hypothesis
Ho: there is no difference in the distributions
Ha: there is a diff in distributions
89
cal chi square test for independence
-make matriz and store
-menu 678
90
equation for x^2
sum of (observed-expected)^2/ expected
91
bigger x^2 value means------
more likely to reject Ho
92
conditions for confidence interval for b slope
L: linear; resideual=scattered
I: <10%
N: normal; no strong skewness/outliers
E: equal SD
R: random/ can make inference
93
calc for confidence interval for b slope
menu 667
94
degrees of freedom for slope
n-2=df
95
make conclusion about confidence slope
does include O: no evidence of linear relationship
doesn't include zero: evidence for linerar realtionship
96
calc population regression line
-list + spreafsheet
menu 667
97
hypothesis for significance test for slope of pop regression line
Ho: B=0
Ha: B<,>,=)
98
df of chi square
#of categories-1
