3.1. Cross-validation_ evaluating estimator performance — scikit-learn 1.4.1 documentation
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3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
1/12
X_test, y_test
train_test_split
C
>>> import
numpy
as
np
>>> from
sklearn.model_selection
import
train_test_split
>>> from
sklearn
import
datasets
>>> from
sklearn
import
svm
>>> X, y = datasets
.
load_iris(return_X_y
=
True
)
>>> X
.
shape, y
.
shape
((150, 4), (150,))
>>> X_train, X_test, y_train, y_test = train_test_split(
... X, y, test_size
=
0.4
, random_state
=
0
)
>>> X_train
.
shape, y_train
.
shape
((90, 4), (90,))
>>> X_test
.
shape, y_test
.
shape
((60, 4), (60,))
>>> clf = svm
.
SVC(kernel
=
'linear'
, C
=
1
)
.
fit(X_train, y_train)
>>> clf
.
score(X_test, y_test)
0.96...
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
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cross_val_score
score
cv
cross_val_score
KFold
StratifiedKFold
ClassifierMixin
>>> from
sklearn.model_selection
import
cross_val_score
>>> clf = svm
.
SVC(kernel
=
'linear'
, C
=
1
, random_state
=
42
)
>>> scores = cross_val_score(clf, X, y, cv
=
5
)
>>> scores
array([0.96..., 1. , 0.96..., 0.96..., 1. ])
>>> print
(
"
%0.2f
accuracy with a standard deviation of %0.2f
" % (scores
.
mean(), scores
.
std()))
0.98 accuracy with a standard deviation of 0.02
>>> from
sklearn
import
metrics
>>> scores = cross_val_score(
... clf, X, y, cv
=
5
, scoring
=
'f1_macro'
)
>>> scores
array([0.96..., 1. ..., 0.96..., 0.96..., 1. ])
>>> from
sklearn.model_selection
import
ShuffleSplit
>>> n_samples = X
.
shape[
0
]
>>> cv = ShuffleSplit(n_splits
=
5
, test_size
=
0.3
, random_state
=
0
)
>>> cross_val_score(clf, X, y, cv
=
cv)
array([0.977..., 0.977..., 1. ..., 0.955..., 1. ])
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
3/12
Pipeline
cross_validate
cross_val_score
['test_score', 'fit_time',
'score_time']
['test_<scorer1_name>', 'test_<scorer2_name>',
'test_<scorer...>', 'fit_time', 'score_time']
return_train_score
False
True
return_estimator=True
re‐
turn_indices=True
>>> def
custom_cv_2folds
(X):
... n = X
.
shape[
0
]
... i = 1
... while
i <= 2
:
... idx = np
.
arange(n * (i - 1
) / 2
, n * i / 2
, dtype
=
int
)
... yield
idx, idx
... i += 1
...
>>> custom_cv = custom_cv_2folds(X)
>>> cross_val_score(clf, X, y, cv
=
custom_cv)
array([1. , 0.973...])
>>> from
sklearn
import
preprocessing
>>> X_train, X_test, y_train, y_test = train_test_split(
... X, y, test_size
=
0.4
, random_state
=
0
)
>>> scaler = preprocessing
.
StandardScaler()
.
fit(X_train)
>>> X_train_transformed = scaler
.
transform(X_train)
>>> clf = svm
.
SVC(C
=
1
)
.
fit(X_train_transformed, y_train)
>>> X_test_transformed = scaler
.
transform(X_test)
>>> clf
.
score(X_test_transformed, y_test)
0.9333...
>>> from
sklearn.pipeline
import
make_pipeline
>>> clf = make_pipeline(preprocessing
.
StandardScaler(), svm
.
SVC(C
=
1
))
>>> cross_val_score(clf, X, y, cv
=
cv)
array([0.977..., 0.933..., 0.955..., 0.933..., 0.977...])
>>> from
sklearn.model_selection
import
cross_validate
>>> from
sklearn.metrics
import
recall_score
>>> scoring = [
'precision_macro'
, 'recall_macro'
]
>>> clf = svm
.
SVC(kernel
=
'linear'
, C
=
1
, random_state
=
0
)
>>> scores = cross_validate(clf, X, y, scoring
=
scoring)
>>> sorted
(scores
.
keys())
['fit_time', 'score_time', 'test_precision_macro', 'test_recall_macro']
>>> scores[
'test_recall_macro'
]
array([0.96..., 1. ..., 0.96..., 0.96..., 1. ])
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3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
4/12
cross_validate
cross_val_predict
cross_val_score
cross_val_predict
cross_val_score
cross_val_score
cross_val_predict
cross_val_predict
cross_val_predict
KFold
>>> from
sklearn.metrics
import
make_scorer
>>> scoring = {
'prec_macro'
: 'precision_macro'
,
... 'rec_macro'
: make_scorer(recall_score, average
=
'macro'
)}
>>> scores = cross_validate(clf, X, y, scoring
=
scoring,
... cv
=
5
, return_train_score
=
True
)
>>> sorted
(scores
.
keys())
['fit_time', 'score_time', 'test_prec_macro', 'test_rec_macro',
'train_prec_macro', 'train_rec_macro']
>>> scores[
'train_rec_macro'
]
array([0.97..., 0.97..., 0.99..., 0.98..., 0.98...])
>>> scores = cross_validate(clf, X, y,
... scoring
=
'precision_macro'
, cv
=
5
,
... return_estimator
=
True
)
>>> sorted
(scores
.
keys())
['estimator', 'fit_time', 'score_time', 'test_score']
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
5/12
KFold
RepeatedKFold
KFold
RepeatedStratifiedKFold
LeaveOneOut
>>> import
numpy
as
np
>>> from
sklearn.model_selection
import
KFold
>>> X = [
"a"
, "b"
, "c"
, "d"
]
>>> kf = KFold(n_splits
=
2
)
>>> for
train, test in
kf
.
split(X):
... print
(
"
%s
%s
" % (train, test))
[2 3] [0 1]
[0 1] [2 3]
>>> X = np
.
array([[
0.
, 0.
], [
1.
, 1.
], [
-
1.
, -
1.
], [
2.
, 2.
]])
>>> y = np
.
array([
0
, 1
, 0
, 1
])
>>> X_train, X_test, y_train, y_test = X[train], X[test], y[train], y[test]
>>> import
numpy
as
np
>>> from
sklearn.model_selection
import
RepeatedKFold
>>> X = np
.
array([[
1
, 2
], [
3
, 4
], [
1
, 2
], [
3
, 4
]])
>>> random_state = 12883823
>>> rkf = RepeatedKFold(n_splits
=
2
, n_repeats
=
2
, random_state
=
random_state)
>>> for
train, test in
rkf
.
split(X):
... print
(
"
%s
%s
" % (train, test))
...
[2 3] [0 1]
[0 1] [2 3]
[0 2] [1 3]
[1 3] [0 2]
>>> from
sklearn.model_selection
import
LeaveOneOut
>>> X = [
1
, 2
, 3
, 4
]
>>> loo = LeaveOneOut()
>>> for
train, test in
loo
.
split(X):
... print
(
"
%s
%s
" % (train, test))
[1 2 3] [0]
[0 2 3] [1]
[0 1 3] [2]
[0 1 2] [3]
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
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LeavePOut
LeaveOneOut
LeaveOneOut
KFold
ShuffleSplit
random_state
ShuffleSplit
ShuffleSplit
KFold
>>> from
sklearn.model_selection
import
LeavePOut
>>> X = np
.
ones(
4
)
>>> lpo = LeavePOut(p
=
2
)
>>> for
train, test in
lpo
.
split(X):
... print
(
"
%s
%s
" % (train, test))
[2 3] [0 1]
[1 3] [0 2]
[1 2] [0 3]
[0 3] [1 2]
[0 2] [1 3]
[0 1] [2 3]
>>> from
sklearn.model_selection
import
ShuffleSplit
>>> X = np
.
arange(
10
)
>>> ss = ShuffleSplit(n_splits
=
5
, test_size
=
0.25
, random_state
=
0
)
>>> for
train_index, test_index in
ss
.
split(X):
... print
(
"
%s
%s
" % (train_index, test_index))
[9 1 6 7 3 0 5] [2 8 4]
[2 9 8 0 6 7 4] [3 5 1]
[4 5 1 0 6 9 7] [2 3 8]
[2 7 5 8 0 3 4] [6 1 9]
[4 1 0 6 8 9 3] [5 2 7]
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3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
7/12
StratifiedKFold
StratifiedShuffleSplit
StratifiedKFold
KFold
StratifiedKFold
RepeatedStratifiedKFold
StratifiedShuffleSplit
>>> from
sklearn.model_selection
import
StratifiedKFold, KFold
>>> import
numpy
as
np
>>> X, y = np
.
ones((
50
, 1
)), np
.
hstack(([
0
] * 45
, [
1
] * 5
))
>>> skf = StratifiedKFold(n_splits
=
3
)
>>> for
train, test in
skf
.
split(X, y):
... print
(
'train - {}
| test - {}
'
.
format(
... np
.
bincount(y[train]), np
.
bincount(y[test])))
train - [30 3] | test - [15 2]
train - [30 3] | test - [15 2]
train - [30 4] | test - [15 1]
>>> kf = KFold(n_splits
=
3
)
>>> for
train, test in
kf
.
split(X, y):
... print
(
'train - {}
| test - {}
'
.
format(
... np
.
bincount(y[train]), np
.
bincount(y[test])))
train - [28 5] | test - [17]
train - [28 5] | test - [17]
train - [34] | test - [11 5]
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
8/12
groups
GroupKFold
GroupKFold
StratifiedGroupKFold
KFold
GroupKFold
KFold
GroupKFold
KFold
shuffle=True
StratifiedGroupKFold
StratifiedKFold
GroupKFold
GroupKFold
>>> from
sklearn.model_selection
import
GroupKFold
>>> X = [
0.1
, 0.2
, 2.2
, 2.4
, 2.3
, 4.55
, 5.8
, 8.8
, 9
, 10
]
>>> y = [
"a"
, "b"
, "b"
, "b"
, "c"
, "c"
, "c"
, "d"
, "d"
, "d"
]
>>> groups = [
1
, 1
, 1
, 2
, 2
, 2
, 3
, 3
, 3
, 3
]
>>> gkf = GroupKFold(n_splits
=
3
)
>>> for
train, test in
gkf
.
split(X, y, groups
=
groups):
... print
(
"
%s
%s
" % (train, test))
[0 1 2 3 4 5] [6 7 8 9]
[0 1 2 6 7 8 9] [3 4 5]
[3 4 5 6 7 8 9] [0 1 2]
>>> from
sklearn.model_selection
import
StratifiedGroupKFold
>>> X = list
(
range
(
18
))
>>> y = [
1
] * 6 + [
0
] * 12
>>> groups = [
1
, 2
, 3
, 3
, 4
, 4
, 1
, 1
, 2
, 2
, 3
, 4
, 5
, 5
, 5
, 6
, 6
, 6
]
>>> sgkf = StratifiedGroupKFold(n_splits
=
3
)
>>> for
train, test in
sgkf
.
split(X, y, groups
=
groups):
... print
(
"
%s
%s
" % (train, test))
[ 0 2 3 4 5 6 7 10 11 15 16 17] [ 1 8 9 12 13 14]
[ 0 1 4 5 6 7 8 9 11 12 13 14] [ 2 3 10 15 16 17]
[ 1 2 3 8 9 10 12 13 14 15 16 17] [ 0 4 5 6 7 11]
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
9/12
GroupKFold
LeaveOneGroupOut
LeavePGroupsOut
n_groups=1
GroupKFold
n_splits
groups
LeaveOneGroupOut
LeavePGroupsOut
LeaveOneGroupOut
>>> from
sklearn.model_selection
import
LeaveOneGroupOut
>>> X = [
1
, 5
, 10
, 50
, 60
, 70
, 80
]
>>> y = [
0
, 1
, 1
, 2
, 2
, 2
, 2
]
>>> groups = [
1
, 1
, 2
, 2
, 3
, 3
, 3
]
>>> logo = LeaveOneGroupOut()
>>> for
train, test in
logo
.
split(X, y, groups
=
groups):
... print
(
"
%s
%s
" % (train, test))
[2 3 4 5 6] [0 1]
[0 1 4 5 6] [2 3]
[0 1 2 3] [4 5 6]
>>> from
sklearn.model_selection
import
LeavePGroupsOut
>>> X = np
.
arange(
6
)
>>> y = [
1
, 1
, 1
, 2
, 2
, 2
]
>>> groups = [
1
, 1
, 2
, 2
, 3
, 3
]
>>> lpgo = LeavePGroupsOut(n_groups
=
2
)
>>> for
train, test in
lpgo
.
split(X, y, groups
=
groups):
... print
(
"
%s
%s
" % (train, test))
[4 5] [0 1 2 3]
[2 3] [0 1 4 5]
[0 1] [2 3 4 5]
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3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
10/12
GroupShuffleSplit
ShuffleSplit
LeavePGroupsOut
LeavePGroupsOut
GroupShuffleSplit
LeavePGroupsOut
PredefinedSplit
test_fold
train_test_split
ShuffleSplit
split()
>>> from
sklearn.model_selection
import
GroupShuffleSplit
>>> X = [
0.1
, 0.2
, 2.2
, 2.4
, 2.3
, 4.55
, 5.8
, 0.001
]
>>> y = [
"a"
, "b"
, "b"
, "b"
, "c"
, "c"
, "c"
, "a"
]
>>> groups = [
1
, 1
, 2
, 2
, 3
, 3
, 4
, 4
]
>>> gss = GroupShuffleSplit(n_splits
=
4
, test_size
=
0.5
, random_state
=
0
)
>>> for
train, test in
gss
.
split(X, y, groups
=
groups):
... print
(
"
%s
%s
" % (train, test))
...
[0 1 2 3] [4 5 6 7]
[2 3 6 7] [0 1 4 5]
[2 3 4 5] [0 1 6 7]
[4 5 6 7] [0 1 2 3]
>>> import
numpy
as
np
>>> from
sklearn.model_selection
import
GroupShuffleSplit
>>> X = np
.
array([
0.1
, 0.2
, 2.2
, 2.4
, 2.3
, 4.55
, 5.8
, 0.001
])
>>> y = np
.
array([
"a"
, "b"
, "b"
, "b"
, "c"
, "c"
, "c"
, "a"
])
>>> groups = np
.
array([
1
, 1
, 2
, 2
, 3
, 3
, 4
, 4
])
>>> train_indx, test_indx = next
(
... GroupShuffleSplit(random_state
=
7
)
.
split(X, y, groups)
... )
>>> X_train, X_test, y_train, y_test = \
... X[train_indx], X[test_indx], y[train_indx], y[test_indx]
>>> X_train
.
shape, X_test
.
shape
((6,), (2,))
>>> np
.
unique(groups[train_indx]), np
.
unique(groups[test_indx])
(array([1, 2, 4]), array([3]))
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
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KFold
ShuffleSplit
TimeSeriesSplit
TimeSeriesSplit
KFold
cv=some_integer
cross_val_score
train_test_split
random_state
None
KFold(..., shuffle=True)
GridSearchCV
fit
random_state
>>> from
sklearn.model_selection
import
TimeSeriesSplit
>>> X = np
.
array([[
1
, 2
], [
3
, 4
], [
1
, 2
], [
3
, 4
], [
1
, 2
], [
3
, 4
]])
>>> y = np
.
array([
1
, 2
, 3
, 4
, 5
, 6
])
>>> tscv = TimeSeriesSplit(n_splits
=
3
)
>>> print
(tscv)
TimeSeriesSplit(gap=0, max_train_size=None, n_splits=3, test_size=None)
>>> for
train, test in
tscv
.
split(X):
... print
(
"
%s
%s
" % (train, test))
[0 1 2] [3]
[0 1 2 3] [4]
[0 1 2 3 4] [5]
3/25/24, 9:05 PM
3.1. Cross-validation: evaluating estimator performance — scikit-learn 1.4.1 documentation
https://scikit-learn.org/stable/modules/cross_validation.html
12/12
permutation_test_score
permutation_test_score
n_permutations
n_permutations
cv
permutation_test_score
permutation_test_score
(n_permutations + 1) * n_cv
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E MASTER - AP Statistics 20-21, section APStats (Sem 2) SP21,
* INSTRUCTOR
Section 8.2: Testing Mean with Known SD (Homework)
Justin Thomas
Springfield Public Schools.MO
A Print Assignment
Current Score
Due Date
QUESTION
5
SAT, MAY 29, 2021
1
3
4
TOTAL SCORE
11:59 PM CDT
POINTS
1/1
1/1
1/1
6/8
4/4
13/15
86.7%
+ Request Extension
Assignment Submission & Scoring
Assignment Submission
For this assignment, you submit answers by question parts. The number of submissions…
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Benign prostatic hyperplasia is a noncancerous
enlargement of the prostate gland that adversely
affects the quality of life (QoL) of millions of men. A
study of minimally invasive procedures for the
treatment for this condition looked at pretreatment
QoL (qol_base) and quality of life after 3 month on
treatment (qol_3mo)
The baseline data for 10 patients and their 3 month
follow-up data is presented below:
MAXFLO_B = maximum urine flow at baseline (urine
flow measurement scale misplaced)
MAXFLO3M = maximum urine flow after 3 months of
treatment
maxflo_b
maxflo3m
7
8
18
8
13
9.
16
11
8
4
12
10
8
14
10
13
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Define Efficiency of GLS (generalized least squares) estimator?
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The average time period that spouses survive after the death of their partner is said to be greater for widows than for widowers. A sociologist who wishes to evaluate such claim obtained survival data (in years) following the death of a spouse on randomly selected 10 widows and 8 widowers.
Software Output:
Shapiro-Wilk TestFor Widows: data: survival p-value = 0.57For Widowers: data: survival p-value = 0.98Levene’s Test for Homogeneity of Variance p-value = 0.98
What is the most appropriate test procedure to test the claim?
a. Z-test
b. Pooled t-test
c. Non-pooled t-test
d. Mann-Whitney Test
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Define Estimation of Dynamic Multipliers Under Strict Exogeneity?
arrow_forward
3association/independencegoodness of fithomogeneity
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Gaydar Exercise 10 in Chapter 1 describes a study thatshowed that heterosexual women, during ovulation, weresignificantly better at correctly identifying the sexualorientation of a man from a photograph of his face than
women who were not ovulating. In other words, ovula-tion improves a woman’s “gaydar.” Near ovulation, on
average women correctly identified the orientation ofabout 65% of the 100 men shown to them. If this is theprobability of correctly identifying the orientation of aman in any given photograph, what is the probability awoman would correctly classify 80 or more of the men(as two women in the study did)?
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Question 3 have many parts please answer them for me!
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Part: 4 / 6
Part 5 of 6
(e) Compute the mean μy. Round the answer to two decimal places.
Mx =
Next Part
X
Ś
WW
Su
2022 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy c
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By using r studio please answer parts a-g
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You can perform this exercise in Minitab? the exercise is already solved!
But I just want to see how to do it in Minitab, please!
Note:
the exercise is already solved! as seen in the image I attached, I just want to see it solved in minitab!
(with pictures, or sreenshots) however, but I want to see the procedure :)
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Benign prostatic hyperplasia is a noncancerous enlargement of the prostate gland that adversely affects the quality of life of millions of men. A study of a minimally invasive procedure for the treatment for this condition looked at pretreatment quality of life (QOL_BASE) and quality of life after three months on treatment (QOL_3MO). The data is shown on the following table: Participant QOL_BASE QOL__3MO 1 2 1 2 4 1 3 3 1 4 4 3 5 5 2 6 6 2 7 4 2 8 4 5 9 3 3 10 3 1 Calculate differences in quality of life scores (delta) for each participant.
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A restaurant chain introduces new dishes on its menu and wants to know if this will increase the number of customers in its branches. According to the data of the chain, 20 students did not suffer changes in their affluence, 6 had a drop in their number of clients and 12 presented An increase with a significance level of 0.05 uses a non-parametric test to reject or accept the null hypothesis that the instruction of new dishes does not generate changes in the flow of diners
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n
texsu.blackboard.com/ultra/courses/ 3217867_1/grades/assessment/ 2701735_1/overvipw/attempt/_1129
Examity::Start Ex...
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Review Example 8.5 and answer the following questions.
EXAMPLE 8.5
Problem
Maps M Gmail
Do you
Additional content
Content
Drag and drop files here or click to add text.
Questions Filter (2) ▼
General Psycholog...
The average earnings per share (EPS) for 10 industrial stocks randomly selected from those listed on
the Dow Jones Industrial Average was found to be X = 1.85 with a standard deviation of s=0.395.
Calculate a 99% confidence interval for the average EPS of all the industrials listed on the DJIA.
a. A confidence interval is a type of estimator. What are we estimating in this example?
b. Name each symbol in the confidence interval given and explain the use of the symbol.
c Suppose you were conducting a two-tailed hypothesis test: Ho: μ = 3.0 at a = .01. What do you conclude?
a. A confidence interval is a type of estimator. What are we estimating in this…
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Female and Smoke are binary indicator variables ( 0 and 1). Age and Education is measured continuously in years. Loan Default is a binary indicator variable (0 = no default, 1=default).
Are the predictor variables in Models A statistically significant at the 5% significance level?
Carefully interpret the coefficients for Female and Education in Models A
Generate a forecast from all 3 models when Age=10, Educ=10, Sex=1, Smoke=0
You are unsure if education has a linear effect on loan default. What are possible transformations you can do to check for non-linearity?
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+
O Standard Checkpoint 7.EE.3: Ma x
didentity
.com/courses/81365/quizzes/126315?module_item_id%3D1353537
G ExploreLearning Gi.
The new generatio
C Cever | Portal
Sp Make Images, Vide.
Test - Google Slides
ntity
Submitted Feb 22 at 2:33pm
. This attempt took 5 minutes.
1/1 pts
Question 1
7.EE.3
Mike's monthly pay after taxes is $2,300.
. He spends 30% of his pay on rent and 10% on
utilities.
He saves 5% of his pay each month.
. He gives of his pay to a local charity.
How much of Mike's pay is left for other purposes?
Correct!
$1035
OMO
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By using r studio answer parts a-g
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- Benign prostatic hyperplasia is a noncancerous enlargement of the prostate gland that adversely affects the quality of life (QoL) of millions of men. A study of minimally invasive procedures for the treatment for this condition looked at pretreatment QoL (qol_base) and quality of life after 3 month on treatment (qol_3mo) The baseline data for 10 patients and their 3 month follow-up data is presented below: MAXFLO_B = maximum urine flow at baseline (urine flow measurement scale misplaced) MAXFLO3M = maximum urine flow after 3 months of treatment maxflo_b maxflo3m 7 8 18 8 13 9. 16 11 8 4 12 10 8 14 10 13arrow_forwardDefine Efficiency of GLS (generalized least squares) estimator?arrow_forwardThe average time period that spouses survive after the death of their partner is said to be greater for widows than for widowers. A sociologist who wishes to evaluate such claim obtained survival data (in years) following the death of a spouse on randomly selected 10 widows and 8 widowers. Software Output: Shapiro-Wilk TestFor Widows: data: survival p-value = 0.57For Widowers: data: survival p-value = 0.98Levene’s Test for Homogeneity of Variance p-value = 0.98 What is the most appropriate test procedure to test the claim? a. Z-test b. Pooled t-test c. Non-pooled t-test d. Mann-Whitney Testarrow_forward
- Define Estimation of Dynamic Multipliers Under Strict Exogeneity?arrow_forward3association/independencegoodness of fithomogeneityarrow_forwardGaydar Exercise 10 in Chapter 1 describes a study thatshowed that heterosexual women, during ovulation, weresignificantly better at correctly identifying the sexualorientation of a man from a photograph of his face than women who were not ovulating. In other words, ovula-tion improves a woman’s “gaydar.” Near ovulation, on average women correctly identified the orientation ofabout 65% of the 100 men shown to them. If this is theprobability of correctly identifying the orientation of aman in any given photograph, what is the probability awoman would correctly classify 80 or more of the men(as two women in the study did)?arrow_forward
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