Module3Assignment
pdf
School
Washington State University *
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Course
319
Subject
Statistics
Date
Apr 3, 2024
Type
Pages
8
Uploaded by SargentMaskHawk21
Module 3 Assignment
DATA 319
Courtney Wilkinson WSU ID: 011685226 10/2/2023
1.
(a) Evaluate T^2 for testing H0: μ = (7, 11)′ using the data X = [[2 12] [8 9] [6 9] [8 10]]
(b) What is the distribution of T^2 for the situation in (a).
(c) Using (a) and (b), test H0 at α
= 0.01 level.
In [52]:
import
numpy as
np
from
scipy.stats import
f
# data
x =
np
.
array
([[
2
, 12
], [
8
, 9
], [
6
, 9
], [
8
, 10
]])
# hypothesized mean vect.
mu =
np
.
array
([
7
, 11
])
# calculate sample mean & covar. matrix
xbar =
np
.
mean
(
x
, axis
=
0
)
S =
np
.
cov
(
x
, rowvar
=
False
)
# number of obs.
n =
x
.
shape
[
0
]
# number of variables
p =
x
.
shape
[
1
]
# calculate Hotelling's T^2 stat.
Tsquared =
n *
np
.
dot
(
np
.
dot
(
xbar -
mu
, np
.
linalg
.
inv
(
S
)), (
xbar -
mu
))
# degrees of freedom
df1 =
p
df2 =
n -
p
# dalculate critical value from F-dist. at alpha = 0.01
alpha =
0.01
critical_value =
f
.
ppf
(
1 -
alpha
, df1
, df2
)
# calculate the p-value
p_value =
1 -
f
.
cdf
(
Tsquared
, df1
, df2
)
# print
print
(
"Hotelling's T^2 statistic:"
, Tsquared
)
print
(
"Critical value:"
, critical_value
)
print
(
"P-value:"
, p_value
)
# perform hypothesis test
if
Tsquared >=
critical_value
:
print
(
"Reject H0: The means are not equal at 0.01 significance level."
)
else
:
print
(
"Fail to reject H0: The means are equal at 0.01 significance level."
)
Hotelling's T^2 statistic: 13.63636363636363
Critical value: 98.99999999999991
P-value: 0.06832298136645965
Fail to reject H0: The means are equal at 0.01 significance level.
2. For the following two variables [X Y] [2 2] [0 0] [−1 3] [0 1] [0 1] [0 1] [1 −1] [1 0]
Perform a hypothesis test to check whether the population means of the two variables are
the same.
t-stat: -0.9142324078276749
p-value: 0.37607193633458214
Fail to reject H0. The means are not different.
In [53]:
import
scipy.stats as
stats
# data
x =
[
2
, 0
, -
1
, 0
, 0
, 0
, 1
, 1
]
y =
[
2
, 0
, 3
, 1
, 1
, 1
, -
1
, 0
]
# perform two-sample t-test
t_stat
, p_value =
stats
.
ttest_ind
(
x
, y
)
# significance level
alpha =
0.05
# print
print
(
"t-stat:"
, t_stat
)
print
(
"p-value:"
, p_value
)
# pompare p-value to alpha & make decision
if
p_value <
alpha
:
print
(
"Reject H0. The means are different."
)
else
:
print
(
"Fail to reject H0. The means are not different."
)
3. In the first phase of a study of the cost of transporting milk from farms to dairy plant, a
survey was taken of firms engaged in milk transportation. Cost data on X1=fuel, X2=repair,
and X3=capital, all measured on a per-mile basis, are presented in the attached data
dairy.csv, for n1 = 36 gasoline and n2 = 23 diesel trucks.
(a) Perform Hotellings T2 test on the dairy data, for the three vectors: μ0 = (12, 8, 10) μ1 =
(10, 18, 10) μ2 = (11, 9, 13)
(b) Compute individual confidence intervals for the mean of each of the three variables at a
confidence level of 95%. Then compute the simultaneous T2 confidence intervals and
summarize the results of these tests in your own words.
(c) Using the Bonferroni correction, compute confidence intervals for the three variables so
that the simultaneous coverage of these intervals is 95%. Describe in your own words how
these intervals differ from those computed in Part (B).
(d) Check for normality of the data set and comment on what you observe. Are there any
concerning outliers? What if you distinguish between the gasoline and diesel trucks?
In [54]:
import
numpy as
np
import
pandas as
pd
from
scipy.stats import
t
, f
, shapiro
import
matplotlib.pyplot as
plt
# data from dairy.csv
data =
pd
.
read_csv
(
'dairy.csv'
)
# define means for each group
mu0 =
np
.
array
([
12
, 8
, 10
])
mu1 =
np
.
array
([
10
, 18
, 10
])
mu2 =
np
.
array
([
11
, 9
, 13
])
# separate data
gasoline_data =
data
[
data
[
'Type'
] ==
'gasoline'
]
.
iloc
[:, :
-
1
]
.
values
diesel_data =
data
[
data
[
'Type'
] ==
'diesel'
]
.
iloc
[:, :
-
1
]
.
values
# sample means and covar. matrices
sample_means_gasoline =
np
.
mean
(
gasoline_data
, axis
=
0
)
sample_means_diesel =
np
.
mean
(
diesel_data
, axis
=
0
)
sample_cov_matrix_gasoline =
np
.
cov
(
gasoline_data
, rowvar
=
False
)
sample_cov_matrix_diesel =
np
.
cov
(
diesel_data
, rowvar
=
False
)
## Step A
# Hotelling's T^2 statistic
d =
len
(
mu0
)
n1 =
len
(
gasoline_data
)
n2 =
len
(
diesel_data
)
T2_gasoline =
(
n1 -
d
) /
((
n1 +
n2 -
2
) *
d
) *
np
.
matmul
(
np
.
matmul
((
sample_means_gasolin
T2_diesel =
(
n2 -
d
) /
((
n1 +
n2 -
2
) *
d
) *
np
.
matmul
(
np
.
matmul
((
sample_means_diesel -
# critical F-value for alpha = 0.05 and degrees of freedom
F_critical =
f
.
ppf
(
0.95
, d
, n1 +
n2 -
d -
1
)
print
(
"Step A"
)
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print
(
"Hotelling's T^2 statistic (Gasoline):"
, T2_gasoline
)
print
(
"Hotelling's T^2 statistic (Diesel):"
, T2_diesel
)
print
(
"Critical F-value:"
, F_critical
)
print
()
## Step B
# individual confidence intervalsat a confidence level of 95%
alpha =
0.05
conf_intervals_gasoline =
[]
conf_intervals_diesel =
[]
for
i in
range
(
d
):
std_error_gasoline =
np
.
sqrt
(
sample_cov_matrix_gasoline
[
i
][
i
] /
n1
)
margin_error_gasoline =
t
.
ppf
(
1 -
alpha /
2
, n1 -
1
) *
std_error_gasoline
conf_intervals_gasoline
.
append
((
sample_means_gasoline
[
i
] -
margin_error_gasoline
, sa
std_error_diesel =
np
.
sqrt
(
sample_cov_matrix_diesel
[
i
][
i
] /
n2
)
margin_error_diesel =
t
.
ppf
(
1 -
alpha /
2
, n2 -
1
) *
std_error_diesel
conf_intervals_diesel
.
append
((
sample_means_diesel
[
i
] -
margin_error_diesel
, sample_m
# simultaneous T^2 confidence intervals
conf_intervals_simultaneous_gasoline =
[]
conf_intervals_simultaneous_diesel =
[]
for
i in
range
(
d
):
radius_gasoline =
np
.
sqrt
(
F_critical *
(
sample_cov_matrix_gasoline
[
i
][
i
] /
n1 +
samp
conf_intervals_simultaneous_gasoline
.
append
((
sample_means_gasoline
[
i
] -
radius_gasol
radius_diesel =
np
.
sqrt
(
F_critical *
(
sample_cov_matrix_gasoline
[
i
][
i
] /
n1 +
sample
conf_intervals_simultaneous_diesel
.
append
((
sample_means_diesel
[
i
] -
radius_diesel
, s
print
(
"Step B"
)
print
(
"Individual Confidence Intervals (Gasoline):"
)
for
i in
range
(
d
):
print
(
f"Variable {
i +
1
}: {
conf_intervals_gasoline
[
i
]
}"
)
print
()
print
(
"Simultaneous T^2 Confidence Intervals (Gasoline):"
)
for
i in
range
(
d
):
print
(
f"Variable {
i +
1
}: {
conf_intervals_simultaneous_gasoline
[
i
]
}"
)
print
()
print
(
"Individual Confidence Intervals (Diesel):"
)
for
i in
range
(
d
):
print
(
f"Variable {
i +
1
}: {
conf_intervals_diesel
[
i
]
}"
)
print
()
print
(
"Simultaneous T^2 Confidence Intervals (Diesel):"
)
for
i in
range
(
d
):
print
(
f"Variable {
i +
1
}: {
conf_intervals_simultaneous_diesel
[
i
]
}"
)
print
()
## Step C
# Bonferroni correction
conf_intervals_bonferroni_gasoline =
[]
conf_intervals_bonferroni_diesel =
[]
for
i in
range
(
d
):
alpha_bonferroni =
alpha /
d
std_error_gasoline =
np
.
sqrt
(
sample_cov_matrix_gasoline
[
i
][
i
] /
len
(
gasoline_data
))
margin_error_gasoline =
t
.
ppf
(
1 -
alpha_bonferroni /
2
, len
(
gasoline_data
) -
1
) *
st
conf_intervals_bonferroni_gasoline
.
append
((
sample_means_gasoline
[
i
] -
margin_error_g
std_error_diesel =
np
.
sqrt
(
sample_cov_matrix_diesel
[
i
][
i
] /
len
(
diesel_data
))
margin_error_diesel =
t
.
ppf
(
1 -
alpha_bonferroni /
2
, len
(
diesel_data
) -
1
) *
std_er
conf_intervals_bonferroni_diesel
.
append
((
sample_means_diesel
[
i
] -
margin_error_diese
print
(
"Step C"
)
print
(
"Bonferroni Confidence Intervals (Gasoline):"
)
for
i in
range
(
d
):
print
(
f"Variable {
i +
1
}: {
conf_intervals_bonferroni_gasoline
[
i
]
}"
)
print
()
print
(
"Bonferroni Confidence Intervals (Diesel):"
)
for
i in
range
(
d
):
print
(
f"Variable {
i +
1
}: {
conf_intervals_bonferroni_diesel
[
i
]
}"
)
print
()
print
(
"Step D"
)
# check normality and outliers
for
col in
data
.
columns
[:
-
1
]:
_
, p_gasoline =
shapiro
(
gasoline_data
[:, data
.
columns
.
get_loc
(
col
)])
_
, p_diesel =
shapiro
(
diesel_data
[:, data
.
columns
.
get_loc
(
col
)])
print
(
f"Shapiro-Wilk Test {
col
} (Gasoline):"
)
print
(
f"p-value: {
p_gasoline
}"
)
print
(
f"Shapiro-Wilk Test {
col
} (Diesel):"
)
print
(
f"p-value: {
p_diesel
}"
)
print
()
Step A
Hotelling's T^2 statistic (Gasoline): 0.0032994325402454757
Hotelling's T^2 statistic (Diesel): 0.32655143827401384
Critical F-value: 2.772536907836251
Step B
Individual Confidence Intervals (Gasoline):
Variable 1: (10.59546393420547, 13.84175828801675)
Variable 2: (6.695292126144336, 9.529707873855669)
Variable 3: (8.325941954456459, 10.854613601099098)
Simultaneous T^2 Confidence Intervals (Gasoline):
Variable 1: (10.702620402883486, 13.734601819338735)
Variable 2: (5.998878093440173, 10.226121906559833)
Variable 3: (7.001968782535609, 12.178586773019948)
Individual Confidence Intervals (Diesel):
Variable 1: (9.202466906896884, 11.008837440929202)
Variable 2: (8.563509689529011, 12.960838136557948)
Variable 3: (15.21413851159266, 21.121513662320382)
Simultaneous T^2 Confidence Intervals (Diesel):
Variable 1: (8.58966146568542, 11.621642882140666)
Variable 2: (8.648552006483651, 12.875795819603308)
Variable 3: (15.57951709171435, 20.75613508219869)
Step C
Bonferroni Confidence Intervals (Gasoline):
Variable 1: (10.208140888536283, 14.229081333685938)
Variable 2: (6.357111289644602, 9.867888710355402)
Variable 3: (8.024240168687994, 11.156315386867563)
Bonferroni Confidence Intervals (Diesel):
Variable 1: (8.977162616266776, 11.23414173155931)
Variable 2: (8.015041394813283, 13.509306431273677)
Variable 3: (14.477325691218773, 21.858326482694267)
Step D
Shapiro-Wilk Test Fuel (Gasoline):
p-value: 9.55536961555481e-05
Shapiro-Wilk Test Fuel (Diesel):
p-value: 0.5117290019989014
Shapiro-Wilk Test Repair (Gasoline):
p-value: 0.2623240351676941
Shapiro-Wilk Test Repair (Diesel):
p-value: 0.5000005960464478
Shapiro-Wilk Test Capital (Gasoline):
p-value: 0.45318958163261414
Shapiro-Wilk Test Capital (Diesel):
p-value: 0.6583071947097778
Question #3 Write Up
Step A: For Gasoline: The T^2 statistic is very low (0.0033). For Diesel: The
T^2 statistic is higher (0.3266). The Critical F-value (2.7725) is a value used to determine
significance. If the T^2 statistic exceeds this value, it indicates a significant difference between the
groups.
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Step B: Individual Confidence Intervals: These intervals provide a 95% confidence range for each
cost variable for both types of truck separately. Simultaneous T^2 Confidence Intervals: These
intervals provide a 95% confidence range for the combined variation of all three cost variables for
the types of trucks.
Step C: With the Bonferroni correction, the intervals are more conservative to account for the
increased probability of making a false positive when comparing multiple intervals. The intervals
in Part C tend to be wider than those in Part B because they are adjusted to be more cautious.
Step D: Outliers are more apparent in the gasoline truck group, particularly for the "Fuel" and
"Repair" variables. The low p-values in these tests indicate that the data may have outliers. The
diesel truck group generally shows data that is more likely to follow a normal distribution, as
indicated by the higher p-values in the Shapiro-Wilk tests.
4. Consider the following multivariate dataset X = [2 2 3] [0 0 2] [−1 3 2] [0 1 1] [0 1 5] [0 1
3] [1 −1 3] [1 0 5] and test the hypothesis H0 : (μ1 − μ2, μ2 − μ3)′ = (0, 0)′ using
differences.
In [51]:
import
numpy as
np
from
scipy import
stats
# data
x =
np
.
array
([[
2
, 2
, 3
],
[
0
, 0
, 2
],
[
-
1
, 3
, 2
],
[
0
, 1
, 1
],
[
0
, 1
, 5
],
[
0
, 1
, 3
],
[
1
, -
1
, 3
],
[
1
, 0
, 5
]])
# split the data
group1 =
x
[:, 0
]
group2 =
x
[:, 1
]
group3 =
x
[:, 2
]
# differences between means
diff_mean_1_2 =
np
.
mean
(
group1
) -
np
.
mean
(
group2
)
diff_mean_2_3 =
np
.
mean
(
group2
) -
np
.
mean
(
group3
)
# t-tests on the differences
t_stat_1_2
, p_value_1_2 =
stats
.
ttest_ind
(
group1
, group2
)
t_stat_2_3
, p_value_2_3 =
stats
.
ttest_ind
(
group2
, group3
)
# significance level
alpha =
0.05
# print
print
(
"Group 1 & Group 2:"
)
print
(
f'Difference of means: {
diff_mean_1_2
}'
)
print
(
f't-stat: {
t_stat_1_2
}'
)
print
(
f'p-value: {
p_value_1_2
}'
)
if
p_value_1_2 <
alpha
:
print
(
'Reject H0 for Group 1 & Group 2'
)
Group 1 & Group 2:
Difference of means: -0.5
t-stat: -0.9142324078276749
p-value: 0.37607193633458214
Fail to reject H0 for Group & Group 2
Group 2 & Group 3:
Difference of means: -2.125
t-stat: -3.1883897418177476
p-value: 0.006570509597450871
Reject H0 for Group 2 & Group 3
else
:
print
(
'Fail to reject H0 for Group & Group 2'
)
print
()
print
(
'Group 2 & Group 3:'
)
print
(
f'Difference of means: {
diff_mean_2_3
}'
)
print
(
f't-stat: {
t_stat_2_3
}'
)
print
(
f'p-value: {
p_value_2_3
}'
)
if
p_value_2_3 <
alpha
:
print
(
'Reject H0 for Group 2 & Group 3'
)
else
:
print
(
'Fail to reject H0 for Group 2 & Group 3'
)
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Author:HOUGHTON MIFFLIN HARCOURT
Publisher:Houghton Mifflin Harcourt
Related Questions
- Every year, the students at a school are given a musical aptitude test that rates them from 0 (no musical aptitude) to 5 (high musical aptitude). This year's results were: Aptitude Score 0 1 2 3 4 5 Frequency 4 1 1 4 4 1 The mean (T) aptitude score: The median aptitude score: (Please show your answer to 1 decimal place.) (Please separate your answers by ',' in bimodal situation. Enter DNE The mode aptitude score: if there is no mode.) Textbook Measures of Centerarrow_forwardEvery year, the students at a school are given a musical aptitude test that rates them from 0 (no musical aptitude) to 5 (high musical aptitude). This year's results were: Aptitude Score Frequency 1 2 3 4 4 1 5 The mean (x) aptitude score: (Please show your answer to 1 decimal place.) The median aptitude score: The mode aptitude score: (Please separate your answers by ',' in bimodal situation. Enter DNE if there ie no medearrow_forwardTotal plasma volume is important in determining the required plasma component in blood replacement therapy for a person undergoing surgery. Plasma volume is influenced by the overall health and physical activity of an individual. Suppose that a random sample of 46 male firefighters are tested and that they have a plasma volume sample mean of x = 37.5 ml/kg (milliliters plasma per kilogram body weight). Assume that ? = 7.80 ml/kg for the distribution of blood plasma. (b) What conditions are necessary for your calculations? (Select all that apply.)arrow_forward
- 5. Oxygen demand is a term blologists use to describe the axygen needed by fish and other aquatic organisms for survval. The Environmental Protection Agency conducted a study of a wetland area in Marin County, Californla. In this wetland environment, the mean oxygen demand was u=9.9 mg/L, and a= 1.7 mg/L. (Reference: EPA Report 832-R- 93- 005). Let x be a random varlable that represents oxygen demand in this wetland environment. Assume x has a probability distribution that is approximately normal. a. An oxygen demand below 8 mg/L Indicates that some organisms In the wetland environment may be dying. What is the probability that the oxygen demand will fall below 8 mg/L? b. A high oxygen demand can also indicate trouble. An oxygen demand above 12 mg/L may indicate an overabundance of organisms that endanger some types of plant life. What is the probability that the oxygen demand will exceed 12 mg/L?arrow_forwardThe variation of Y around its mean can be decomposed into which two parts?arrow_forwardQ3arrow_forward
- 10) A regression was run to determine if there is a relationship between hours of TV watched per day (x) and number of situps a person can do (y).The results of the regression were:y=ax+b a=-0.767 b=31.009 r2=0.609961 r=-0.781 Use this to predict the number of situps a person who watches 7.5 hours of TV can do (to one decimal place)arrow_forwardApproximate the population mean and standard deviation of age for females. By the way, we find the midpoints by adding the first numbers before the hyphens in the age column and then dividing by 2 (ex: 0+10/2=5, 10+20/2=15, 20+30/2=25).arrow_forward) You are considering the relationship between annual returns on the S&P 500 index (January 31 to January 31) and annual changes in the unemployment rate. You define:S = annual % change in the S&P500 (SPX)U = annual % change in the unemployment rateYou consider the following univariate relationship:Si = b0 + b1Ui + εiYou have data on annual changes in the unemployment rate and the S&P500 (SPX) from 2002 through 2020 (19 observations) and you want to use the data to estimate the regression coefficients (intercept b0 and slope b1)You are given the following statistics: Statistic Value Cov(U,S) -0.025043 Var(U) 0.034590 E(U) -0.001530 E(S) 0.064267 TSS 0.627907 RSS 0.326368 Compute the estimated coefficients for the intercept (b0) and slope (b1) and the R2 So far in 2021, the unemployment rate has decreased by 24% (from 6.3 percent to 4.8 percent). If the unemployment rate holds steady until the end of the year, resulting in an…arrow_forward
- The Variance Inflationary Factor (VIF) measures the O standard deviation of the slope. correlation of the X variables with the Y variable. contribution of each X variable with the Y variable after all other X variables are included in the model. O correlation of the X variables with each other.arrow_forwardFind mean and standard deviation for each uniform continuous model Mean place and deviation answers to 4 decimal places Standard deviation A.(3, 13); B.(90, 190); C.(3,99)arrow_forward
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