T-test
Traditionally it is known that a green gram cultivation
yields 12.0 quintals per hectare on an average.
To increase crop yields, scientists have
developed a new variety of green grams, that can supposedly produce more than
the expected average yield of 12 quintals per hectare.
To test the same, this variety of green grams was tested
on 10 randomly selected farmer's fields.
The yield (quintals/hectare) was recorded as: [14.3,12.6,13.7,10.9,13.7,12.0,11.4,12.0,12.6,13.1]
With a 5% significance level, can we
conclude that the average yield of this variety of green gra
ms is more than the expected
yield (12 quintals/hectare)?
Perform an appropriate test and choose the correct option
below :
A.
p_value = 0.033, Average sample yield is
same as expected
B.
p_value = 0.049, Average sample yield is
same as expected
C.
p_value = 0.049, Average sample yield is
greater than expected
D. p_value
= 0.033, Average sample yield is greater than expected
Correct
Option: p_value
= 0.049, Average sample yield is greater than expected
Explanation:
Based on the question, we define our hypothesis as:
We can
solve this using One sample T-test
Since
we will be performing Right Tail Test, we will set alternative =
"greater"
Code:
from scipy.stats import
ttest_1samp
yield_data =
[14.3,12.6,13.7,10.9,13.7,12.0,11.4,12.0,12.6,13.1]
alpha = 0.05 #5% significance level
tstat, pvalue =
ttest_1samp(yield_data, 12.0, alternative = "greater")
print(pvalue)
if pvalue < alpha:
print('Reject H0 ; Yield will be more than 12
quintals per hectare')
else:
print('Fail to reject H0 ; Yield will be 12
quintals per hectare')
Output:
0.04979938002326665
Reject H0 ; Yield will be more than 12 quintals per hectare
Q2. Gym body fat percentage
Samples of Body fat percentages of few gym going men
and women are recorded.
men = [13.3, 6.0, 20.0, 8.0, 14.0, 19.0, 18.0, 25.0,
16.0, 24.0, 15.0, 1.0, 15.0]
women = [22.0, 16.0, 21.7, 21.0, 30.0, 26.0, 12.0,
23.2, 28.0, 23.0]
Perform an appropriate test to check if the mean body
fat percentage of men and women is statistically different.
Assume the significance level to be 5%.
Choose the correct option below :
A.
p_value = 0.03, Body fat percentage of
gym going men and women are same
B.
p_value = 0.01, Body fat percentage of
gym going men and women are same
C.
p_value = 0.03, Body fat percentage of
gym going men and women are statistically different
D. p_value
= 0.01, Body fat percentage of gym going men and women are statistically
different
Correct
Option: p_value
= 0.01, Body fat percentage of gym going men and women are statistically
different
Explanation:
Based on the given problem, we define our hypothesis as:
Since
we need to compare two different samples, whose sample size is small (<30),
we will use Two Sample T-test
Further,
in order to test for the proposed alternate hypothesis, we will have to perform
a Two Tailed Test.
Code:
from scipy.stats import
ttest_ind
# Given samples of gym-going
men and women body fat percentages
men = [13.3, 6.0, 20.0, 8.0,
14.0, 19.0, 18.0, 25.0, 16.0, 24.0, 15.0, 1.0, 15.0]
women = [22.0, 16.0, 21.7,
21.0, 30.0, 26.0, 12.0, 23.2, 28.0, 23.0]
alpha = 0.05 # Significance
level
# Performing 2 Sample
Two-Tailed T-test
t_stats, pvalue =
ttest_ind(men, women, alternative ='two-sided')
print(pvalue)
if pvalue < alpha:
print('Reject H0 ; Body fat percentage of men
and women are different')
else:
print('Fail to reject H0 ; Body fat
percentage of men and women are the same')
Output:
0.010730607904197957
Reject H0 ; Body fat percentage of men and women are different
Q3. Quality assurance
The quality assurance department claims that on average
the non-fat milk contains more than 190 mg of
Calcium per 500 ml packet.
To check this claim 45 packets
of milk are collected and the content of calcium is recorded.
Perform an appropriate test to check the claim with
a 90% confidence level.
data = [193, 321, 222, 158, 176,
149, 154, 223, 233, 177, 280, 244, 138, 210, 167, 129, 254, 167, 194, 191, 128,
191, 144, 184, 330, 216, 212, 142, 216, 197, 231, 133, 205, 192, 195, 243, 224,
137, 234, 171, 176, 249, 222, 234, 191]
Note: Round off the answer to four decimal places.
A. Test
statistic: 1.3689 , Reject null hypothesis
B. Test
statistic: 1.3689 , Fail to reject null hypothesis
C. Test
statistic: 1.2851, Reject null hypothesis
D.
Test statistic: 1.2851 , Fail to reject
null hypothesis
Correct option: Test statistic: 1.3689 , Reject null
hypothesis
Explanation:
Since we need to test the claim made by company, we define our hypothesis as:
We perform one sample
t-test.
Solution Approach code:
from scipy.stats import ttest_1samp
data = pd.Series([193, 321, 222, 158, 176, 149, 154, 223, 233, 177, 280, 244,
138, 210, 167, 129, 254, 167, 194, 191, 128, 191, 144, 184, 330, 216, 212, 142,
216, 197, 231, 133, 205, 192, 195, 243, 224, 137, 234, 171, 176, 249, 222, 234,
191])
print("Observed sample mean = ", round(data.mean(), 2))
t_stat, p_value = ttest_1samp(data, popmean=190,
alternative="greater")
print("Test statistic = ", round(t_stat,4))
print("P-value = ", round(p_value,4))
if p_value < 0.10:
print("Reject H0")
else:
print("Fail to reject H0")
Output:
Observed sample mean = 199.49
Test statistic = 1.3689
P-value = 0.089
Reject H0
Therefore, we conclude that the claim that on average the non-fat milk contains
more than 190 mg of Calcium per 500 ml packet.
Another way to find the T-statistic is using the formula: t = nsx−μ
Q4. Coaching class
There are 8 females and 12
males in a coaching class.
After a practice test, the coach wants to know whether the
average score of females is greater than the average
score of males.
Given data describes the scores of
females and males in his class.
female_scores=[25,30,45,49,47,35,32,42]
male_scores=[45,47,25,22,29,32,27,28,40,49,50,33]
Use an appropriate test to check whether the assumption of
the coach is significant or not, at a 2% significance level?
A.
P_value = 0.580, There is significant
evidence that the average score of females is greater than the average score of
males.
B.
P_value = 0.285, There is no significant
evidence that the average score of females is greater than the average score of
males.
C.
P_value = 0.285, There is significant
evidence that the average score of females is greater than the average score of
males.
D. P_value
= 0.580, There is no significant evidence that the average score of females is
greater than the average score of males.
Correct
option: P_value
= 0.285, There is no significant evidence that the average score of females is
greater than the average score of males.
Explanation:
Based on the given problem, we define our hypothesis as:
import pandas as pd
from scipy.stats import
ttest_ind
female_scores=pd.Series([25,30,45,49,47,35,32,42])
male_scores=pd.Series([45,47,25,22,29,32,27,28,40,49,50,33])
t_stat, p_value =
ttest_ind(female_scores, male_scores, alternative="greater")
print("Test statistic =
", round(t_stat,3))
print("P_value
=",p_value )
alpha=0.02
if p_value < alpha:
print("Reject the null hypothesis.
There is significant evidence that the average score of females is greater than
the average score of males.")
else:
print("Fail to reject the null
hypothesis. There is no significant evidence that the average score of females
is greater than the average score of males.")
Output:
Test statistic = 0.58
P_value = 0.2847023809445894
Fail to reject the null hypothesis. There is no significant evidence that the
average score of females is greater than the average score of males.
Q5. Effectiveness on yield
An experiment was performed to compare the effectiveness
of Ammonium Chloride and urea on
the grain yield (in quintal per hectare) and the results are given in the
arrays below:
Ammonium_chloride = [13.4, 10.9, 11.2, 11.8, 14, 15.3, 14.2, 12.6, 17, 16.2, 16.5, 15.7]
Urea = [12, 11.7, 10.7, 11.2, 14.8, 14.4, 13.9, 13.7, 16.9, 16, 15.6, 16]
Conduct an appropriate test to compare the same with
a 95% confidence level and choose the appropriate
option below.
A. p_val=0.002,
The effect of ammonium chloride and urea on grain yield is not equal
B. p_val=0.855,
The effect of ammonium chloride and urea on grain yield is equal
C. p_val=0.855,The
effect of ammonium chloride and urea on grain yield is not equal
D.
p_val=0.002, The effect of ammonium
chloride and urea on grain yield is equal
Correct
option: p_val=0.855,
its greater than alpha, therefore the null hypothesis that effects are equal is
accepted.
Explanation:
We
define our hypothesis as:
Since
we need to compare two different samples, whose sample size is small (<30),
we will use Two Sample T-test
Further,
in order to test for the proposed alternate hypothesis, we will have to perform
a Two Tailed Test.
Code:
from scipy.stats import
ttest_ind
# Given samples of farm yields
with following chemicals
ammonium_chloride = [13.4,
10.9, 11.2, 11.8, 14, 15.3, 14.2, 12.6, 17, 16.2, 16.5, 15.7]
urea = [12, 11.7, 10.7, 11.2,
14.8, 14.4, 13.9, 13.7, 16.9, 16, 15.6, 16]
alpha = 0.05 # Significance
level
# Performing 2 Sample
Two-Tailed T-test
t_stats, pvalue =
ttest_ind(ammonium_chloride, urea, alternative ='two-sided')
print(pvalue)
if pvalue < alpha:
print('Reject H0 ; The effect of ammonium
chloride and urea on grain yield is not equal')
else:
print('Fail to reject H0 ; The effect of
ammonium chloride and urea on grain yield is equal')
Output:
0.8551954147800473
Fail to reject H0 ; The effect of ammonium chloride and urea on grain yield is
equal
Q6. Zumba trainer
The Zumba trainer claims to the
customers, that their new dance routine helps to reduce more weight.
Weight of 8 people were
recorded before and after following the new Zumba training for a month:
wt_before = [85, 74, 63.5, 69.4, 71.6, 65,90,78]
wt_after = [82, 71, 64, 65.2, 67.8, 64.7,95,77]
Test the trainer's claim
with 90% confidence. Further, what would be the pvalue?
A. P
value: 0.854, Customers did not reduce their weight
B. P
value: 0.145, Customers did not reduce their weight
C. P
value: 0.854, Customers have reduced their weight
D.
P value: 0.145, Customers have reduced
their weight
Correct
option: P
value: 0.145, Customers did not reduce their weight
Explanation:
Based on the given problem, we define our hypothesis as:
The
nature of data here is in the form of “Before” and “After” the new Zumba dance
routine. Hence we will use Paired T-test using ttest_rel()
Further,
in order to test our proposed alternate hypothesis, we would have to perform
Right Tail test
Code:
from scipy.stats import
ttest_rel
# Given samples of weight
before and after 1 month of following new Zumba routine
wt_before = [85, 74, 63.5,
69.4, 71.6, 65,90,78]
wt_after = [82, 71, 64, 65.2,
67.8, 64.7,95,77]
alpha = 0.10 # Significance
level
# Performing Paired
Right-Tailed T-test
t_stat, pvalue =
ttest_rel(wt_before, wt_after, alternative="greater" )
print('Test Statistic:',
t_stat)
print('P value:', pvalue)
if pvalue < alpha:
print('Reject H0 ; Customers have reduced
their weight after 1 month of new Zumba routine')
else:
print('Fail to reject H0 ; Customers did not
reduce their weight after 1 month of new Zumba routine')
Output:
Test Statistic: 1.1421853793555032
P value: 0.14546808501326386
Fail to reject H0 ; Customers did not reduce their weight after 1 month of new
Zumba routine
Q7. The correct test
A certain company decided to roll out a new
training regime for its employees.
To test which regime (old or new) would be preferred by
the employees, they made 5 employees (who had
earlier cleared the old regime) take part in the new training regime, and then
score them both, out of 100.
Which of the following statistical procedures would
be most appropriate to test the
claim that employee overall scores are the same in
both training regimes?
A. Two-tailed
two-sample paired/dependent t-test of means
B. Two-tailed
two-sample independent t-test of means
C. Two-tailed
two-sample independent z-test of means
D. One-tailed
two-sample z-test of means
E.
One-tailed two-sample paired/dependent
t-test of means
Correct
option: Two-tailed
two-sample paired/dependent t-test of means
Explanation:
Null and alternative hypotheses:
We know
that two sample test is done when we are comparing the mean of two groups with
each other. Here the condition for this satisfies.
Similarly, two sided tailed test is used when we evaluate on both sides, and
here again, the condition satisfies to evaluate both the regimes.
And finally, since the same sample group takes part in both the before and
after regimes, we can conclude that we need a Two-tailed two-sample
dependent/paired t-test of means.
Q8. Test the Training
Program
You are appointed as a Data Analyst for a training program
deployed by the Government of India.
The participants’ skills were tested before and after the
training using some metrics on a scale of 10.
before =[2.45, 0.69, 1.80, 2.80, 0.07, 1.67, 2.93, 0.47, 1.45, 1.34]
after =[7.71, 2.17, 5.65, 8.79, 0.23, 5.23, 9.19, 1.49, 4.56, 4.20]
Conduct an appropriate test to assess a
statistically significant increase in the average
skill score after the training program, and then answer the below questions
accordingly.
Note: Perform the test at alpha =
5%.
A.
T statistic = -5.11
B.
There is no effect of the training.
C. There is a
positive effect of the training.
D. We fail to
reject the null hypothesis.
E.
We reject the null hypothesis.
Correct
options:
Explanation:
We propose the Hypothesis as follows:
We can
test this using Paired T-test
Code:
import scipy.stats as stats
# Sample of participants’
skills tested before and after the training
before= [2.45, 0.69, 1.80,
2.80, 0.07, 1.67, 2.93, 0.47, 1.45, 1.34]
after = [7.71, 2.17, 5.65,
8.79, 0.23, 5.23, 9.19, 1.49, 4.56, 4.20]
before_mean =
np.mean(np.array(before))
after_mean =
np.mean(np.array(after))
print("before
mean:",before_mean)
print("after
mean:",after_mean)
# Performing the paired t-test
t_obs, p =
stats.ttest_rel(before, after, alternative="less")
print(" T statistic =
", round(t_obs,2))
print(" p-value = ",
round(p,4))
if(p < 0.05):
print("Since, p-value < alpha, we
reject the null hypothesis. Positive effect / improvement in skills after
training")
else:
print("Since, p-value > alpha, we
fail to reject the null hypothesis. No effect / improvement in skills after
training")
Output:
before mean: 1.567
after mean: 4.922000000000001
T statistic = -5.11
p-value = 0.0003
Since, p-value < alpha, we reject the null hypothesis. Positive effect /
improvement in skills after training.
Q1. Average British man
height
The average British man is 175.3 cm tall.
A survey recorded the heights of 10 UK men and we
want to know whether the mean of the sample is different from the population
mean.
survey_height = [177.3, 182.7,
169.6, 176.3, 180.3, 179.4, 178.5, 177.2, 181.8, 176.5]
Perform an appropriate test and choose the correct option
below, that we can conclude with a 5% significance.
A. p_value
= 0.031, sample mean height is same as the population mean.
B. p_value
= 0.047, sample mean height is same as the population mean.
C. p_value
= 0.047, sample mean height is different from the population mean.
D.
p_value = 0.031, sample mean height is
different from the population mean.
Correct Answer:p_value = 0.047, sample mean height is different
from the population mean.
Explanation:
import numpy as npfrom scipy import stats '''Null Hypothesis (H0): The sample mean height is equal to the population mean (μ = 175.3 cm).Alternate Hypothesis (H1): The sample mean height is different from the population mean (μ ≠ 175.3 cm).''' # Population meanpopulation_mean = 175.3 # Heights of 10 UK mensurvey_height = [177.3, 182.7, 169.6, 176.3, 180.3, 179.4, 178.5, 177.2, 181.8, 176.5] # Perform one-sample t-testt_statistic, p_value = stats.ttest_1samp(survey_height, population_mean,alternative = "two-sided") # Set the significance levelalpha = 0.05 # Print the resultsprint("t-statistic:", t_statistic)print("p-value:", p_value) # Determine the conclusionif p_value < alpha: print("Reject the null hypothesis") print("There is enough evidence to conclude that the sample mean height is different from the population mean.")else: print("Fail to reject the null hypothesis") print("There is no enough evidence to conclude that the sample mean height is different from the population mean.")
Output:
t-statistic: 2.295568968083183
p-value: 0.04734137339747034
Reject the null hypothesis
There is enough evidence to conclude that the sample mean height is different
from the population mean.
Q2. Two Schools IQ
Samples of IQ scores are collected from two competing
schools, as follows:
school_1 = [115, 111, 112, 101, 95, 98, 100, 90, 89, 108]
school_2 = [107, 103, 91, 99, 104, 98, 117, 113, 92, 96, 108, 115, 116, 88]
Perform an appropriate test with a 5% significance level
to check if there is any statistically significant difference in the mean IQ's
of these schools.
Choose the correct option below
:
A. p_value
= 0.56, IQ scores of the two schools are same
B. p_value
= 0.71, IQ scores of the two schools are same
C. p_value
= 0.56, IQ scores of the two schools are statistically different
D.
p_value = 0.71, IQ scores of the two
schools are statistically different
Correct
Option: p_value
= 0.71, IQ of two schools are same
Explanation:
Based on the given problem, we define our hypothesis as:
Since
we need to compare two different samples, whose sample size is small (<30),
we will use Two Sample T-test
Further,
in order to test for the proposed alternate hypothesis, we will have to perform
a Two Tailed Test.
Code:
from scipy.stats import
ttest_ind
# Sample IQ scores
school_1 = [115, 111, 112,
101, 95, 98, 100, 90, 89, 108]
school_2 = [107, 103, 91, 99,
104, 98, 117, 113, 92, 96, 108, 115, 116, 88]
alpha = 0.05 # Significance
level
# Performing Two sample Twp
Tiled T-test
t_stat, pvalue =
ttest_ind(school_1, school_2, alternative ='two-sided')
print(pvalue)
if pvalue < alpha:
print('Reject H0 ; IQ scores of the two
schools are different')
else:
print('Fail to reject H0 ; IQ scores of the
two schools are same')
Output:
0.7154458095186707
Fail to reject H0 ; IQ scores of the two schools are same
Q3. T/Z
Test
The distributions of two variables x and y are
given below in the form of boxplots.
Mark the correct option
regarding variables x and y.
A.
We can use One sample t-test for variable x.
B. We can use
One sample t-test for variable y.
C. We can't
use One sample z-test for variable y.
D. We can’t
use both z-test or t-test for variable x.
E.
We can’t use both z-test or t-test for variable y.
Correct
options:
Explanation:
Distribution
of variable x:
Distribution
of variable y:
Q4. Select
the Test
A pharmaceutical company is developing a new drug to
decrease the blood pressure. In a clinical trial, they measured the blood
pressure of 30 participants before administering the new drug.
After a month of treatment, they measured the blood
pressure again for the same group. The company wants to determine if there is a
significant decrease in blood pressure after using the new drug.
Which hypothesis test should the company use, and what are
the characteristics of the test?
A.
One-tailed,
Paired T-test
B. One-tailed,
One-sample, Z-test
C. One-tailed,
Two-sample, T-test
D. Two-tailed,
Paired T-test
E.
Two-tailed,
Two-sample, T-test
Correct
Option: One-tailed,
Paired T-test
Explanation:
In this scenario, the company is interested in assessing whether there is a
significant decrease in blood pressure after using the new drug.
Null
and Alternative Hypotheses:
The
test is one-tailed because the focus is in a particular direction of the
distribution.