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Step 1: Data preparation in a excel table:
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Step 2: how to conduct and where to quit, and Should perform
Welch’s t-test
!
#pip install scipy
#pip install openpyxl
import pandas as pd
import numpy as np
# read the data from excel
group1 = np.array(pd.read_excel('C:/Users/Mr.R/Desktop/excels/zm.xlsx', sheet_name='1', usecols='C'))
group2 = np.array(pd.read_excel('C:/Users/Mr.R/Desktop/excels/zm.xlsx', sheet_name='1', usecols='E'))
print(group1, group2)
import scipy.stats as stats
#find variance for each group
# As a rule of thumb, we can assume the populations have equal variances
# if the ratio of the larger sample variance to the smaller sample variance is less than 4:1.
print("\nVariances of the groups are:\nnp.var(group1):", np.var(group1), "\nnp.var(group2):", np.var(group2))
# calculate the ratio of large variance to smaller variance
if np.var(group1) > np.var(group2):
ratio = np.var(group1)/np.var(group2)
else:
ratio = np.var(group2)/np.var(group1)
#calculate whether can do two-sample t-test
if ratio < 4.0:
print("\n Calculated large/small ratio is :", ratio)
print("\n Yeah! The population variances are equal.\n Let us perform the two sample t-test!") # change the line
else:
print("\n Calculated large/small ratio is :", ratio)
print("\nOops! The population variances are not equal,can not proceed the two sample t-test!\n Should perform Welch’s t-test!")
#The ratio of the larger sample variance to the smaller sample variance
# is 17.333333333333332/11.209876543209877 = 1.5462555066079293, which is less than 4.
#This means we can assume that the population variances are equal.
#so when it is right to do the 2-sample t-test then we do!
if ratio < 4.0:
import scipy.stats as stats
#perform two sample t-test with equal variances
two_sample_t_test_result = stats.ttest_ind(a=group1, b=group2, equal_var=True)
print("\n Here is the two sample t-test result:\n", two_sample_t_test_result)
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Step3: Interpret the results.:
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The two hypotheses for this particular two sample t-test are as follows:
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H0: µ1 = µ2 (the two population means are equal)
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HA: µ1 ≠µ2 (the two population means are not equal)
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Because the
p-value of our test (0.64428611) is greater than alpha = 0.05
, we fail to reject the null hypothesis of the test. Wedo not have sufficient evidence
to say that themean height of plants between the two populations is different
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