I need help interpreting the results of four hypothesis tests. An explanation of
ID: 3375288 • Letter: I
Question
I need help interpreting the results of four hypothesis tests. An explanation of what the results mean and addressing the null hypothesis, alternative hypothesis, and level of significance.
Results:
Hypothesis test for the difference of two population proportions - Step 2
(1.0561177090573837, 0.29091444062700333)
Hypothesis test for the difference of two population proportions - Step 3
(-2.1328430809234744, 0.032937600869161199)
Hypothesis test for the difference of two population means - Step 4
Ttest_indResult(statistic=1.2930625541846945, pvalue=0.20130524110032363)
Hypothesis test for the difference of two population means - Step 5
Ttest_indResult(statistic=-2.2438440532488544, pvalue=0.02990948832379026)
Problems that led to these reults:
Step 2: Perform hypothesis test for the difference of two population proportions (EMXT)
It is claimed that the proportion of Extreme Maximum Temperature (EMXT) with temperatures over 32.5c (EMXT = 325) is the same for the month of July (Month=7) and August (Month=8). Test this claim using a hypothesis test at 1% level of significance.
Step 3: Perform hypothesis test for the difference of two population proportions (EMXP)
It is claimed that the proportion of Extreme Maximum Precipitation (EMXP) with precipitation over 20.0mm (EMXP = 200) is the same for the month of February (Month=2) and August (Month=8). Test this claim using a hypothesis test at 5% level of significance.
Step 4: Perform hypothesis test for the difference of two population means (EMXT)
It is claimed that the average Extreme Maximum Temperature (EMXT) for July is not the same as for August. Test this claim using a hypothesis test at 5% level of significance.
Step 5: Perform hypothesis test for the difference of two population means (EMXP)
It is claimed that the average Extreme Maximum Precipitation (EMXP) for February is less than August. Test this claim using a hypothesis test at 5% level of significance.
The python code that was used to get the results:
import pandas as pd
import scipy.stats as st
from statsmodels.stats.proportion import proportions_ztest
##Step 1: Import your data set
manchesterweather = pd.read_csv('ManchesterWeather.csv')
####### Step 2: Perform hypothesis test for the difference of two population proportions
print ('Hypothesis test for the difference of two population proportions - Step 2')
n1 = manchesterweather.loc[manchesterweather['Month'] == 7]['EMXT'].count()
n2 = manchesterweather.loc[manchesterweather['Month'] == 8]['EMXT'].count()
x1 = (manchesterweather.loc[manchesterweather['Month'] == 7]['EMXT'] > 325).values.sum()
x2 = (manchesterweather.loc[manchesterweather['Month'] == 8]['EMXT'] > 325).values.sum()
counts = [x1, x2]
n = [n1, n2]
print (proportions_ztest(counts, n))
print ('')
####### Step 3: Perform hypothesis test for the difference of two population proportions
print ('Hypothesis test for the difference of two population proportions - Step 3')
n1 = manchesterweather.loc[manchesterweather['Month'] == 2]['EMXP'].count()
n2 = manchesterweather.loc[manchesterweather['Month'] == 8]['EMXP'].count()
x1 = (manchesterweather.loc[manchesterweather['Month'] == 2]['EMXP'] > 200).values.sum()
x2 = (manchesterweather.loc[manchesterweather['Month'] == 8]['EMXP'] > 200).values.sum()
counts = [x1, x2]
n = [n1, n2]
print (proportions_ztest(counts, n))
print ('')
####### Step 4: Perform hypothesis test for the difference of two population means
print ('Hypothesis test for the difference of two population means - Step 4')
jul_data = manchesterweather.loc[manchesterweather['Month'] == 7]['EMXT']
aug_data = manchesterweather.loc[manchesterweather['Month'] == 8]['EMXT']
print (st.ttest_ind(jul_data, aug_data, equal_var=False))
print ('')
####### Step 5: Perform hypothesis test for the difference of two population means
print ('Hypothesis test for the difference of two population means - Step 5')
feb_data = manchesterweather.loc[manchesterweather['Month'] == 2]['EMXP']
aug_data = manchesterweather.loc[manchesterweather['Month'] == 8]['EMXP']
print (st.ttest_ind(feb_data, aug_data, equal_var=False))
print ('')
Explanation / Answer
The claim is the considered as the alternative hypothesis and opposite of that is the null hypothesis. If the p-value is less than the level of significance we reject the null hypothesis and conclude that statistical significance exists.
Note: You might have p-value for step 2 and step 3 as well, please check. If we have confidence intervals then the test can be interpreted as “Zero is the null value of the parameter (in this case the difference in proportion). If a confidence interval includes the null value, then there is no statistically meaningful or statistically significant difference between the groups.”
Step 2: Perform hypothesis test for the difference of two population proportions (EMXT)
It is claimed that the proportion of Extreme Maximum Temperature (EMXT) with temperatures over 32.5c (EMXT = 325) is the same for the month of July (Month=7) and August (Month=8). Test this claim using a hypothesis test at 1% level of significance.
Interpretation: Since, the confidence interval does not contain 0 that means we can conclude that the proportion of Extreme Maximum Temperature (EMXT) with temperatures over 32.5c (EMXT = 325) is the same for the month of July (Month=7) and August (Month=8).
Step 3: Perform hypothesis test for the difference of two population proportions (EMXP)
It is claimed that the proportion of Extreme Maximum Precipitation (EMXP) with precipitation over 20.0mm (EMXP = 200) is the same for the month of February (Month=2) and August (Month=8). Test this claim using a hypothesis test at 5% level of significance.
Interpretation: Since, the confidence interval contains 0 that means we can conclude that the proportion of Extreme Maximum Precipitation (EMXP) with precipitation over 20.0mm (EMXP = 200) is the same for the month of February (Month=2) and August (Month=8).
Step 4: Perform hypothesis test for the difference of two population means (EMXT)
It is claimed that the average Extreme Maximum Temperature (EMXT) for July is not the same as for August. Test this claim using a hypothesis test at 5% level of significance.
Interpretation: Since the p-value is greater than the level of significance(5% or 0.05) we will conclude that the average Extreme Maximum Temperature (EMXT) for July is same as for August.
Step 5: Perform hypothesis test for the difference of two population means (EMXP)
It is claimed that the average Extreme Maximum Precipitation (EMXP) for February is less than August. Test this claim using a hypothesis test at 5% level of significance.
Interpretation: Since the p-value is less than the level of significance(5% or 0.05) we will conclude that the average Extreme Maximum Precipitation (EMXP) for February is less than August.
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