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统计代写-H0
时间：2021-11-08
Analysis
I: The null hypothesis of this statement will be H0: SHi = Kuj and the
alternative hypothesis H1:SHi ≠ Kuj. The null hypothesis means there is no
significant difference between average air temperature of Shanghai and
Kusatsu in August 2019.
The sample size n is 31 days for both Shanghai and Kusatsu city.
Mean of sample ̅SH ≈ 84.6639, ̅KU ≈ 86.4113
Standard Deviation of sample SSH ≈ 3.7790,SKU ≈ 4.8410
The significant level = 0.05.
Standard Error Mean is SESH ≈ 0.6787,SEKU ≈ 0.8695
Standard Error Difference is SEDiff ≈ 1.1030
Statistic z ≈ (86.4113- 84.6639)/1.1030 ≈ 1.5842
Critical value Z0.025 ≈ 1.9599
1.5842∈[-1.9599, 1.9599], so accept H0
Two tailed test: p-value ≈ 0.1131
In this case, H0 is accepted and H1 is rejected.
II: The null hypothesis of this statement will be H0: = 0 and the
alternative hypothesis H1: ≠ 0 The null hypothesis means there is no
significant difference between average air temperature of Shanghai in
October 2020 and 2018
The sample size n is 31 days.
The ̅ ≈ 0.4632
Standard Deviation of sample Sd ≈ 3.6425
The significant level = 0.05.
Standard Error Mean is SEmean ≈ 0.6542
Statistic z ≈ 0.4632/0.6542 ≈ 0.7081
Critical value Z0.025 ≈ 1.9599
0.7081∈[-1.9599, 1.9599], so accept H0
Two tailed test: p-value ≈ 0.4789
In this case, H0 is accepted and H1 is rejected.
III: The null hypothesis of this statement will be H0: = 0 and the
alternative hypothesis H1: ≠ 0 The null hypothesis means there is no
significant difference between average air temperature of Kusatsu city in
December 2018 and 2019.
The sample size n is 31 days.
The ̅ ≈ 0.3577
Standard Deviation of sample Sd ≈ 7.3113
The significant level = 0.05.
Standard Error Mean is SEmean ≈ 1.3132
Statistic z ≈ 0.3577/1.3132 ≈ 0.2724
Critical value Z0.025 ≈ 1.9599
0.2724∈[-1.9599, 1.9599], so accept H0
Two tailed test: p-value ≈ 0.7853
In this case, H0 is accepted and H1 is rejected.
Experiment on Average Air Temperature in Shanghai and Kusatsu City
WANG Guan 26001904590
Introduction
This experiment focusing on verifying the following statements:
I: On average, there is no difference on air temperature in August 2019
between Shanghai and Kusatsu city
II: On average, October was colder in 2020 than 2018 in Shanghai
III: On average, there is no difference on air temperature in December
2018 and 2019 in Kusatsu City
Method
Three different methods will be used for verifying three statements. Since
the sample size for all the data is bigger than 30, the experiment will use
z-test instead of t-test. Following is the method will be used for testing
three statements, the reasons why this method is used and the sample of
the data.
I: Independent two-sample z-test because the data is from two
populations (Shanghai and Kusatsu) at same time period (August).
The sample consists of two groups of independent data SHi & Kuj where i
= 1……nSH and j = 1……nKU .
II: Paired sample z-test because the data is from same population
(Shanghai) at different time period (October 2020 and October 2018). The
sample consists of two sets of mutually-dependent data SH20i & SH18i
where i = 1……n
III: Paired sample z-test because the data is from same population
(Kusatsu) at different time period (December 2019 and December 2018).
The sample consists of two sets of mutually-dependent dataKU18i & KU19i
where i = 1……n
Result
From the data analysis, it is suggested that all three statements are
accepted. Therefore, it means that for statement:
I: On average, there is no significant difference between temperature of
Shanghai and Kusatsu city in August 2019.
II: On average, there is no significant difference between temperature of
Shanghai in October 2020 and 2018 which means October 2020 was not
colder than October 2018.
III: On average, there is no significant difference between temperature of
Kusatsu city in December 2018 and 2019.
Conclusion
From the experiment, there is no significant difference of temperature between Shanghai and Kusatsu city. This result is expected since two cities have
similar longitude and latitude which the climate of two cities should be similar. Moreover, there is no significant climate change in the same city through
1-2 years. It can be said that there is not any factors, for example global warming, result in climate changes in both cities in past two years.
Data
The data in the graph below shows the average air temperature of each
day in designated month and city. All the following data is from
https://www.wunderground.com/history. (Unit: Fahrenheit)
The box-whisker graph shows the
overall data. The three histograms
show the data necessary for each
statement separately.
nalysis
I: The nu l hypothesis of this state ent i l be H0: SHi = Kuj and the
alternative hypothesis H1:SHi ≠ Kuj. The nu l hypothesis eans there is no
significant di ference bet een average air te perature of Shanghai and
Kusatsu in August 2019.
The sa ple size n is 31 days for both Shanghai and Kusatsu city.
ean of sa ple ̅SH ≈ 84.6639, ̅KU ≈ 86.4113
Standard Deviation of sa ple SSH ≈ 3.7790,SKU ≈ 4.8410
The significant level = 0.05.
Standard Error ean is SESH ≈ 0.6787,SEKU ≈ 0.8695
Standard Error Di ference is SEDi f ≈ 1.1030
Statistic z ≈ (86.4113- 84.6639)/1.1030 ≈ 1.5842
Critical value Z0.025 ≈ 1.9599
1.5842∈[-1.9599, 1.9599], so accept H0
T o tailed test: p-value ≈ 0.1131
In this case, H0 is accepted and H1 is rejected.
I: The nu l hypothesis of this state ent i l be H0: = 0 and the
alternative hypothesis H1: ≠ 0 The nu l hypothesis eans there is no
significant di ference bet een average air te perature of Shanghai in
October 2020 and 2018
The sa ple size n is 31 days.
The ̅ ≈ 0.4632
Standard Deviation of sa ple Sd ≈ 3.6425
The significant level = 0.05.
Standard Error ean is SEmean ≈ 0.6542
Statistic z ≈ 0.4632/0.6542 ≈ 0.7081
Critical value Z0.025 ≈ 1.9599
0.7081∈[-1.9599, 1.9599], so accept H0
T o tailed test: p-value ≈ 0.4789
In this case, H0 is accepted and H1 is rejected.
I: The nu l hypothesis of this state ent i l be H0: = 0 and the
alternative hypothesis H1: ≠ 0 The nu l hypothesis eans there is no
significant di ference bet een average air te perature of Kusatsu city in
Dece ber 2018 and 2019.
The sa ple size n is 31 days.
The ̅ ≈ 0.3577
Standard Deviation of sa ple Sd ≈ 7.3113
The significant level = 0.05.
Standard Error ean is SEmean ≈ 1.3132
Statistic z ≈ 0.3577/1.3132 ≈ 0.2724
Critical value Z0.025 ≈ 1.9599
0.2724∈[-1.9599, 1.9599], so accept H0
T o tailed test: p-value ≈ 0.7853
In this case, H0 is accepted and H1 is rejected.
学霸联盟
I: The null hypothesis of this statement will be H0: SHi = Kuj and the
alternative hypothesis H1:SHi ≠ Kuj. The null hypothesis means there is no
significant difference between average air temperature of Shanghai and
Kusatsu in August 2019.
The sample size n is 31 days for both Shanghai and Kusatsu city.
Mean of sample ̅SH ≈ 84.6639, ̅KU ≈ 86.4113
Standard Deviation of sample SSH ≈ 3.7790,SKU ≈ 4.8410
The significant level = 0.05.
Standard Error Mean is SESH ≈ 0.6787,SEKU ≈ 0.8695
Standard Error Difference is SEDiff ≈ 1.1030
Statistic z ≈ (86.4113- 84.6639)/1.1030 ≈ 1.5842
Critical value Z0.025 ≈ 1.9599
1.5842∈[-1.9599, 1.9599], so accept H0
Two tailed test: p-value ≈ 0.1131
In this case, H0 is accepted and H1 is rejected.
II: The null hypothesis of this statement will be H0: = 0 and the
alternative hypothesis H1: ≠ 0 The null hypothesis means there is no
significant difference between average air temperature of Shanghai in
October 2020 and 2018
The sample size n is 31 days.
The ̅ ≈ 0.4632
Standard Deviation of sample Sd ≈ 3.6425
The significant level = 0.05.
Standard Error Mean is SEmean ≈ 0.6542
Statistic z ≈ 0.4632/0.6542 ≈ 0.7081
Critical value Z0.025 ≈ 1.9599
0.7081∈[-1.9599, 1.9599], so accept H0
Two tailed test: p-value ≈ 0.4789
In this case, H0 is accepted and H1 is rejected.
III: The null hypothesis of this statement will be H0: = 0 and the
alternative hypothesis H1: ≠ 0 The null hypothesis means there is no
significant difference between average air temperature of Kusatsu city in
December 2018 and 2019.
The sample size n is 31 days.
The ̅ ≈ 0.3577
Standard Deviation of sample Sd ≈ 7.3113
The significant level = 0.05.
Standard Error Mean is SEmean ≈ 1.3132
Statistic z ≈ 0.3577/1.3132 ≈ 0.2724
Critical value Z0.025 ≈ 1.9599
0.2724∈[-1.9599, 1.9599], so accept H0
Two tailed test: p-value ≈ 0.7853
In this case, H0 is accepted and H1 is rejected.
Experiment on Average Air Temperature in Shanghai and Kusatsu City
WANG Guan 26001904590
Introduction
This experiment focusing on verifying the following statements:
I: On average, there is no difference on air temperature in August 2019
between Shanghai and Kusatsu city
II: On average, October was colder in 2020 than 2018 in Shanghai
III: On average, there is no difference on air temperature in December
2018 and 2019 in Kusatsu City
Method
Three different methods will be used for verifying three statements. Since
the sample size for all the data is bigger than 30, the experiment will use
z-test instead of t-test. Following is the method will be used for testing
three statements, the reasons why this method is used and the sample of
the data.
I: Independent two-sample z-test because the data is from two
populations (Shanghai and Kusatsu) at same time period (August).
The sample consists of two groups of independent data SHi & Kuj where i
= 1……nSH and j = 1……nKU .
II: Paired sample z-test because the data is from same population
(Shanghai) at different time period (October 2020 and October 2018). The
sample consists of two sets of mutually-dependent data SH20i & SH18i
where i = 1……n
III: Paired sample z-test because the data is from same population
(Kusatsu) at different time period (December 2019 and December 2018).
The sample consists of two sets of mutually-dependent dataKU18i & KU19i
where i = 1……n
Result
From the data analysis, it is suggested that all three statements are
accepted. Therefore, it means that for statement:
I: On average, there is no significant difference between temperature of
Shanghai and Kusatsu city in August 2019.
II: On average, there is no significant difference between temperature of
Shanghai in October 2020 and 2018 which means October 2020 was not
colder than October 2018.
III: On average, there is no significant difference between temperature of
Kusatsu city in December 2018 and 2019.
Conclusion
From the experiment, there is no significant difference of temperature between Shanghai and Kusatsu city. This result is expected since two cities have
similar longitude and latitude which the climate of two cities should be similar. Moreover, there is no significant climate change in the same city through
1-2 years. It can be said that there is not any factors, for example global warming, result in climate changes in both cities in past two years.
Data
The data in the graph below shows the average air temperature of each
day in designated month and city. All the following data is from
https://www.wunderground.com/history. (Unit: Fahrenheit)
The box-whisker graph shows the
overall data. The three histograms
show the data necessary for each
statement separately.
nalysis
I: The nu l hypothesis of this state ent i l be H0: SHi = Kuj and the
alternative hypothesis H1:SHi ≠ Kuj. The nu l hypothesis eans there is no
significant di ference bet een average air te perature of Shanghai and
Kusatsu in August 2019.
The sa ple size n is 31 days for both Shanghai and Kusatsu city.
ean of sa ple ̅SH ≈ 84.6639, ̅KU ≈ 86.4113
Standard Deviation of sa ple SSH ≈ 3.7790,SKU ≈ 4.8410
The significant level = 0.05.
Standard Error ean is SESH ≈ 0.6787,SEKU ≈ 0.8695
Standard Error Di ference is SEDi f ≈ 1.1030
Statistic z ≈ (86.4113- 84.6639)/1.1030 ≈ 1.5842
Critical value Z0.025 ≈ 1.9599
1.5842∈[-1.9599, 1.9599], so accept H0
T o tailed test: p-value ≈ 0.1131
In this case, H0 is accepted and H1 is rejected.
I: The nu l hypothesis of this state ent i l be H0: = 0 and the
alternative hypothesis H1: ≠ 0 The nu l hypothesis eans there is no
significant di ference bet een average air te perature of Shanghai in
October 2020 and 2018
The sa ple size n is 31 days.
The ̅ ≈ 0.4632
Standard Deviation of sa ple Sd ≈ 3.6425
The significant level = 0.05.
Standard Error ean is SEmean ≈ 0.6542
Statistic z ≈ 0.4632/0.6542 ≈ 0.7081
Critical value Z0.025 ≈ 1.9599
0.7081∈[-1.9599, 1.9599], so accept H0
T o tailed test: p-value ≈ 0.4789
In this case, H0 is accepted and H1 is rejected.
I: The nu l hypothesis of this state ent i l be H0: = 0 and the
alternative hypothesis H1: ≠ 0 The nu l hypothesis eans there is no
significant di ference bet een average air te perature of Kusatsu city in
Dece ber 2018 and 2019.
The sa ple size n is 31 days.
The ̅ ≈ 0.3577
Standard Deviation of sa ple Sd ≈ 7.3113
The significant level = 0.05.
Standard Error ean is SEmean ≈ 1.3132
Statistic z ≈ 0.3577/1.3132 ≈ 0.2724
Critical value Z0.025 ≈ 1.9599
0.2724∈[-1.9599, 1.9599], so accept H0
T o tailed test: p-value ≈ 0.7853
In this case, H0 is accepted and H1 is rejected.
学霸联盟
