Mixing
How do I compare means when I have a sample and the whole population?
Clash Royale CLAN TAG#URR8PPP
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2
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Scenario
I have the avg room price for all the hotels of my chain (32 observations of the 32 hotels). Then I have the avg room price for a sample of competitors (60 observations taken from a larger population).
Problem
I would like to understand whether the avg room price of the hotels of my chain is equal to that of the competitors
Proposed solution
First, I computed the average room price across all the hotels of my chain $p_a$. Since the entire population is known, I would say that there is no uncertainty here, this is the exact average.
Then I computed the average room price of the sample of competitors and the related std deviation ($barp_c$ and $tildesigma_c$).
I performed a hypothesis testing with the null $H_0: p_c = p_a$ against the alternative $H_1: p_c neq p_a$. The test statistic is then ($n=60$):
$$
t = fracbarp_c-p_afractildesigma_csqrtn
$$
If the associated p-value is sufficently low, I reject the null hypothesis.
Basically here I'm considering that the average room price of the hotels of my chain is known and well-established, hence I'm doing the hypothesis testing for a single population mean (the competitors' population). Do you think this is the right approach or shall I test a hypothesis about two population mean (this is the alternative that comes to my mind)
Thanks for any help, T.
hypothesis-testing mean sample population
asked 5 hours ago
tuspazio
1133
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up vote
2
down vote
favorite
Scenario
I have the avg room price for all the hotels of my chain (32 observations of the 32 hotels). Then I have the avg room price for a sample of competitors (60 observations taken from a larger population).
Problem
I would like to understand whether the avg room price of the hotels of my chain is equal to that of the competitors
Proposed solution
First, I computed the average room price across all the hotels of my chain $p_a$. Since the entire population is known, I would say that there is no uncertainty here, this is the exact average.
Then I computed the average room price of the sample of competitors and the related std deviation ($barp_c$ and $tildesigma_c$).
I performed a hypothesis testing with the null $H_0: p_c = p_a$ against the alternative $H_1: p_c neq p_a$. The test statistic is then ($n=60$):
$$
t = fracbarp_c-p_afractildesigma_csqrtn
$$
If the associated p-value is sufficently low, I reject the null hypothesis.
Basically here I'm considering that the average room price of the hotels of my chain is known and well-established, hence I'm doing the hypothesis testing for a single population mean (the competitors' population). Do you think this is the right approach or shall I test a hypothesis about two population mean (this is the alternative that comes to my mind)
Thanks for any help, T.
hypothesis-testing mean sample population
asked 5 hours ago
tuspazio
1133
New contributor
tuspazio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
The hypothesis testing for a single population mean is good. Need to pay attention to the calculation of std deviation if the total number of hotels from competitors is limited, for example less than 500.
– a_statistician
2 hours ago
Thanks a lot, @a_statistician. What should I exactly pay attention to if the population is limited? How does this affect the computation of the std deviation? Thx, T.
– tuspazio
2 hours ago
It is called The finite population correction. See eq. 3.19 on page 3-15 of ph.ucla.edu/epi/rapidsurveys/RScourse/RSbook_ch3.pdf
– a_statistician
2 hours ago
Ah understood, thanks for the suggestion!
– tuspazio
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Scenario
I have the avg room price for all the hotels of my chain (32 observations of the 32 hotels). Then I have the avg room price for a sample of competitors (60 observations taken from a larger population).
Problem
I would like to understand whether the avg room price of the hotels of my chain is equal to that of the competitors
Proposed solution
First, I computed the average room price across all the hotels of my chain $p_a$. Since the entire population is known, I would say that there is no uncertainty here, this is the exact average.
Then I computed the average room price of the sample of competitors and the related std deviation ($barp_c$ and $tildesigma_c$).
I performed a hypothesis testing with the null $H_0: p_c = p_a$ against the alternative $H_1: p_c neq p_a$. The test statistic is then ($n=60$):
$$
t = fracbarp_c-p_afractildesigma_csqrtn
$$
If the associated p-value is sufficently low, I reject the null hypothesis.
Basically here I'm considering that the average room price of the hotels of my chain is known and well-established, hence I'm doing the hypothesis testing for a single population mean (the competitors' population). Do you think this is the right approach or shall I test a hypothesis about two population mean (this is the alternative that comes to my mind)
Thanks for any help, T.
hypothesis-testing mean sample population
asked 5 hours ago
tuspazio
1133
New contributor
tuspazio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Scenario
I have the avg room price for all the hotels of my chain (32 observations of the 32 hotels). Then I have the avg room price for a sample of competitors (60 observations taken from a larger population).
Problem
I would like to understand whether the avg room price of the hotels of my chain is equal to that of the competitors
Proposed solution
First, I computed the average room price across all the hotels of my chain $p_a$. Since the entire population is known, I would say that there is no uncertainty here, this is the exact average.
Then I computed the average room price of the sample of competitors and the related std deviation ($barp_c$ and $tildesigma_c$).
I performed a hypothesis testing with the null $H_0: p_c = p_a$ against the alternative $H_1: p_c neq p_a$. The test statistic is then ($n=60$):
$$
t = fracbarp_c-p_afractildesigma_csqrtn
$$
If the associated p-value is sufficently low, I reject the null hypothesis.
Basically here I'm considering that the average room price of the hotels of my chain is known and well-established, hence I'm doing the hypothesis testing for a single population mean (the competitors' population). Do you think this is the right approach or shall I test a hypothesis about two population mean (this is the alternative that comes to my mind)
Thanks for any help, T.
hypothesis-testing mean sample population
hypothesis-testing mean sample population
asked 5 hours ago
tuspazio
1133
New contributor
tuspazio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
tuspazio
1133
New contributor
tuspazio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
tuspazio
1133
New contributor
tuspazio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 hours ago
tuspazio
1133
asked 5 hours ago
tuspazio
1133
1133
New contributor
tuspazio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
tuspazio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
tuspazio is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
The hypothesis testing for a single population mean is good. Need to pay attention to the calculation of std deviation if the total number of hotels from competitors is limited, for example less than 500.
– a_statistician
2 hours ago
Thanks a lot, @a_statistician. What should I exactly pay attention to if the population is limited? How does this affect the computation of the std deviation? Thx, T.
– tuspazio
2 hours ago
It is called The finite population correction. See eq. 3.19 on page 3-15 of ph.ucla.edu/epi/rapidsurveys/RScourse/RSbook_ch3.pdf
– a_statistician
2 hours ago
Ah understood, thanks for the suggestion!
– tuspazio
1 hour ago
add a comment |Â
1
The hypothesis testing for a single population mean is good. Need to pay attention to the calculation of std deviation if the total number of hotels from competitors is limited, for example less than 500.
– a_statistician
2 hours ago
Thanks a lot, @a_statistician. What should I exactly pay attention to if the population is limited? How does this affect the computation of the std deviation? Thx, T.
– tuspazio
2 hours ago
It is called The finite population correction. See eq. 3.19 on page 3-15 of ph.ucla.edu/epi/rapidsurveys/RScourse/RSbook_ch3.pdf
– a_statistician
2 hours ago
Ah understood, thanks for the suggestion!
– tuspazio
1 hour ago
1
1
The hypothesis testing for a single population mean is good. Need to pay attention to the calculation of std deviation if the total number of hotels from competitors is limited, for example less than 500.
– a_statistician
2 hours ago
The hypothesis testing for a single population mean is good. Need to pay attention to the calculation of std deviation if the total number of hotels from competitors is limited, for example less than 500.
– a_statistician
2 hours ago
Thanks a lot, @a_statistician. What should I exactly pay attention to if the population is limited? How does this affect the computation of the std deviation? Thx, T.
– tuspazio
2 hours ago
Thanks a lot, @a_statistician. What should I exactly pay attention to if the population is limited? How does this affect the computation of the std deviation? Thx, T.
– tuspazio
2 hours ago
It is called The finite population correction. See eq. 3.19 on page 3-15 of ph.ucla.edu/epi/rapidsurveys/RScourse/RSbook_ch3.pdf
– a_statistician
2 hours ago
It is called The finite population correction. See eq. 3.19 on page 3-15 of ph.ucla.edu/epi/rapidsurveys/RScourse/RSbook_ch3.pdf
– a_statistician
2 hours ago
Ah understood, thanks for the suggestion!
– tuspazio
1 hour ago
Ah understood, thanks for the suggestion!
– tuspazio
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
This seems completely reasonable to me. It is what I would have done.
answered 4 hours ago
Peter Flom♦
72.8k11103197
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
This seems completely reasonable to me. It is what I would have done.
answered 4 hours ago
Peter Flom♦
72.8k11103197
add a comment |Â
up vote
3
down vote
accepted
This seems completely reasonable to me. It is what I would have done.
answered 4 hours ago
Peter Flom♦
72.8k11103197
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
This seems completely reasonable to me. It is what I would have done.
answered 4 hours ago
Peter Flom♦
72.8k11103197
This seems completely reasonable to me. It is what I would have done.
answered 4 hours ago
Peter Flom♦
72.8k11103197
answered 4 hours ago
Peter Flom♦
72.8k11103197
answered 4 hours ago
Peter Flom♦
72.8k11103197
answered 4 hours ago
Peter Flom♦
72.8k11103197
72.8k11103197
add a comment |Â
add a comment |Â
tuspazio is a new contributor. Be nice, and check out our Code of Conduct.
tuspazio is a new contributor. Be nice, and check out our Code of Conduct.
tuspazio is a new contributor. Be nice, and check out our Code of Conduct.
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1
The hypothesis testing for a single population mean is good. Need to pay attention to the calculation of std deviation if the total number of hotels from competitors is limited, for example less than 500.
– a_statistician
2 hours ago
Thanks a lot, @a_statistician. What should I exactly pay attention to if the population is limited? How does this affect the computation of the std deviation? Thx, T.
– tuspazio
2 hours ago
It is called The finite population correction. See eq. 3.19 on page 3-15 of ph.ucla.edu/epi/rapidsurveys/RScourse/RSbook_ch3.pdf
– a_statistician
2 hours ago
Ah understood, thanks for the suggestion!
– tuspazio
1 hour ago