Paired T Test Hypothesis Interpretation
up vote
0
down vote
favorite
With a paired T test I understand the first part of the hypothesis:
$H0: mu_d = 0$
The alternative hypothesis I get a little mixed up and I am trying to figure out when to appropriately use the alternatives.
If the alternative hypothesis is:
$H1: mu_dneq0$
I think this means we are just using the paired T test to see if there was a change between test 1 and test 2.
If the alternative hypothesis is:
$H1: mu_d>0$
Is this setting up that the alternative hypothesis is checking that the results of test 2 were lower than that result of test one?
If so then by extension an alternative hypothesis of:
$H1: mu _{d} < 0$
This is checking that test 2 had higher results then test 1?
Any insight would be greatly appreciated.
hypothesis-testing
add a comment |
up vote
0
down vote
favorite
With a paired T test I understand the first part of the hypothesis:
$H0: mu_d = 0$
The alternative hypothesis I get a little mixed up and I am trying to figure out when to appropriately use the alternatives.
If the alternative hypothesis is:
$H1: mu_dneq0$
I think this means we are just using the paired T test to see if there was a change between test 1 and test 2.
If the alternative hypothesis is:
$H1: mu_d>0$
Is this setting up that the alternative hypothesis is checking that the results of test 2 were lower than that result of test one?
If so then by extension an alternative hypothesis of:
$H1: mu _{d} < 0$
This is checking that test 2 had higher results then test 1?
Any insight would be greatly appreciated.
hypothesis-testing
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
With a paired T test I understand the first part of the hypothesis:
$H0: mu_d = 0$
The alternative hypothesis I get a little mixed up and I am trying to figure out when to appropriately use the alternatives.
If the alternative hypothesis is:
$H1: mu_dneq0$
I think this means we are just using the paired T test to see if there was a change between test 1 and test 2.
If the alternative hypothesis is:
$H1: mu_d>0$
Is this setting up that the alternative hypothesis is checking that the results of test 2 were lower than that result of test one?
If so then by extension an alternative hypothesis of:
$H1: mu _{d} < 0$
This is checking that test 2 had higher results then test 1?
Any insight would be greatly appreciated.
hypothesis-testing
With a paired T test I understand the first part of the hypothesis:
$H0: mu_d = 0$
The alternative hypothesis I get a little mixed up and I am trying to figure out when to appropriately use the alternatives.
If the alternative hypothesis is:
$H1: mu_dneq0$
I think this means we are just using the paired T test to see if there was a change between test 1 and test 2.
If the alternative hypothesis is:
$H1: mu_d>0$
Is this setting up that the alternative hypothesis is checking that the results of test 2 were lower than that result of test one?
If so then by extension an alternative hypothesis of:
$H1: mu _{d} < 0$
This is checking that test 2 had higher results then test 1?
Any insight would be greatly appreciated.
hypothesis-testing
hypothesis-testing
edited Nov 15 at 5:35
asked Nov 15 at 5:29
AxGryndr
1125
1125
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
That depends entirely on how the test statistic is specified, or equivalently, how $mu_d$ relates to the means of each group. If you define $mu_d = mu_2 - mu_1$, that is, the mean of the second tests minus the mean of the first tests, then $mu_d > 0$ implies $mu_2 > mu_1$; conversely, if $mu_d = mu_1 - mu_2$, then $mu_d > 0$ implies $mu_2 < mu_1$. Correspondingly, if you use the first definition, your test statistic must be constructed as $$T = frac{bar x_2 - bar x_1}{s/sqrt{n}},$$ whereas if using the second definition of $mu_d$, the order of the difference must also be interchanged.
Ah, that makes sense. Thank you!
– AxGryndr
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
That depends entirely on how the test statistic is specified, or equivalently, how $mu_d$ relates to the means of each group. If you define $mu_d = mu_2 - mu_1$, that is, the mean of the second tests minus the mean of the first tests, then $mu_d > 0$ implies $mu_2 > mu_1$; conversely, if $mu_d = mu_1 - mu_2$, then $mu_d > 0$ implies $mu_2 < mu_1$. Correspondingly, if you use the first definition, your test statistic must be constructed as $$T = frac{bar x_2 - bar x_1}{s/sqrt{n}},$$ whereas if using the second definition of $mu_d$, the order of the difference must also be interchanged.
Ah, that makes sense. Thank you!
– AxGryndr
2 days ago
add a comment |
up vote
0
down vote
accepted
That depends entirely on how the test statistic is specified, or equivalently, how $mu_d$ relates to the means of each group. If you define $mu_d = mu_2 - mu_1$, that is, the mean of the second tests minus the mean of the first tests, then $mu_d > 0$ implies $mu_2 > mu_1$; conversely, if $mu_d = mu_1 - mu_2$, then $mu_d > 0$ implies $mu_2 < mu_1$. Correspondingly, if you use the first definition, your test statistic must be constructed as $$T = frac{bar x_2 - bar x_1}{s/sqrt{n}},$$ whereas if using the second definition of $mu_d$, the order of the difference must also be interchanged.
Ah, that makes sense. Thank you!
– AxGryndr
2 days ago
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
That depends entirely on how the test statistic is specified, or equivalently, how $mu_d$ relates to the means of each group. If you define $mu_d = mu_2 - mu_1$, that is, the mean of the second tests minus the mean of the first tests, then $mu_d > 0$ implies $mu_2 > mu_1$; conversely, if $mu_d = mu_1 - mu_2$, then $mu_d > 0$ implies $mu_2 < mu_1$. Correspondingly, if you use the first definition, your test statistic must be constructed as $$T = frac{bar x_2 - bar x_1}{s/sqrt{n}},$$ whereas if using the second definition of $mu_d$, the order of the difference must also be interchanged.
That depends entirely on how the test statistic is specified, or equivalently, how $mu_d$ relates to the means of each group. If you define $mu_d = mu_2 - mu_1$, that is, the mean of the second tests minus the mean of the first tests, then $mu_d > 0$ implies $mu_2 > mu_1$; conversely, if $mu_d = mu_1 - mu_2$, then $mu_d > 0$ implies $mu_2 < mu_1$. Correspondingly, if you use the first definition, your test statistic must be constructed as $$T = frac{bar x_2 - bar x_1}{s/sqrt{n}},$$ whereas if using the second definition of $mu_d$, the order of the difference must also be interchanged.
answered 2 days ago
heropup
61.8k65997
61.8k65997
Ah, that makes sense. Thank you!
– AxGryndr
2 days ago
add a comment |
Ah, that makes sense. Thank you!
– AxGryndr
2 days ago
Ah, that makes sense. Thank you!
– AxGryndr
2 days ago
Ah, that makes sense. Thank you!
– AxGryndr
2 days ago
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999266%2fpaired-t-test-hypothesis-interpretation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown