Mixing
T distribution hypothesis testing
$begingroup$
I'm trying to solve the following problem:
I don't quite get how to solve the hypotheses testing part.
Here is how I tried tackling the problem:
$$H_0 : x_t = 0 $$
$$H_1 : x_t ne 0 $$
I then checked my T distribution table and got that:
$$begin{align}&t_L = -1.3304\&t_U = 1.3304end{align}$$
Now the problem accures when I try to calculate $t$:
$$t = frac {0-overline x}{se(x_t )}$$
as I don't know $overline x$ and $se(x_t)$ I can't calculate it.
I'm thinking of doing the following:
$$t = frac {0-0.38}{0.084} = -4.524$$
But not sure if it would be correct or not. Thank you very much for your help.
(Just to be sure, is x important in determining y due to the fact that they are positively correlated)
statistics normal-distribution regression hypothesis-testing
asked Jan 24 at 12:15
FozoroFozoro
1265
$endgroup$
add a comment |
$begingroup$
I'm trying to solve the following problem:
I don't quite get how to solve the hypotheses testing part.
Here is how I tried tackling the problem:
$$H_0 : x_t = 0 $$
$$H_1 : x_t ne 0 $$
I then checked my T distribution table and got that:
$$begin{align}&t_L = -1.3304\&t_U = 1.3304end{align}$$
Now the problem accures when I try to calculate $t$:
$$t = frac {0-overline x}{se(x_t )}$$
as I don't know $overline x$ and $se(x_t)$ I can't calculate it.
I'm thinking of doing the following:
$$t = frac {0-0.38}{0.084} = -4.524$$
But not sure if it would be correct or not. Thank you very much for your help.
(Just to be sure, is x important in determining y due to the fact that they are positively correlated)
statistics normal-distribution regression hypothesis-testing
asked Jan 24 at 12:15
FozoroFozoro
1265
$endgroup$
2
$begingroup$
You want to test for the coefficient value ($0.38$), rather than the regressor $x_{t}$ (ie. you're interested to know whether the slope of the line is non-zero). To get started, set up with the hypotheses $H_{0}: beta_{1} = 0$ and $H_{0}: beta_{1} neq 0$, where your regression equation is given by $hat{y}_{t} = beta_{0} + beta_{1}x_{t}$.
$endgroup$
– rzch
Jan 24 at 13:56
$begingroup$
Thank you very much for your reply. So t will be equall to -4.524? $$t = frac {0-0.38}{0.084} = -4.524$$ @rzch
$endgroup$
– Fozoro
Jan 24 at 14:03
1
$begingroup$
The coefficient estimate should come first in the formula, so it's $t = frac{0.38 - 0}{0.084}$.
$endgroup$
– rzch
Jan 24 at 14:05
$begingroup$
@rzch oops my bad, are the upper and lower bounds correct? and is my reason for the importance of x to determine y correct? Thanks a ton for your help!
$endgroup$
– Fozoro
Jan 24 at 14:08
1
$begingroup$
It's asking whether $x$ is 'important', but doesn't specify any preference for which direction. So you should take this to mean any correlation whatsoever (positive or negative slope). This gives you a two-tailed test, and looking up the t-table at 18 degrees of freedom for 10% two-tailed, (eg. sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf) it gives you an upper and lower bound of -1.734 and 1.734 respectively.
$endgroup$
– rzch
Jan 24 at 14:15
add a comment |
$begingroup$
I'm trying to solve the following problem:
I don't quite get how to solve the hypotheses testing part.
Here is how I tried tackling the problem:
$$H_0 : x_t = 0 $$
$$H_1 : x_t ne 0 $$
I then checked my T distribution table and got that:
$$begin{align}&t_L = -1.3304\&t_U = 1.3304end{align}$$
Now the problem accures when I try to calculate $t$:
$$t = frac {0-overline x}{se(x_t )}$$
as I don't know $overline x$ and $se(x_t)$ I can't calculate it.
I'm thinking of doing the following:
$$t = frac {0-0.38}{0.084} = -4.524$$
But not sure if it would be correct or not. Thank you very much for your help.
(Just to be sure, is x important in determining y due to the fact that they are positively correlated)
statistics normal-distribution regression hypothesis-testing
asked Jan 24 at 12:15
FozoroFozoro
1265
$endgroup$
I'm trying to solve the following problem:
I don't quite get how to solve the hypotheses testing part.
Here is how I tried tackling the problem:
$$H_0 : x_t = 0 $$
$$H_1 : x_t ne 0 $$
I then checked my T distribution table and got that:
$$begin{align}&t_L = -1.3304\&t_U = 1.3304end{align}$$
Now the problem accures when I try to calculate $t$:
$$t = frac {0-overline x}{se(x_t )}$$
as I don't know $overline x$ and $se(x_t)$ I can't calculate it.
I'm thinking of doing the following:
$$t = frac {0-0.38}{0.084} = -4.524$$
But not sure if it would be correct or not. Thank you very much for your help.
(Just to be sure, is x important in determining y due to the fact that they are positively correlated)
statistics normal-distribution regression hypothesis-testing
statistics normal-distribution regression hypothesis-testing
asked Jan 24 at 12:15
FozoroFozoro
1265
asked Jan 24 at 12:15
FozoroFozoro
1265
edited Jan 24 at 17:31
Fozoro
asked Jan 24 at 12:15
FozoroFozoro
1265
asked Jan 24 at 12:15
FozoroFozoro
1265
asked Jan 24 at 12:15
FozoroFozoro
1265
1265
2
$begingroup$
You want to test for the coefficient value ($0.38$), rather than the regressor $x_{t}$ (ie. you're interested to know whether the slope of the line is non-zero). To get started, set up with the hypotheses $H_{0}: beta_{1} = 0$ and $H_{0}: beta_{1} neq 0$, where your regression equation is given by $hat{y}_{t} = beta_{0} + beta_{1}x_{t}$.
$endgroup$
– rzch
Jan 24 at 13:56
$begingroup$
Thank you very much for your reply. So t will be equall to -4.524? $$t = frac {0-0.38}{0.084} = -4.524$$ @rzch
$endgroup$
– Fozoro
Jan 24 at 14:03
1
$begingroup$
The coefficient estimate should come first in the formula, so it's $t = frac{0.38 - 0}{0.084}$.
$endgroup$
– rzch
Jan 24 at 14:05
$begingroup$
@rzch oops my bad, are the upper and lower bounds correct? and is my reason for the importance of x to determine y correct? Thanks a ton for your help!
$endgroup$
– Fozoro
Jan 24 at 14:08
1
$begingroup$
It's asking whether $x$ is 'important', but doesn't specify any preference for which direction. So you should take this to mean any correlation whatsoever (positive or negative slope). This gives you a two-tailed test, and looking up the t-table at 18 degrees of freedom for 10% two-tailed, (eg. sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf) it gives you an upper and lower bound of -1.734 and 1.734 respectively.
$endgroup$
– rzch
Jan 24 at 14:15
add a comment |
2
$begingroup$
You want to test for the coefficient value ($0.38$), rather than the regressor $x_{t}$ (ie. you're interested to know whether the slope of the line is non-zero). To get started, set up with the hypotheses $H_{0}: beta_{1} = 0$ and $H_{0}: beta_{1} neq 0$, where your regression equation is given by $hat{y}_{t} = beta_{0} + beta_{1}x_{t}$.
$endgroup$
– rzch
Jan 24 at 13:56
$begingroup$
Thank you very much for your reply. So t will be equall to -4.524? $$t = frac {0-0.38}{0.084} = -4.524$$ @rzch
$endgroup$
– Fozoro
Jan 24 at 14:03
1
$begingroup$
The coefficient estimate should come first in the formula, so it's $t = frac{0.38 - 0}{0.084}$.
$endgroup$
– rzch
Jan 24 at 14:05
$begingroup$
@rzch oops my bad, are the upper and lower bounds correct? and is my reason for the importance of x to determine y correct? Thanks a ton for your help!
$endgroup$
– Fozoro
Jan 24 at 14:08
1
$begingroup$
It's asking whether $x$ is 'important', but doesn't specify any preference for which direction. So you should take this to mean any correlation whatsoever (positive or negative slope). This gives you a two-tailed test, and looking up the t-table at 18 degrees of freedom for 10% two-tailed, (eg. sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf) it gives you an upper and lower bound of -1.734 and 1.734 respectively.
$endgroup$
– rzch
Jan 24 at 14:15
2
2
$begingroup$
You want to test for the coefficient value ($0.38$), rather than the regressor $x_{t}$ (ie. you're interested to know whether the slope of the line is non-zero). To get started, set up with the hypotheses $H_{0}: beta_{1} = 0$ and $H_{0}: beta_{1} neq 0$, where your regression equation is given by $hat{y}_{t} = beta_{0} + beta_{1}x_{t}$.
$endgroup$
– rzch
Jan 24 at 13:56
$begingroup$
You want to test for the coefficient value ($0.38$), rather than the regressor $x_{t}$ (ie. you're interested to know whether the slope of the line is non-zero). To get started, set up with the hypotheses $H_{0}: beta_{1} = 0$ and $H_{0}: beta_{1} neq 0$, where your regression equation is given by $hat{y}_{t} = beta_{0} + beta_{1}x_{t}$.
$endgroup$
– rzch
Jan 24 at 13:56
$begingroup$
Thank you very much for your reply. So t will be equall to -4.524? $$t = frac {0-0.38}{0.084} = -4.524$$ @rzch
$endgroup$
– Fozoro
Jan 24 at 14:03
$begingroup$
Thank you very much for your reply. So t will be equall to -4.524? $$t = frac {0-0.38}{0.084} = -4.524$$ @rzch
$endgroup$
– Fozoro
Jan 24 at 14:03
1
1
$begingroup$
The coefficient estimate should come first in the formula, so it's $t = frac{0.38 - 0}{0.084}$.
$endgroup$
– rzch
Jan 24 at 14:05
$begingroup$
The coefficient estimate should come first in the formula, so it's $t = frac{0.38 - 0}{0.084}$.
$endgroup$
– rzch
Jan 24 at 14:05
$begingroup$
@rzch oops my bad, are the upper and lower bounds correct? and is my reason for the importance of x to determine y correct? Thanks a ton for your help!
$endgroup$
– Fozoro
Jan 24 at 14:08
$begingroup$
@rzch oops my bad, are the upper and lower bounds correct? and is my reason for the importance of x to determine y correct? Thanks a ton for your help!
$endgroup$
– Fozoro
Jan 24 at 14:08
1
1
$begingroup$
It's asking whether $x$ is 'important', but doesn't specify any preference for which direction. So you should take this to mean any correlation whatsoever (positive or negative slope). This gives you a two-tailed test, and looking up the t-table at 18 degrees of freedom for 10% two-tailed, (eg. sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf) it gives you an upper and lower bound of -1.734 and 1.734 respectively.
$endgroup$
– rzch
Jan 24 at 14:15
$begingroup$
It's asking whether $x$ is 'important', but doesn't specify any preference for which direction. So you should take this to mean any correlation whatsoever (positive or negative slope). This gives you a two-tailed test, and looking up the t-table at 18 degrees of freedom for 10% two-tailed, (eg. sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf) it gives you an upper and lower bound of -1.734 and 1.734 respectively.
$endgroup$
– rzch
Jan 24 at 14:15
add a comment |
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2
$begingroup$
You want to test for the coefficient value ($0.38$), rather than the regressor $x_{t}$ (ie. you're interested to know whether the slope of the line is non-zero). To get started, set up with the hypotheses $H_{0}: beta_{1} = 0$ and $H_{0}: beta_{1} neq 0$, where your regression equation is given by $hat{y}_{t} = beta_{0} + beta_{1}x_{t}$.
$endgroup$
– rzch
Jan 24 at 13:56
$begingroup$
Thank you very much for your reply. So t will be equall to -4.524? $$t = frac {0-0.38}{0.084} = -4.524$$ @rzch
$endgroup$
– Fozoro
Jan 24 at 14:03
1
$begingroup$
The coefficient estimate should come first in the formula, so it's $t = frac{0.38 - 0}{0.084}$.
$endgroup$
– rzch
Jan 24 at 14:05
$begingroup$
@rzch oops my bad, are the upper and lower bounds correct? and is my reason for the importance of x to determine y correct? Thanks a ton for your help!
$endgroup$
– Fozoro
Jan 24 at 14:08
1
$begingroup$
It's asking whether $x$ is 'important', but doesn't specify any preference for which direction. So you should take this to mean any correlation whatsoever (positive or negative slope). This gives you a two-tailed test, and looking up the t-table at 18 degrees of freedom for 10% two-tailed, (eg. sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf) it gives you an upper and lower bound of -1.734 and 1.734 respectively.
$endgroup$
– rzch
Jan 24 at 14:15