T distribution hypothesis testing












0












$begingroup$


I'm trying to solve the following problem:





Question





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)










share|cite|improve this question











$endgroup$








  • 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


















0












$begingroup$


I'm trying to solve the following problem:





Question





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)










share|cite|improve this question











$endgroup$








  • 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
















0












0








0





$begingroup$


I'm trying to solve the following problem:





Question





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)










share|cite|improve this question











$endgroup$




I'm trying to solve the following problem:





Question





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 17:31







Fozoro

















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
















  • 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












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