9 added 1 character in body edited May 29 '15 at 11:22 Tony 21822 silver badges99 bronze badges So suppose after a simpple OLS regression we want to know what the chance(P-value) is that the Beta coefficient is 0 . First we assume that many random processes caused the errors ($$\epsilon\!_i$$), meaning that the residual is normally distributed white noise with mean of μ and standard deviation of σ. How can we deduct the shape and standard deviation of the beta coefficient's probability distribution using this information. IE. How does the firstsecond formula presented below (also equaling the standard deviation of the errors) arise from this fact? How does increasing the sample size affect the confidence that the beta coefficient is 0, when the population mean and variance are known (that is Z distribution is used)? Here are the relevant mathematical formulas: Simple regression: $$X=Y+\epsilon\!_i$$ B's standard error(=standard deviation of B): $$\sqrt{\dfrac{\sum_{1}^{n}(\epsilon\!_i)^2}{N}}$$ = $$\sqrt{\dfrac{\sum_{1}^{n}(Y_i-Ý_i)^2}{N}}$$ Z-value =$$B/S(B)$$ = $$B\sqrt{\dfrac{N}{\sum_{1}^{n}(Y_i-Ý_i)^2}}$$ If $$(Y-Ý)$$ is constant the standard error won't change as a result of increasing N and thus the p-value won't change either. Is this reasoning correct? So suppose after a simpple OLS regression we want to know what the chance(P-value) is that the Beta coefficient is 0 . First we assume that many random processes caused the errors ($$\epsilon\!_i$$), meaning that the residual is normally distributed white noise with mean of μ and standard deviation of σ. How can we deduct the shape and standard deviation of the beta coefficient's probability distribution using this information. IE. How does the first formula presented below (also equaling the standard deviation of the errors) arise from this fact? How does increasing the sample size affect the confidence that the beta coefficient is 0, when the population mean and variance are known (that is Z distribution is used)? Here are the relevant mathematical formulas: Simple regression: $$X=Y+\epsilon\!_i$$ B's standard error(=standard deviation of B): $$\sqrt{\dfrac{\sum_{1}^{n}(\epsilon\!_i)^2}{N}}$$ = $$\sqrt{\dfrac{\sum_{1}^{n}(Y_i-Ý_i)^2}{N}}$$ Z-value =$$B/S(B)$$ = $$B\sqrt{\dfrac{N}{\sum_{1}^{n}(Y_i-Ý_i)^2}}$$ If $$(Y-Ý)$$ is constant the standard error won't change as a result of increasing N and thus the p-value won't change either. Is this reasoning correct? So suppose after a simpple OLS regression we want to know what the chance(P-value) is that the Beta coefficient is 0 . First we assume that many random processes caused the errors ($$\epsilon\!_i$$), meaning that the residual is normally distributed white noise with mean of μ and standard deviation of σ. How can we deduct the shape and standard deviation of the beta coefficient's probability distribution using this information. IE. How does the second formula presented below (also equaling the standard deviation of the errors) arise from this fact? How does increasing the sample size affect the confidence that the beta coefficient is 0, when the population mean and variance are known (that is Z distribution is used)? Here are the relevant mathematical formulas: Simple regression: $$X=Y+\epsilon\!_i$$ B's standard error(=standard deviation of B): $$\sqrt{\dfrac{\sum_{1}^{n}(\epsilon\!_i)^2}{N}}$$ = $$\sqrt{\dfrac{\sum_{1}^{n}(Y_i-Ý_i)^2}{N}}$$ Z-value =$$B/S(B)$$ = $$B\sqrt{\dfrac{N}{\sum_{1}^{n}(Y_i-Ý_i)^2}}$$ If $$(Y-Ý)$$ is constant the standard error won't change as a result of increasing N and thus the p-value won't change either. Is this reasoning correct? 8 added 20 characters in body edited May 29 '15 at 11:15 Tony 21822 silver badges99 bronze badges So suppose after a simpple OLS regression we want to know what the chance(P-value) is that the Beta coefficient is 0 . First we assume that many random processes caused the errors ($$\epsilon\!_i$$), meaning that the residual is normally distributed white noise with mean of μ and standard deviation of σ. How can we deduct the shape and standard deviation of the beta coefficient's probability distribution using this information. IE. How does the first formula presented below (also equaling the standard deviation of the errors) arise from this fact? How does increasing the sample size affect the confidence that the beta coefficient is 0, when the population mean and variance are known (that is Z distribution is used)? Here are the relevant mathematical formulas: Simple regression: $$X=Y+\epsilon\!_i$$ B's standard error(=standard deviation of B): $$\dfrac{\sum_{1}^{n}(Y_i-Ý_i)^2}{N}$$$$\sqrt{\dfrac{\sum_{1}^{n}(\epsilon\!_i)^2}{N}}$$ = $$\sqrt{\dfrac{\sum_{1}^{n}(Y_i-Ý_i)^2}{N}}$$ Z-value =$$B/S(B)$$ = $$B\dfrac{N}{\sum_{1}^{n}(Y_i-Ý_i)^2}$$$$B\sqrt{\dfrac{N}{\sum_{1}^{n}(Y_i-Ý_i)^2}}$$ If $$(Y-Ý)$$ is constant the standard error won't change as a result of increasing N and thus the p-value won't change either. Is this reasoning correct? So suppose after a simpple OLS regression we want to know what the chance(P-value) is that the Beta coefficient is 0 . First we assume that many random processes caused the errors ($$\epsilon\!_i$$), meaning that the residual is normally distributed white noise with mean of μ and standard deviation of σ. How can we deduct the shape and standard deviation of the beta coefficient's probability distribution using this information. IE. How does the first formula presented below (also equaling the standard deviation of the errors) arise from this fact? How does increasing the sample size affect the confidence that the beta coefficient is 0, when the population mean and variance are known (that is Z distribution is used)? Here are the relevant mathematical formulas: Simple regression: $$X=Y+\epsilon\!_i$$ B's standard error(=standard deviation of B): $$\dfrac{\sum_{1}^{n}(Y_i-Ý_i)^2}{N}$$ Z-value =$$B/S(B)$$ = $$B\dfrac{N}{\sum_{1}^{n}(Y_i-Ý_i)^2}$$ If $$(Y-Ý)$$ is constant the standard error won't change as a result of increasing N and thus the p-value won't change either. Is this reasoning correct? So suppose after a simpple OLS regression we want to know what the chance(P-value) is that the Beta coefficient is 0 . First we assume that many random processes caused the errors ($$\epsilon\!_i$$), meaning that the residual is normally distributed white noise with mean of μ and standard deviation of σ. How can we deduct the shape and standard deviation of the beta coefficient's probability distribution using this information. IE. How does the first formula presented below (also equaling the standard deviation of the errors) arise from this fact? How does increasing the sample size affect the confidence that the beta coefficient is 0, when the population mean and variance are known (that is Z distribution is used)? Here are the relevant mathematical formulas: Simple regression: $$X=Y+\epsilon\!_i$$ B's standard error(=standard deviation of B): $$\sqrt{\dfrac{\sum_{1}^{n}(\epsilon\!_i)^2}{N}}$$ = $$\sqrt{\dfrac{\sum_{1}^{n}(Y_i-Ý_i)^2}{N}}$$ Z-value =$$B/S(B)$$ = $$B\sqrt{\dfrac{N}{\sum_{1}^{n}(Y_i-Ý_i)^2}}$$ If $$(Y-Ý)$$ is constant the standard error won't change as a result of increasing N and thus the p-value won't change either. Is this reasoning correct? 7 added 60 characters in body edited May 29 '15 at 9:27 Tony 21822 silver badges99 bronze badges So suppose after ana simpple OLS regression we want to know what the chance(P-value) is that the Beta coefficient is 0 . First we assume that many random processes caused the errors ($$\epsilon\!_t$$$$\epsilon\!_i$$), meaning that the residual is normally distributed white noise with mean of μ and standard deviation of σ. How can we deduct the shape and standard deviation of the beta coefficient's probability distribution using this information. IE. How does the first formula presented below (also equaling the standard deviation of the errors) arise from this fact? How does increasing the sample size affect the confidence that the beta coefficient is 0, when the population mean and variance are known (that is Z distribution is used)? Here are the relevant mathematical formulas: Simple regression: $$X=Y+\epsilon\!_i$$ B's standard error(=standard deviation of B): $$\dfrac{\sum_{1}^{n}(Y_i-Ý_i)^2}{N}$$ Z-value =$$B/S(B)$$ = $$B\dfrac{N}{\sum_{1}^{n}(Y_i-Ý_i)^2}$$ If $$(Y-Ý)$$ is constant the standard error won't change as a result of increasing N and thus the p-value won't change either. Is this reasoning correct? So suppose after an OLS regression we want to know what the chance(P-value) is that the Beta coefficient is 0 . First we assume that many random processes caused the errors ($$\epsilon\!_t$$), meaning that the residual is normally distributed white noise with mean of μ and standard deviation of σ. How can we deduct the shape and standard deviation of the beta coefficient's probability distribution using this information. IE. How does the first formula presented below (also equaling the standard deviation of the errors) arise from this fact? How does increasing the sample size affect the confidence that the beta coefficient is 0, when the population mean and variance are known (that is Z distribution is used)? Here are the mathematical formulas: B's standard error(=standard deviation of B): $$\dfrac{\sum_{1}^{n}(Y_i-Ý_i)^2}{N}$$ Z-value =$$B/S(B)$$ = $$B\dfrac{N}{\sum_{1}^{n}(Y_i-Ý_i)^2}$$ If $$(Y-Ý)$$ is constant the standard error won't change as a result of increasing N and thus the p-value won't change either. Is this reasoning correct? So suppose after a simpple OLS regression we want to know what the chance(P-value) is that the Beta coefficient is 0 . First we assume that many random processes caused the errors ($$\epsilon\!_i$$), meaning that the residual is normally distributed white noise with mean of μ and standard deviation of σ. How can we deduct the shape and standard deviation of the beta coefficient's probability distribution using this information. IE. How does the first formula presented below (also equaling the standard deviation of the errors) arise from this fact? How does increasing the sample size affect the confidence that the beta coefficient is 0, when the population mean and variance are known (that is Z distribution is used)? Here are the relevant mathematical formulas: Simple regression: $$X=Y+\epsilon\!_i$$ B's standard error(=standard deviation of B): $$\dfrac{\sum_{1}^{n}(Y_i-Ý_i)^2}{N}$$ Z-value =$$B/S(B)$$ = $$B\dfrac{N}{\sum_{1}^{n}(Y_i-Ý_i)^2}$$ If $$(Y-Ý)$$ is constant the standard error won't change as a result of increasing N and thus the p-value won't change either. Is this reasoning correct? 6 added 53 characters in body edited May 29 '15 at 1:49 Tony 21822 silver badges99 bronze badges 5 added 314 characters in body edited May 29 '15 at 1:39 Tony 21822 silver badges99 bronze badges 4 added 10 characters in body edited May 29 '15 at 0:56 Tony 21822 silver badges99 bronze badges 3 added 10 characters in body edited May 28 '15 at 21:42 Tony 21822 silver badges99 bronze badges 2 added 30 characters in body edited May 28 '15 at 21:36 Tony 21822 silver badges99 bronze badges 1 asked May 28 '15 at 21:27 Tony 21822 silver badges99 bronze badges