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ttnphns
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This has been asserted multiple times in CV (herehere, herehere, herehere and others).

These two tests statistics are clearly different, and result in different results for the actual test statistics, as well as for the p-valuesresult in different results for the actual test statistics, as well as for the p-values: 5.8481 for the $\chi^2$ and 2.4183 for the z-test, where $\small 2.4183^2=5.84817$ (thank you, @mark999). The p-value for the $\chi^2$ test is 0.01559, while for the z-test is 0.0077. The difference explained by two-tailed versus one-tailed: $\small 0.01559/2=0.007795$ (thank you @amoeba).

This has been asserted multiple times in CV (here, here, here and others).

These two tests statistics are clearly different, and result in different results for the actual test statistics, as well as for the p-values: 5.8481 for the $\chi^2$ and 2.4183 for the z-test, where $\small 2.4183^2=5.84817$ (thank you, @mark999). The p-value for the $\chi^2$ test is 0.01559, while for the z-test is 0.0077. The difference explained by two-tailed versus one-tailed: $\small 0.01559/2=0.007795$ (thank you @amoeba).

This has been asserted multiple times in CV (here, here, here and others).

These two tests statistics are clearly different, and result in different results for the actual test statistics, as well as for the p-values: 5.8481 for the $\chi^2$ and 2.4183 for the z-test, where $\small 2.4183^2=5.84817$ (thank you, @mark999). The p-value for the $\chi^2$ test is 0.01559, while for the z-test is 0.0077. The difference explained by two-tailed versus one-tailed: $\small 0.01559/2=0.007795$ (thank you @amoeba).

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And indeed we can proveprove that $\chi^2_{1\,df}$ is equivalent to $X^2$ with $X\sim N(0,1)$:

And indeed we can prove that $\chi^2_{1\,df}$ is equivalent to $X^2$ with $X\sim N(0,1)$:

And indeed we can prove that $\chi^2_{1\,df}$ is equivalent to $X^2$ with $X\sim N(0,1)$:

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Antoni Parellada
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