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I am stuck with a small issue (compared to others 1, 2 I am still stuck with) in generalizing CI for a good (i.e normally distributed) sampling distribution

If my normal approximation is described by $x$ (with sample mean $\overline{X}$ and sample SD $S = \frac{\sigma}{\sqrt{n}}$), I have,

$$ Pr(x-1.96\dfrac{\sigma}{\sqrt{n}} \leq \mu \leq x + 1.96\dfrac{\sigma}{\sqrt{n}}) = 0.95 \tag{1} $$

Now, the probability area 0.025 in standard normal distribution Z, corresponds to $Z = -1.96$

So $$ Z_{0.025} = -1.96 \\ Z_{\frac{0.05}{2}} = -1.96 $$

With $\alpha$ as significance level, $1-\alpha = 0.95$, then $\alpha = 0.05$, So

$$ Z_{\frac{\alpha}{2}} = -1.96 \tag{2} $$

Substituting $1$ in $2$, we get

$$ Pr(x+Z_{\frac{\alpha}{2}}\dfrac{\sigma}{\sqrt{n}} \leq \mu \leq x - Z_{\frac{\alpha}{2}}\dfrac{\sigma}{\sqrt{n}}) = 0.95 \tag{3} $$

But books define the other way. For eg, in "Probability and Statistical Inference" by Hogg $et. al$, it is defined as below (page 310)

$$ Pr(x-Z_{\frac{\alpha}{2}}\dfrac{\sigma}{\sqrt{n}} \leq \mu \leq x + Z_{\frac{\alpha}{2}}\dfrac{\sigma}{\sqrt{n}}) = 0.95 \tag{4} $$

which implies

$$ Z_{\frac{\alpha}{2}} = 1.96 \tag{5} $$

which is not true. The probability area covered at $Z = 1.96$ is $0.975$, That is,

$$ Z_{1-\frac{\alpha}{2}} = 1.96 \tag{6} $$

So what am I missing?

My take for now:
Somehow unanimously the book authors (and many others), assume, the normal curve area for right tail, the alternate convention of how to read area. Like here. Then $Z_{0.025} = 1.96$ and all fall in place I suppose. I have been used reading area with left tail probabilities so far which I thought was the convention.

Parthiban Rajendran
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  • do you have a typo here $$Pr(x+Z_{\frac{\alpha}{2}}\dfrac{\sigma}{\sqrt{n}} \leq \mu \leq x - Z_{\frac{\alpha}{2}}\dfrac{\sigma}{\sqrt{n}}) = 0.95 \tag{3}$$ ? – Ahmad Bazzi Sep 11 '18 at 18:01
  • I could not spot (its midnight, am already sleepy), can you please be more specific? – Parthiban Rajendran Sep 11 '18 at 18:25
  • lower bound is greater than upper bound ? – Ahmad Bazzi Sep 11 '18 at 18:26
  • yeah, thats part of the doubt, because in my derivation, I find, $Z_{\frac{\alpha}{2}}$ to be negative. Check equation $2$ in my post – Parthiban Rajendran Sep 11 '18 at 18:27
  • No hold on. You're doing a two sided (tailed) test. $Z_{\frac{\alpha}{2}} > 0$ is a convention. – Ahmad Bazzi Sep 11 '18 at 18:28
  • But for area of 0.025, the respective Z value is $Z = -1.96$, that is, the area covered from $Z=-\infty$ to $-1.96$ is 0.025, so $Z_{0.025} = -1.96$ which is $< 0$? For any Z value, the respective area is always from $-\infty$ to that value which is again a convention? By that convention, $Z=1.96$ covers an area of $0.975$, so $Z_{0.975} = 1.96$? – Parthiban Rajendran Sep 12 '18 at 06:00
  • yeah $z = -1.96$ if you're looking at the left tail, i.e. $Pr( Z < -1.96) = 0.025$. Similarly $Pr(Z > 1.96) = 0.025$... In short, you're saying that $Pr(Z < 1.96) = 0.975$. It should be $Pr(Z > 1.96)$. – Ahmad Bazzi Sep 12 '18 at 12:53

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