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I have this exercise where enter image description here is the answer. Using this annex(text at the top means tenths of the x), enter image description herethe value becomes enter image description here . What I don't understand is how it becomes 2,58. Our first answer is 0,495, so I'm looking on the left searching the 0,4 table. After that, I'm looking at the tenths of our answer, but there is no number equivalent to 2,58. Is the exercise wrong or am I not understanding the annex right?

noirangel110
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  • There is a 2.58 value in the margins of the table, what's not there is an exact 0.495 in the body of the table. I'm not sure whether this is your issue, but see the answer here and the two linked answers for details of how to read the tables. Typically once you find the z-values with areas either side of the desired probability, you'd want to use interpolation to back out a more accurate z-value, but here you're stymied by the old bugbear, ... ctd – Glen_b May 28 '23 at 22:52
  • ctd ... severe loss of accuracy when subtracting two almost-equal quantities, reducing the significant figures in the difference to a single digit at best (keeping in mind the table values are rounded off, it's less than a whole digit), which makes the interpolation unhelpful. If you have no way to get a more accurate answer, the usual approach is to err on the more conservative side (which is more conservative depends on the circumstances - wider CIs and larger p-values are generally 'more conservative'). – Glen_b May 28 '23 at 22:52

1 Answers1

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Note to readers: On this forum, $\Phi(x)$ usually denotes the cumulative distribution function of the standard normal random variable, and thus $\Phi(x) = \Phi_{\text{CDF}}(x) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}}e^{-x^2/2} \mathrm dx$, which takes on values increasing from $0$ at $-\infty$ to $1$ at $\infty$ as $x$ increases from $-\infty$ to $\infty$. In contrast, the OP's notation uses $\Phi(x)$ (I will call it $\Phi_{\text{OP}}(x))$ to mean $\int_0^x \frac{1}{\sqrt{2\pi}}e^{-x^2/2} \mathrm dx$, which takes on values increasing from $0$ at $x=0$ to $0.5$ at $x=\infty$ as $x$ increases from $0$ to $\infty$. Thus, $\Phi_{\text{CDF}}(x) = 0.5+ \Phi_{\text{OP}}(x)$, The table shows the values of $\Phi_{\text{OP}}(x)$.

You need to find the value of $x$ for which the "table entry" is $0.4950$. Unfortunately, there is no such entry; the closest that we can come is $x=2.57$ for which the table entry is $0.4949$ or $x=2.58$ for which the table entry is $0.4951$. What is really needed is the smallest value of $x$ for which we are guaranteed that the "table entry" is $0.495$ or larger. By the way, that's an exact number, no round-off etc: if you write the number as a decimal, all the digits after that $5$ are 0; $0.495000000000000\cdots$. Now, since $x=2.58$ clearly satisfies the desired criterion $(0.4951 > 0.495)$, we choose that value as the answer.

An alternative would be to do linear interpolation and say that $x=2.575$, midway between $x=2.57$ and $x=2.58$, is the right answer, but this would be incorrect. The value of $x$ at which $\Phi_{\text{CDF}}(x) = 0.9950$, which value of $x$ is denoted by $x_{0.9950}$, is just a little smaller than $2.58$. According to the tables on pages 968-871 of Abramowitz and Stegun, Handbook of Mathematical Functions, $\Phi_{\text{CDF}}(2.58) = 0.9950599642\cdots$ (and so $\Phi_{\text{OP}}(2.58) = 0.4950599642\cdots$), and so the exact value of $x_{0.9950}$ is definitely larger than the $2.575$ value obtained by linear interpolation.

Dilip Sarwate
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