In a sample of $15$ normally distributed random variables with unknown expectation $µ$ and variance $σ^2$, the sample mean was $10.3$ and the sample variance $s^2$ was $0.13.$ Make a symmetric confidence interval for $µ$ at the $99$% confidence level.

Well, the confidence interval is given by

$$\mu=\overline{X}\pm z\frac{s}{\sqrt{n}},$$

where $\Phi(z)=(1+q)/2.$ I have that

$$\Phi(z)=\frac{1+0.99}{2}=0.995 \implies z=\Phi^{-1}(0.995)=??$$

Which one of the ones I've marked should I choose?

The book says $z=2.977,$ which doesn't seem close to what I had in mind. Why this deviation?

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  • $\begingroup$ You need to consult the inverse-phi tables. I dont have them in front of me but i think its 2.5758 from memory... $\endgroup$ – David Quinn May 15 '18 at 17:34
  • $\begingroup$ Have you seen the picture I've inserted? That's the exact table I'm refering to and there are 7 different values of 0.995 there. $\endgroup$ – Parseval May 15 '18 at 17:37
  • $\begingroup$ Yes this is the phi table, but you need the inverse phi table! $\endgroup$ – David Quinn May 15 '18 at 17:38
  • $\begingroup$ I don't have such a table, it's not in the book. And the book refers to the phi-table only. Also, the answer you got as 2.5758 is wrong according to the book. $\endgroup$ – Parseval May 15 '18 at 17:42
  • 1
    $\begingroup$ @Parseval At this "normal" table we have $\Phi(2.57)=0.99492$ and $\Phi(2.58)= .99506$. Thus $\Phi^{-1}(0.995)\approx 2.575$. You can obtain a better result by applying $\text{linear interpolation}$. $\endgroup$ – callculus May 15 '18 at 18:49

The discrepancy is related to the $t$-distribution. Recall that the cost of replacing an unknown $\sigma^2$ with its estimate $S^2$ is that the resulting quantity $$\frac{\overline x - \mu}{S / \sqrt n}$$ has a student's $t$-distribution, not a normal distribution. (If you replaced $S$ with $\sigma$, then the quantity would have a normal distribution; also, as $n \to \infty$, the student's $t$-distribution approximates a standard normal distribution.)

As a result, you are consulting the wrong table. You instead need a table of values for a $t$-distribution with 14 df. You can verify that the book's calculator is correct with, for instance, this calculator (df = 14, $P(T \leq t) = 0.995$).

  • $\begingroup$ Ah I see now. So everytime my true variance is unknown, but is approximated by a sample, then I have to consult the degrees of fredom table? $\endgroup$ – Parseval May 15 '18 at 18:26
  • $\begingroup$ Correct.${}{}{}$ $\endgroup$ – Aaron Montgomery May 15 '18 at 18:26
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    $\begingroup$ I'll keep this in mind, thank you Aaron! $\endgroup$ – Parseval May 15 '18 at 18:30

I am reliably informed by my calculator that $$\Phi^{-1}(0.995)=2.57582936$$

  • $\begingroup$ You have a great calculator. It can be checked by an online calculator as well. $\endgroup$ – callculus May 15 '18 at 17:57
  • $\begingroup$ I have a Ti-84 Texas instrument. I can also confirm this now. $\endgroup$ – Parseval May 15 '18 at 18:17

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