0
$\begingroup$

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?

enter image description here

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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, $P(T \leq t) = 0.995$).

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ I'll keep this in mind, thank you Aaron! $\endgroup$ – Parseval May 15 '18 at 18:30
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.