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I was trying to solve the following exercise:

"Given a population $\Omega$, normally distributed, and its sample $C=\left\{5,10,6,14\right\}$, calculate the confidence interval of the population mean with a $90\%$ significance level. Afterwards, test with a $1\%$ significance level if the population mean might be supposed equal to 10.$

First of all, I calculated the sample mean $$\bar{X}=35/4,$$ then the unbiased sample variance $$S^2=\frac{203}{16},$$ and finally I wrote the confidence interval $$\left|\frac{\bar{X}-\mu}{\frac{S}{\sqrt{n}}}\right|<t_{0.90}(3)\Rightarrow-\frac{\sqrt{n}}{S}t_{0.90}+\bar{X}<\mu<\frac{\sqrt{n}}{S}t_{0.90}+\bar{X}.$$ Is that correct? I'm wondering if the "smallness" of the sample requires an altnernative procedure to solve the esercise. Then I completed the latter calculating the statistics $$T=\frac{\bar{X}-\mu_0}{\frac{S}{\sqrt{n}}}\simeq -0.68$$ and I stated the the null hypotesis should not be rejected since $$|-0.68|<t_{0.975}(3)=3.182.$$ Any suggestion would be highly appreciated. Thanks in advance!

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  • $\begingroup$ You're almost there. Just think what the "accepted region" will look like if you reject 10% on each side. This will create you an $80%$ confidence interval. $\endgroup$
    – MRobinson
    Sep 5, 2018 at 15:52
  • $\begingroup$ I think I've got what you meant. The quantile I should consider is $t_{0.95}$, isn't it? And what about the rest? Is everything ok? $\endgroup$ Sep 5, 2018 at 16:07
  • $\begingroup$ You've got $\dfrac{\sqrt n} S$ where you need $\dfrac S {\sqrt n}. \qquad$ $\endgroup$ Sep 5, 2018 at 16:36
  • $\begingroup$ Yes, that's a typo! Thank you! $\endgroup$ Sep 5, 2018 at 18:03

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The smallness of the sample is precisely the reason why it's important to use the t-distribution rather than the normal distribution, in this case with $3$ degrees of freedom.

$\require{cancel}$ \begin{align} \text{wrong: } & \xcancel{-\frac{\sqrt n}S t_{0.95}+\bar{X}<\mu<\frac{\sqrt n}S t_{0.95}+\bar{X}. \vphantom{\frac{\displaystyle\int}{\displaystyle\int}} } \\[10pt] \text{right: } & -\frac S {\sqrt n} t_{0.95}+\bar{X}<\mu< \frac S {\sqrt n} t_{0.95}+\bar{X}. \end{align}

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  • $\begingroup$ May I ask you why $2$ degrees of freedom instead of $3$? The length of the sample is $4$. $\endgroup$ Sep 5, 2018 at 18:05
  • $\begingroup$ @GianoRugge : Sorry -- typo. It's $3. \qquad$ $\endgroup$ Sep 6, 2018 at 1:02
  • $\begingroup$ @ MichaelHardy Moreover, the quantile should be $t_{0.95}$ since $5%$ is both in the right tail and in the left one, shouldn't it? $\endgroup$ Sep 6, 2018 at 7:43
  • $\begingroup$ @GianoRugge : Correct. $\qquad$ $\endgroup$ Sep 6, 2018 at 16:07

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