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Olive oil is bottled. The filling quantity is normally distributed. A sample of $n=27$ bottlings had a mean value of $198.3$ ml and a sample standard deviation of $10.3$ ml.

I would like to find the lower limit for a $0.95$ confidence interval for the filling quantity to be expected on average.

So, we have that $\bar{X}=198.3, n=27, \sigma=10.3$

Now, we know that $$0.95=P\left(\frac{\sqrt{n}|\bar{X}-\mu|}{\sigma}\leq 1.96\right)\\=P\left(\bar{X}-1.96\left(\frac{\sigma}{\sqrt{n}}\right)\leq \mu\leq \bar{X}+1.96\left(\frac{\sigma}{\sqrt{n}}\right)\right)$$

So, the lower limit it $$198.3-1.96(1.98)=198.3-3.89=194.41$$

I have the following answers in my textbook which I'm not sure exactly how to choose correctly.

$(A) (193.90,194.30]$

$(B) (194.70,195.10]$

$(C) (194.30,194.70]$

$(D) (A)-(C) \text{false}$

I would choose $C$ since the value I calculated lies in that interval. However I am not sure if there is any more rigorous way to calculate this interval or is it just that I'm supposed to pick an answer which contains my calculated value?

Slim Shady
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1 Answers1

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Your answer is not correct because your population's distribution is $N(\mu;\sigma^2)$ with both parameters unknown. Thus the satistic to be used is the following

$$t=\frac{\overline{X}_n-\mu}{s}\sqrt{n}\sim\mathcal{T}_{(n-1)}$$

tommik
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