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A machine is set up such that the average content of juice per bottle equals u. Assume that the population standard deviation is $5$cl.

  1. A sample of 100 bottles yields to an average of $48$cl. Calculate a $90\%$ and $95\%$ confidence interval for the average content.

  2. Suppose the sample size is unknown. What sample size is required to estimate the average contents to be within $0.5$cl at the $95\%$ confidence level?

For the first question I found that:

  • $\alpha=10\%$ gives $\text{CI} = \bar x\pm t_{1-\alpha/2}\frac\sigma{\sqrt n}=48\pm t_{0.05}\frac5{\sqrt{100}}=(47.175,48.825)$ and similarly
  • $\alpha=5\%$ gives $\text{CI} =48\pm t_{0.025}\frac5{\sqrt{100}}= (47.02,48.98)$.

I have difficulty regarding the second question. I have never faced such a question and don't really know how to tackle the problem.

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  • $\begingroup$ For the first question, CI = [X - t * sd/sqrt(n) ; X + t * sd/sqrt(n)] with X as the sample mean, sd the standard deviation, t the value of the normal distribution given a confidence level and n the sample size. $\endgroup$
    – user90379
    Dec 27 '19 at 19:48
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At the $95\%$ confidence level, the confidence interval is given by $\mu\pm1.96\sigma/\sqrt n$ where $\mu$ is the mean, $\sigma$ is the standard deviation and $n$ is the sample size. From the question, we want $1.96\sigma/\sqrt n=0.5$ and since $\sigma=5$, the sample size required is $n=(1.96\cdot5/0.5)^2=384.16$, or $385$ after rounding up.

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