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.

  • $\begingroup$ 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

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.


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.