Find sample size given standard deviation, sample mean, confidence interval

Question

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.

Answer 1

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.