1. ## A random sample of grades of 50

A random sample of grades of 50 students in physics out of 200 students showed a mean of 75 and a standard deviation of 10. What are the 95% confidence limits for estimates of mean of the 200 grades? Since the population size is not very large compared with the sample size, make adjustments for this?

2. Here we have an unknown population mean $\displaystyle \mu$, and a known standard deviation $\displaystyle \sigma$ based on a simple random sample of size n, which we are also given.

$\displaystyle \bar{X} = 75$

$\displaystyle n = 50$

$\displaystyle \sigma = 10$

Since our sample is large enough (usually n > 30 will suffice), we consider the sample mean $\displaystyle \bar{X}$ normally distributed, with the same expectation for the population mean $\displaystyle \mu$, but with standard error $\displaystyle \frac{\sigma}{\sqrt{n}}$. We standardize and get a random variable

$\displaystyle Z = \frac{\bar{X}-\mu}{\sigma / \sqrt{n}} = \frac{75-\mu}{1.4142}$

that depends on the population mean parameter $\displaystyle \mu$ (which is to be estimated), but with a normal distribution independent of $\displaystyle \mu$. Thus, we can find the range endpoints -z and z, independent of the population mean parameter, which is the range that Z falls somewhere in between with 95% probability. So we set the following equation up:

$\displaystyle P(-z \leq Z \leq z) = 1 - \alpha = 0.95$

$\displaystyle \implies P(Z \leq z) = 1 - \frac{\alpha}{2} =0.975$

$\displaystyle \implies z = 1.96$ (which you could have looked up easily enough (or had it memorized)

So now we have

$\displaystyle 0.95 = 1 - \alpha = P(-z \leq Z \leq z) = P(-1.96 \leq \frac{\bar{X}-\mu}{\sigma / \sqrt{n}} \leq 1.96)$

$\displaystyle = P\bigg{(}75 - 1.96(1.4142) \leq \mu \leq \bar{x} + 1.96 (1.4142 )\bigg{)}$

$\displaystyle = P(75 - 2.7719 \leq \mu \leq 75 + 2.7719)$

$\displaystyle CI = (72.2281 ; 77.7719)$