Question

the marks scored by five students in a test of statistics carrying 100 marks are 50, 60, 50, 60 and 40. A simple random sample of size 4 draws without replacement construct the sampling distribution of sample mean and find the standard error of the sample mean?

Answer 1

We have population values $50,60,50,60,40,$ population size $N=5,$ and sample size $n=4.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{5}{4}=5$$

$$mean=\mu=\dfrac{50+60+50+60+40}{5}=52$$

$$Variance=\sigma^2$$

$$=\dfrac{1}{5}((50-52)^2+(60-52)^2+(50-52)^2$$

$$+(60-52)^2+(40-52)^2)=56$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c} No & Sample & Mean \\ \hline 1 & (50, 60, 50, 60) & 55 \\ \hdashline 2 & (50, 60, 50, 40) & 50 \\ \hdashline 3 & (50, 60, 60, 40) & 52.5 \\ \hdashline 4 & (50, 50, 60, 40) & 50 \\ \hdashline 5 & (60, 50,60,40) & 52.5 \\ \hdashline \end{array}$$

The sampling distribution of the sample mean $\bar{x}$ and its mean and standard deviation are:

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:} \bar{x} & f & f(\bar{x})& \bar{x}f(\bar{x})& \bar{x}^2f(\bar{x}) \\ \hline 50 & 2 & 2/5 & 20 & 1000 \\ \hdashline 52.5 & 2 & 2/5 & 21 & 1102.5 \\ \hdashline 55 & 1 & 1/5 & 11 & 605 \\ \hdashline Total & 5 & 1 & 52 & 2707.5 \\ \hdashline \end{array}$$

$$E(\bar{X})=\sum\bar{x}f(\bar{x})=52$$

$$Var(\bar{X})=\sum\bar{x}^2f(\bar{x})-(\sum\bar{x}f(\bar{x}))^2$$

$$=2707.5-52^2=3.5$$

$$E(\bar{X})=52=\mu$$

$$Var(\bar{X})=3.5=\dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})=\dfrac{56}{4}(\dfrac{5-4}{5-1})$$

$$SE=\dfrac{s}{\sqrt{n}}=\dfrac{\sqrt{3.5}}{\sqrt{4}}\approx0.9354$$