## Related questions with answers

Consider the data set 7, 9, 14, 15, 22. The standard error of the sample mean is: A. 2.53 B. 2.56 C. 2.59 D. 2.62 E. 2.65

Solution

VerifiedLet $X_1, X_2, ... , X_n$ n observations from population with mean $\mu$ and variance $\sigma^2$. Then $\textbf{sample mean}$ is defined by:

$\begin{align*} \overline{X}=\frac{1}{n}\sum_{i=1}^nX_i, \end{align*}$

and it has normal distribution $N(\mu, \frac{\sigma^2}{n})$ and it is point estimate of $\mu$, e.c. $\widehat{\mu}=\overline{X}$.

$\textbf{Standard error}$ of $\overline{x}$ is $\frac{\sigma}{\sqrt{n}}$. But in many cases standard deviation is unknown and we can replace it with sample standard deviation s, where $s=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\overline{X})^2$ is sample standard deviation.

The sample is 7, 9, 14, 15, 22.

$\begin{align*} \overline{x}&=\frac{7+9+14+15+22}{5}=13.4,\\ s&=\sqrt{\frac{(7-13.4)^2+(9-13.4)^2+(14-13.4)^2+(15-13.4)^2+(22-13.4)^2}{5-1}}\\ &=\sqrt{\frac{137.2}{4}}=\sqrt{34.3}=5.86,\\ \text{Standard error}(\overline{x})&=\frac{5.86}{\sqrt{5}}=\fbox{2.62}. \end{align*}$

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