Chapter 11, Problem 32CCSR

Solution

To calculate: The mean, standard deviation, and margin of error and interpret the margin of error.

The mean and standard deviation of the given data is 14.4 and 2.28 respectively. The margin of error with a 95% confidence level is 0.82. The margin of error is the deviation from the mean with 95% confidence where the population mean lies.

Given:The data of a sample of ages of students in an art school.

13	12	15	18	14	16	18	12	13	14	14	17	15	8	17
16	12	16	14	15	13	13	17	15	14	18	16	12	12	13

Concept used:

The mean and standard deviation are calculated using the formula

  $\begin{array}{l}\text{Mean}\stackrel{\_\_}{X}=\frac{{\displaystyle \sum X}}{n}\\ \text{Standard deviation s}=\sqrt{\frac{{{\displaystyle \sum \left(X-\stackrel{\_\_}{X}\right)}}^2}{n-1}}\\ \text{where n is sample size}\end{array}$

The margin of error for 95% confidence level is calculated using formula $ME=1.96\times \frac{s}{\sqrt{n}}$ . The margin of error is the deviation away from mean where with 95% confidence population mean lies.

Calculation:

The mean and standard deviation are calculated as shown below

  $\begin{array}{c}\text{Mean}\stackrel{\_\_}{X}=\frac{{\displaystyle \sum X}}{n}\\ =\frac{13+12+15+\mathrm{...}+12+13}{30}\\ =14.4\end{array}$

  $\begin{array}{l}\text{s}=\sqrt{\frac{{{\displaystyle \sum \left(X-\stackrel{\_\_}{X}\right)}}^2}{n-1}}\\ s=\sqrt{\frac{{\left(13-14.4\right)}^2+{\left(12-14.4\right)}^2+\mathrm{..}{\left(13-14.4\right)}^2}{30-1}}\\ s=\sqrt{\frac{151.2}{29}}\\ s=2.283373\end{array}$

The margin of error is calculated for 95% confidence level is

  $\begin{array}{c}ME=1.96\times \frac{s}{\sqrt{n}}\\ =1.96\times \frac{2.283373}{\sqrt{30}}\\ =0.817095\end{array}$

Interpretation:The mean and standard deviation for sample of students in an art school is 14.4 and 2.28 rounded to the hundredth place. The margin of error for 95% confidence level is 0.82. Thus at 95% confidence level the population mean is within 0.82 years away from the sample mean 14 which is $\left(14.4\pm 0.82\right)$ .