91Ó°ÊÓ

The amount of caffeine obtained in one can of soft drink from each of the 20 brands is recorded. The confidence interval is given as\(99\% \).

Let\(n = 20\)be the number of brands.

Themean value is given below:

\(\begin{array}{c}\bar x = \frac{{\sum\limits_{i = 1}^n {{x_i}} }}{n}\\ = \frac{{0 + 0 + .... + 41 + 47}}{{20}}\\ = \frac{{651}}{{20}}\\ = 32.55\end{array}\)

The mean value is 32.55 mg per 12oz.

The standard deviation of the amount of caffeine is:

\(\begin{array}{c}{s_{}} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {0 - 32.55} \right)}^2} + {{\left( {0 - 32.55} \right)}^2} + ... + {{\left( {41 - 32.55} \right)}^2} + {{\left( {47 - 32.55} \right)}^2}}}{{20 - 1}}} \\ = \sqrt {\frac{{7854.95}}{{19}}} \\ = 20.3327\end{array}\)

Therefore, the standard deviation is 20.3327 mg per 12oz.

T-distribution would be used as the population standard deviation is unknown. Moreover, it is assumed that the population is normally distributed.

The degrees of freedom are as follows:

\(\begin{array}{c}{\rm{df}} = n - 1\\ = 20 - 1\\ = 19\end{array}\)

At\(99\% \)confidence interval and\(\alpha = 0.01\)with 19 degrees of freedom, the critical value can be obtained using the t-table.

\(\begin{array}{c}{t_{crit}} = {t_{\frac{\alpha }{2},df}}\\ = {t_{\frac{{0.05}}{2},19}}\\ = 2.86\end{array}\)

The margin of error is given as follows:

\(\begin{array}{c}E = {t_{crit}} \times \frac{s}{{\sqrt n }}\\ = 2.86 \times \frac{{20.3327}}{{\sqrt {20} }}\\ = 13.0073\end{array}\)

The confidence interval is given as follows:

\(\begin{array}{c}CI = \bar x - E < \mu < \bar x + E\\ = \left( {32.55 - 13.0073 < \mu < 32.55 + 13.0073} \right)\\ = (19.5427 < \mu < 45.5573)\end{array}\)

Thus, with 99% confidence it can be expressed that the mean amount of caffeine obtained in one can from each of the 20 brands lies between (19.5 mg per 12oz and 45.6 mg per 12oz).

The normal Q-Q plot is sketched using the following steps:

1. Draw two axes; horizontal and vertical.
2. Mark z-scores corresponding to the observations on the change by means of scaling the axes.
3. Thus, the relevant graph along with a tentative line is shown below.

In the graph, the points deviate from the straight line.

Since the plotted points are not close to the straight line, the population is inferred to be non-normal. In non-normal distribution, the parametric confidence interval leads to a bad result.