Question

Caffeine in Soft Drinks Listed below are measured amounts of caffeine (mg per 12oz of drink) obtained in one can from each of 20 brands (7UP, A&W root Beer, Cherry Coke, …TaB). Use a confidence interval 99%. Does the confidence interval give us good information about the population of all cans of the same 20 brands that are consumed? Does the sample appear to be from a normally distributed population? If not, how are the results affected?

0 0 34 34 34 45 41 51 55 36 47 41 0 0 53 54 38 0 41 47

Short answer

The confidence interval ranges from19.5 mg per 12oz to 45.6 mg per 12oz. It provides the limits of the mean amount of caffeine obtained in one can of soft drink from each of the 20 brands.

No, it does not follow a normal distribution. Therefore, the parametric confidence interval method does not provide a good result.

Step-by-step solution

Given information

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.

Calculate the mean 

Let $n=20$be the number of brands.

Themean value is given below:

$$\begin{gathered} \overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n} \\ =\frac{0+0+....+41+47}{20} \\ =\frac{651}{20} \\ =32.55 \end{gathered}$$

The mean value is 32.55 mg per 12oz.

Calculate the standard deviation

The standard deviation of the amount of caffeine is:

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

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

Compute the degrees of freedom

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:

$$\begin{gathered} \text{df}=n-1 \\ =20-1 \\ =19 \end{gathered}$$

Compute the critical value

At confidence interval and with 19 degrees of freedom, the critical value can be obtained using the t-table.

$$\begin{gathered} t_{crit}=t_{\frac{\alpha }{2},df} \\ =t_{\frac{0.05}{2},19} \\ =2.86 \end{gathered}$$

Compute the margin of error

The margin of error is given as follows:

$$\begin{gathered} E=t_{crit}\times \frac{s}{\sqrt{n}} \\ =2.86\times \frac{20.3327}{\sqrt{20}} \\ =13.0073 \end{gathered}$$

Compute the confidence interval

The confidence interval is given as follows:

$$\begin{gathered} CI=\overline{x}-E<\mu <\overline{x}+E \\ =\left(32.55-13.0073<\mu <32.55+13.0073\right) \\ =(19.5427<\mu <45.5573) \end{gathered}$$

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).

Check normality

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.