Chapter 11.1, Problem 16E

To determine

To check: Whether the population of Froot Loops that have been produced contain an equal proportion of each flavor.

Answer to Problem 16E

No, the population of Froot Loops that have been produced do not contain an equal proportion of each flavor.

Explanation of Solution

Given:

Flavour	Count
Orange	28
Lemon	21
Cherry	16
Raspberry	25
Blueberry	14
Lime	16

Formula used:

The formula to compute the test statistic is

  ${\chi }^2={\displaystyle \sum \frac{{\left(O-E\right)}^2}{E}}$

Calculation:

The null and alternative hypotheses are:

  $\begin{array}{l}{H}_0:p_1=p_2=p_3=p_4=p_5=p_6=\frac{1}{6}\\ H_a:\text{ At least one of }{p}_i\text{ is different}\end{array}$

The calculation for test statistic is done as:

Observed value	Expected value	$\left(O-E\right)$	${\left(O-E\right)}^2$	${\left(O-E\right)}^2/E$
28	20	8	64	3.2
21	20	1	1	0.05
16	20	-4	16	0.8
25	20	5	25	1.25
14	20	-6	36	1.8
16	20	-4	16	0.8
				${\displaystyle \sum {\frac{\left(O-E\right)}{E}}^2=7.9}$

The test statistic is:

  $\begin{array}{c}{\chi }^2={\displaystyle \sum {\frac{\left(O-E\right)}{E}}^2}\\ =7.9\end{array}$

The degree of freedom is calculated as:

  $\begin{array}{c}\text{Degree of freedom =Number of categories - 1}\\ =6-1\\ =5\end{array}$

The p-value using chi-square table at 5 degree of freedom is 0.162.

The p -value is above significance level. The null hypothesis does not get rejected. Thus, there is insufficient evidence to conclude that the population of Froot Loops that have been produced do not contain an equal proportion of each flavor.