91Ó°ÊÓ

Given in the question that, Kellogg’s Froot Loops cereal comes in six fruit flavors: orange, lemon, cherry, raspberry, blueberry, and lime. Charise poured out her morning bowl of cereal and methodically counted the number of cereal pieces of each flavor. Here are her data:

Given,

The test statistic will be calculated using the formula below.

${\mathrm{χ}}^2=∑\frac{(O-E{)}^2}{E}$

The following are the null and alternative hypotheses:

localid="1650619689827" $H_0:p_1=\frac{1}{6}$

$H_0:p_2=\frac{1}{6}$

$H_0:p_3=\frac{1}{6}$

$H_0:p_4=\frac{1}{6}$

$H_0:p_5=\frac{1}{6}$

$H_0:p_6=\frac{1}{6}$

$H_a:$At least one of $p_i$is different

The test statistic is calculated as follows:

The following is the test statistic:

localid="1650619811979" $$\begin{gathered} {\mathrm{χ}}^2=∑\frac{(O-E{)}^2}{E} \\ =7.9 \end{gathered}$$

The formula for calculating the degree of freedom is:

$$\begin{gathered} Degreeoffreedom=Numberofcategories-1 \\ =6-1 \\ =5 \end{gathered}$$

At $5$degrees of freedom, the $p$-value using the chi-square table is $0.162.$

Here the $p$-Value will be higher than the level of significance. The null hypothesis is not disproved.