Chapter 11.1, Problem 13E

a.

To determine

To state: The appropriate hypothesis for significance test

Explanation of Solution

Given:

The percentage of each flavor is 20%.

Calculation:

Using the provided information, the null and alternative hypotheses are:

  $\begin{array}{l}{H}_0:p_{Lemon}=p_{Lime}=p_{Orange}=p_{Straberry}=p_{Grape}=0.20\\ H_a:\text{ At least one of the }{p}_i\text{ is incorrect}\end{array}$

b.

To determine

To find: The expected counts.

Explanation of Solution

Calculation:

The expected counts could be calculated as:

Flavour	Expected count
Lemon	60 (0.2) = 12
Lime	60 (0.2) = 12
Orange	60 (0.2) = 12
Strawberry	60 (0.2) = 12
Grape	60 (0.2) = 12

c.

To determine

To find: The value of the test statistic that is required to obtain to have the evidence against the claim of the company

Explanation of Solution

Calculation:

The degree of freedom is calculated as:

Degree of freedom = Number of categories -1 = 5-1 = 4.

The critical values are:

  $\begin{array}{l}\alpha =0.05,{\chi }^2=9.49\\ \alpha =0.01,{\chi }^2=13.28\end{array}$

The values that are above the chi-square test statistic would reject the null hypothesis and significant test.

d.

To determine

To show: The calculation the test statistic.

Explanation of Solution

Calculation:

The calculation for test statistic is done as:

Observed value	Expected value	$\left(O-E\right)$	${\left(O-E\right)}^2$	${\left(O-E\right)}^2/E$
6	10.4	-4.4	19.36	1.8615
6	10.4	-4.4	19.36	1.8615
8	10.4	-2	4	0.4
16	10.4	5.6	31.36	3.0154
16	10.4	5.6	31.36	3.0154
				${\displaystyle \sum {\frac{(}{}}^{}}$ O−E)E2=10.1538

The test statistic is:

  $\begin{array}{c}{\chi }^2={\displaystyle \sum {\frac{\left(O-E\right)}{E}}^2}\\ =10.1538\end{array}$

This value of the test statistic would p-value between 0.01 and 0.05.