Chapter 11.1, Problem 17E

a)

To determine

To state hypotheses.

Answer to Problem 17E

  $H_0:p_{lemon}=0.20,p_{\lim e}=0.20,p_{orange}=0.20,p_{strawberry}=0.20,p_{graph}=0.20$

  $H_a:Atleastoneofthe{p}_i’sisincorrect.$

Explanation of Solution

Given:

The flavor blend is 20% of each flavor.

Calculation:

The null and alternative hypotheses:

  $H_0:p_{lemon}=0.20,p_{\lim e}=0.20,p_{orange}=0.20,p_{strawberry}=0.20,p_{grape}=0.20$

  $H_a:Atleastoneofthe{p}_i’sisincorrect.$

b)

To determine

To find expected count.

Answer to Problem 17E

Expected count for each flavor is 12.

Explanation of Solution

Given:

The flavor blend is 20% of each flavor.

Sample size is 60 candies.

Calculation:

For finding expected count, we need to multiply by 60 to each frequency of flavor. Therefore, expected count is,

Flavor	Frequency	Expected
Lemon	0.2	12
Lime	0.2	12
Orange	0.2	12
Strawberry	0.2	12
Grape	0.2	12

c)

To determine

To explain how large a test statistic would need a significant evidence against claim.

Answer to Problem 17E

If chi square statistic is greater than above critical values at specified level of significance then reject H0.

Explanation of Solution

Given:

The flavor blend is 20% of each flavor.

Sample size is 60 candies.

Flavor	Frequency	Expected
Lemon	0.2	12
Lime	0.2	12
Orange	0.2	12
Strawberry	0.2	12
Grape	0.2	12

Formula:

The degrees of freedom = df = c-1

Calculation:

The df = 5-1 = 4

There will be different critical values for each level of significance.

When significance level = a = 0.05

  ${\chi }^2=9.49$ …Using excel formula, =CHIINV(0.05,4)

  ${\chi }^2=13.28$ …Using excel formula, =CHIINV(0.01,4)

Decision: If chi square statistic is greater than above critical values at specified level of significance then reject H0.

d)

To determine

To test a claim by taking own sample data for 60 candies.

Explanation of Solution

Given:

The flavor blend is 20% of each flavor.

Sample size is 60 candies.

Calculation:

Following table shows the random sample and their frequency:

Flavor	Observed	Frequency
Lemon	20	0.2
Lime	10	0.2
Orange	15	0.2
Strawberry	4	0.2
Grape	11	0.2

Using excel,

Flavor	observed	expected	O - E	(O - E)² / E	% of chisq
Lemon	20	12.000	8.000	5.333	45.07
Lime	10	12.000	-2.000	0.333	2.82
Orange	15	12.000	3.000	0.750	6.34
Strawberry	4	12.000	-8.000	5.333	45.07
Grape	11	12.000	-1.000	0.083	0.70
	60	60.000	0.000	11.833	100.00
	11.83	chi-square
	4	df
	.0186	p-value

Therefore, the p-value is between 0.01 and 0.05