Chapter 11.1, Problem 5E

a)

To determine

To find the degrees of freedom.

Answer to Problem 5E

The df = 3

Explanation of Solution

Given:

Type of nut	Count	Frequency	Expected
Cashew	83	52%	78
Almond	29	27%	40.5
Macadamia	20	13%	19.5
Brazil	18	8%	12
Total	150	100%	150

Chi square test statistic = 6.60

Calculation:

The expected counts are large enough to use a chi-square distribution if all expected counts are at least 5. This condition is satisfied.

The degrees of freedom = df = c-1 = 4-1 = 3

b)

To determine

To find the p-value using table and technology.

Answer to Problem 5E

P-value = 0.0858

Explanation of Solution

Given:

Type of nut	Count	Frequency	Expected
Cashew	83	52%	78
Almond	29	27%	40.5
Macadamia	20	13%	19.5
Brazil	18	8%	12
Total	150	100%	150

Chi square test statistic = 6.60

The degrees of freedom = df = 3

Calculation:

The p-value using table for df = 3 is,

0.05< P < 0.10

The excel formula, =CHIDIST(6.6,3)

P-value = 0.0858

c)

To determine

To explain the conclusion.

Answer to Problem 5E

There is not sufficient evidence to reject the claim about the specified probability distribution.

Explanation of Solution

Given:

Type of nut	Count	Frequency	Expected
Cashew	83	52%	78
Almond	29	27%	40.5
Macadamia	20	13%	19.5
Brazil	18	8%	12
Total	150	100%	150

Chi square test statistic = 6.60

The degrees of freedom = df = c-1 = 4-1 = 3

P-value = 0.0858

Calculation:

Here, p-value > 0.05, fail to reject H₀.

Conclusion: There is not sufficient evidence to reject the claim about the specified probability distribution.