Chapter 11.1, Problem 6E

a)

To determine

To find the degrees of freedom.

Answer to Problem 6E

The df = 2

Explanation of Solution

Given:

Color	Observed Count	Frequency	Expected count
Red	85	0.47	94.74
Black	99	0.47	94.74
Green	16	0.05	10.53
Total	200	1.00	200

The chi-square test statistic = 4.04

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 = 3-1 = 2

b)

To determine

To find the p-value using table and technology.

Answer to Problem 6E

The p-value = 0.1108

Explanation of Solution

Given:

Color	Observed Count	Frequency	Expected count
Red	85	0.47	94.74
Black	99	0.47	94.74
Green	16	0.05	10.53
Total	200	1.00	200

The chi-square test statistic = 4.04

The degrees of freedom = 2

Calculation:

The p-value using table for df = 2 is,

0.10 < P < 0.15

The excel formula, =CHIDIST(4.40,2)

P-value = 0.1108

c)

To determine

To explain the conclusion.

Answer to Problem 6E

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

Explanation of Solution

Given:

Color	Observed Count	Frequency	Expected count
Red	85	0.47	94.74
Black	99	0.47	94.74
Green	16	0.05	10.53
Total	200	1.00	200

The chi-square test statistic = 4.04

The degrees of freedom = 2

The p-value = 0.1108

Calculation:

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

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