Chapter 11.1, Problem 2E

a)

To determine

To state the hypotheses.

Answer to Problem 2E

  $H_0:p_{red}=\frac{18}{38}=0.47,p_{black}=\frac{18}{38}=0.47,p_{green}=\frac{2}{38}=0.05$

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

Explanation of Solution

Given:

Color	Observed Count	Frequency
Red	85	0.47
Black	99	0.47
Green	16	0.05
Total	200	1.00

Calculation:

The null and alternative hypotheses:

  $H_0:p_{red}=\frac{18}{38}=0.47,p_{black}=\frac{18}{38}=0.47,p_{green}=\frac{2}{38}=0.05$

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

b)

To determine

To find expected count for each type of nut.

Explanation of Solution

Given:

Color	Observed Count	Frequency
Red	85	0.47
Black	99	0.47
Green	16	0.05
Total	200	1.00

Formula:

  $Expectedvalue=n\times frequency$

Where, n = 200

Calculation:

Expected count will be:

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

c)

To determine

To find chi-square test statistic.

Answer to Problem 2E

The chi-square test statistic = 4.04

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

Calculation:

Using excel,

Color	observed	Expected	O - E	(O - E)² / E	% of chisq
Red	85	94.737	-9.737	1.001	24.78
Black	99	94.737	4.263	0.192	4.75
Green	16	10.526	5.474	2.846	70.47
Total	200	200.000	0.000	4.039	100.00
	4.04	chi-square
	2	Df
	.1327	p-value