Chapter 11.1, Problem 1E

a)

To determine

To state the hypotheses.

Answer to Problem 1E

  $H_0:p_c=52\%,p_a=27\%,p_m=13\%,p_b=8\%,$

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

Explanation of Solution

Given:

Type of nut	Count	Frequency
Cashew	83	52%
Almond	29	27%
Macadamia	20	13%
Brazil	18	8%

Calculation:

The null and alternative hypotheses:

  $H_0:p_c=52\%,p_a=27\%,p_m=13\%,p_b=8\%,$

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

b)

To determine

To find expected count for each type of nut.

Explanation of Solution

Given:

Type of nut	Count	Frequency
Cashew	83	52%
Almond	29	27%
Macadamia	20	13%
Brazil	18	8%

Formula:

  $Expectedvalue=n\times frequency$

Where, n = 150

Calculation:

Expected count will be:

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

c)

To determine

To find chi-square test statistic.

Answer to Problem 1E

The chi-square test statistic = 6.60

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

Calculation:

Using excel,

Type of nut	observed	expected	O - E	(O - E)² / E	% of chisq
Cashew	83	78.000	5.000	0.321	4.86
Almond	29	40.500	-11.500	3.265	49.49
Macadamia	20	19.500	0.500	0.013	0.19
Brazil	18	12.000	6.000	3.000	45.46
	150	150.000	0.000	6.599	100.00
	6.60	chi-square
	3	Df
	.0858	p-value