Question

Aw, nuts! A company claims that each batch of its deluxe mixed nuts

contains $52\%$ cashews, $27\%$almonds, $13\%$ macadamia nuts, and 8% Brazil nuts. To test this claim, a quality-control inspector takes a random sample of $150$ nuts from the latest batch. The table displays the sample data.

a. State appropriate hypotheses for performing a test of the company’s claim.

b. Calculate the expected count for each type of nut.

c. Calculate the value of the chi-square test statistic.

Short answer

Part (a)$H_0:p_c=52\%,p_a=27\%,p_m=13\%,p_b=8\%,$$H_a:Atleastoneofthe{p}_i’sisincorrect.$

Part (b)

Type of nuts	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

Part (c) The chi-square test statistic $=6.60$

Step-by-step solution

Part (a) Step 1: Given information

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

Part (a) Step 2: Explanation

The null and alternative hypotheses:

$$\begin{gathered} H_0:p_c=52\%,p_a=27\%,p_m=13\%,p_b=8\%, \\ {H}_a:Atleastoneofthe{p}_i’sisincorrect. \end{gathered}$$

Part (b) Step 1: Explanation

$Expectedvalue=n\times frequency$

Where, $n=150$

Expected count will be:

Type of nuts	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

Part (c) Step 1: Explanation

Using excel

Type of nuts	Observed	Expected	$O-E$	${\left(O-E\right)}^2/E$	% of chisq
Cashew	83	78.000	5.000	5.000	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