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Q 1.

Expert-verified

The Practice of Statistics for AP Examination

Found in: Page 712

Book edition 6th

Author(s) Daren Starnes, Josh Tabor

Pages 837 pages

ISBN 9781319113339

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Short Answer

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.

Part (a) $H_{0} : p_{c} = 52 %, p_{a} = 27 %, p_{m} = 13 %, p_{b} = 8 %,$ $H_{a} : A t l e a s t o n e o f t h e p_{i} ’ s i s i n c o r r e c t .$

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 $= 6.60$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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:

$H_{0} : p_{c} = 52 %, p_{a} = 27 %, p_{m} = 13 %, p_{b} = 8 %, H_{a} : A t l e a s t o n e o f t h e p_{i} ’ s i s i n c o r r e c t .$

Part (b) Step 1: Explanation

$E x p e c t e d v a l u e = n \times f r e q u e n c y$

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	$O - E$	${(O - E)}^{2} / E$	% 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

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