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Q 1.

Expert-verified
Found in: Page 712

### The Practice of Statistics for AP Examination

Book edition 6th
Author(s) Daren Starnes, Josh Tabor
Pages 837 pages
ISBN 9781319113339

# Aw, nuts! A company claims that each batch of its deluxe mixed nutscontains $52%$ 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) ${H}_{0}:{p}_{c}=52%,{p}_{a}=27%,{p}_{m}=13%,{p}_{b}=8%,$ ${H}_{a}:Atleastoneofthe{p}_{i}’sisincorrect.$

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 $=6.60$

See the 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:

${H}_{0}:{p}_{c}=52%,{p}_{a}=27%,{p}_{m}=13%,{p}_{b}=8%,\phantom{\rule{0ex}{0ex}}{H}_{a}:Atleastoneofthe{p}_{i}’sisincorrect.$

## Part (b) Step 1: Explanation

$Expectedvalue=n×frequency$

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

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