4. A manufacturer packs a prize in each box ofits cereal and claims that the probabilities ofgetting various prizes are as follows:Probability0.20PrizeRingKey ChainPuzzle0.300.15Pen0.25Flashlight0.10To see if the company's claim is accurate, aconsumer group randomly selects 1,020 boxes ofthis product and obtains the following data:PrizeFrequency186RingKey ChainPuzzle327168Pen225Flashlight114Using a 5% level of significance, test the claimthat the distribution of prizes is what themanufacturer claims.

Question
Asked Nov 20, 2019
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4. A manufacturer packs a prize in each box of
its cereal and claims that the probabilities of
getting various prizes are as follows:
Probability
0.20
Prize
Ring
Key Chain
Puzzle
0.30
0.15
Pen
0.25
Flashlight
0.10
To see if the company's claim is accurate, a
consumer group randomly selects 1,020 boxes of
this product and obtains the following data:
Prize
Frequency
186
Ring
Key Chain
Puzzle
327
168
Pen
225
Flashlight
114
Using a 5% level of significance, test the claim
that the distribution of prizes is what the
manufacturer claims.
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4. A manufacturer packs a prize in each box of its cereal and claims that the probabilities of getting various prizes are as follows: Probability 0.20 Prize Ring Key Chain Puzzle 0.30 0.15 Pen 0.25 Flashlight 0.10 To see if the company's claim is accurate, a consumer group randomly selects 1,020 boxes of this product and obtains the following data: Prize Frequency 186 Ring Key Chain Puzzle 327 168 Pen 225 Flashlight 114 Using a 5% level of significance, test the claim that the distribution of prizes is what the manufacturer claims.

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Expert Answer

Step 1

It is given that the probabilities of getting various prizes.

Step 2

In this case, the chi-square goodness of fit test is appropriate.

The null and alternative hypotheses are shown below:

H0: The true proportions of all 5 categories is equal to their expected value.

H1: At least one proportions is different from its expected value.

The test statistic is obtained as follows:

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- E, i=1

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Step 3

Here, Oi represents observed values and Ei represents frequencies.

The ex...

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Expected fraction (p) Expected frequency (E= np) Observed (O) (O-E) 2/E 0.20 186 204 1.5882 0.30 327 306 1.4412 153 0.15 168 1.4706 225 255 0.25 3.5294 0.10 114 102 1.4118 Total 1,020 1,020 9.4412

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