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aksik [14]
3 years ago
8

A researcher drilled a hole in a die and filled it with a lead​ weight, then proceeded to roll it 209 times. The observed freque

ncies for the outcomes of​ 1, 2,​ 3, 4,​ 5, and 6 respectively are 36​, 30​, 41​, 40​, 23​, and 39. Use a 0.05 significance level to test the claim that the outcomes are not equally likely. Does it appear that the loaded die behaves differently than a fair​ die?
Mathematics
1 answer:
ddd [48]3 years ago
5 0

Answer:

Step-by-step explanation:

Given that the observed frequencies for the outcomes as follows:

To check this we can use chi square goodness of fit test.

H_0: equally likely\\H_a: not equally likely

(Two tailed test at 5% significance level)

Assuming equally likely expected observations are found out and then chi square is calculated as (0-E)^2/E

Df = 6-1 =5

Outcome Frequency Expected frequency (Obs-exp)^2/Exp

   

1 36 34.83333333 0.03907496

2 30 34.83333333 0.670653907

3 41 34.83333333 1.091706539

4 40 34.83333333 0.766347687

5 23 34.83333333 4.019936204

6 39 34.83333333 0.498405104

   

209 209          7.086124402

p value =0.214

Since p >alpha, we accept null hypothesis

It appears that the  loaded die does not behave differently than a fair​ die at 5% level of significance

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  • a) x = 9
  • b) arc JK = 68
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================================

Explanation:

Arcs JK and KL form a semicircle, so they add to 180 degrees

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Then you'll use this x value to find arc JK and arc KL

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Since central angles MNJ and KNL are vertical angles, this means minor arcs MJ and KL are congruent arcs. So arc MJ is also 112 degrees

Arc LMK is basically nearly everything of the full circle, but we exclude out the portion from L to K (the shorter distance)

arc LMK = (full circle) - (measure of minor arc LK)

arc LMK = 360 - 112

arc LMK = 248

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