You play a game of dice at the state fair, but suspect the dice may have been ri
ID: 3311011 • Letter: Y
Question
You play a game of dice at the state fair, but suspect the dice may have been rigged. You observe the game owner rolling the dice 96 times and note the following results: Face "1" "2" "3" "4" "5" "6"
Observed count 17 23 8 11 21 16
(a) Which hypotheses should be used to test if the dice are fair: Identify H0:
H0: p1 = p2 = p3 = p4 = p5 = p6 = 1/6, implying the dice are unfair.
H0: p1 = p2 = p3 = p4 = p5 = p6 = 0, implying the dice are unfair.
H0: p1 = p2 = p3 = p4 = p5 = p6 = 1/6, implying the dice are fair.
H0: p1 = p2 = p3 = p4 = p5 = p6 = 0, implying the dice are fair.
Identify Ha:
Ha: every p 1/6, implying the dice are unfair.
Ha: at least one p 1/6, implying the dice are fair.
Ha: every p 1/6, implying the dice are fair. Ha: at least one p 1/6, implying the dice are unfair.
(b) What is the expected count for each side of the dice?
(c) Is the cell count condition met?
Yes, because the expected cell count is close to the observed cell count for all cells.
No, because some cells have less than 30 observed counts.
Yes, because the expected cell count is at least 5 for all cells.
No, because some cells have less than 10 observed counts.
(d) Calculate the contribution to the test statistic for the dice side with number "2": (Use 4 decimals):
(e) The full test statistic is 2 = 10.250. What is the P-value? (Use 4 decimals.)
(f) What is the appropriate test conclusion, for = 0.05?
There is sufficient evidence the dice are unfair.
There is sufficient evidence the dice are fair.
There is insufficient evidence the dice are fair.
There is insufficient evidence the dice are unfair.
Explanation / Answer
a)H0: p1 = p2 = p3 = p4 = p5 = p6 = 1/6, implying the dice are fair
Ha: at least one p 1/6, implying the dice are unfair.
b) expected count for each side of the dice =total/6 =96/6 =16
c) Yes, because the expected cell count is at least 5 for all cells.
d) applying chi square test on above data:
contribution to the test statistic for the dice side with number "2" =(23-16)2/16 =3.0625
e) P-value =0.0685
f) as p value is greater then 0.05 level
There is insufficient evidence the dice are unfair.
observed Expected Chi square face Probability O E=total*p =(O-E)^2/E "1" 1/6 17.000 16.00 0.06 "2" 1/6 23.000 16.00 3.06 "3" 1/6 8.000 16.00 4.00 "4" 1/6 11.000 16.00 1.56 "5" 1/6 21.000 16.00 1.56 "6" 1/6 16.000 16.00 0.00 1 96 96 10.2500Related Questions
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