Suspecting that a die may be unfair,you want to investigate.To check,you roll it 54 times,recording the number of times each face appears. $19215364125666\begin{array} { l | l } \text { Face } & \text { Count } \\\hline 1 & 9 \\2 & 15 \\3 & 6 \\4 & 12 \\5 & 6 \\6 & 6\end{array}$ Find $x ^ { 2 }$ .

A) 1.000
B) 2.000
C) 4.000
D) 8.000
E) 10.000

Question

Suspecting that a die may be unfair,you want to investigate.To check,you roll it 54 times,recording the number of times each face appears.

19215364125666\begin{array} { l | l } \text { Face } & \text { Count } \\\hline 1 & 9 \\2 & 15 \\3 & 6 \\4 & 12 \\5 & 6 \\6 & 6\end{array}

Find

x ^ { 2 }

.

A) 1.000
B) 2.000
C) 4.000
D) 8.000
E) 10.000

KATHRYN HIGHFIELD · Accepted Answer

Final Answer : D, Explanation :   The chi-square (  x 2 x^2 x 2  ) statistic is used to test the hypothesis that an observed frequency distribution fits a expected distribution. In this case, we expect each face of the die to appear an equal number of times if the die is fair. With 54 rolls, the expected frequency for each face is   54 ÷ 6 = 9 54 \div 6 = 9 54 ÷ 6 = 9  .The formula for   x 2 x^2 x 2   is   ∑ ( O i − E i ) 2 E i \sum \frac{(O_i - E_i)^2}{E_i} ∑ E i ​ ( O i ​ − E i ​ ) 2 ​  , where   O i O_i O i ​   is the observed frequency and   E i E_i E i ​   is the expected frequency for each category.- For face 1:   ( 9 − 9 ) 2 / 9 = 0 (9 - 9)^2 / 9 = 0 ( 9 − 9 ) 2 /9 = 0  - For face 2:   ( 15 − 9 ) 2 / 9 = 36 / 9 = 4 (15 - 9)^2 / 9 = 36 / 9 = 4 ( 15 − 9 ) 2 /9 = 36/9 = 4  - For face 3:   ( 6 − 9 ) 2 / 9 = 9 / 9 = 1 (6 - 9)^2 / 9 = 9 / 9 = 1 ( 6 − 9 ) 2 /9 = 9/9 = 1  - For face 4:   ( 12 − 9 ) 2 / 9 = 9 / 9 = 1 (12 - 9)^2 / 9 = 9 / 9 = 1 ( 12 − 9 ) 2 /9 = 9/9 = 1  - For face 5:   ( 6 − 9 ) 2 / 9 = 9 / 9 = 1 (6 - 9)^2 / 9 = 9 / 9 = 1 ( 6 − 9 ) 2 /9 = 9/9 = 1  - For face 6:   ( 6 − 9 ) 2 / 9 = 9 / 9 = 1 (6 - 9)^2 / 9 = 9 / 9 = 1 ( 6 − 9 ) 2 /9 = 9/9 = 1  Adding these up gives   x 2 = 0 + 4 + 1 + 1 + 1 + 1 = 8 x^2 = 0 + 4 + 1 + 1 + 1 + 1 = 8 x 2 = 0 + 4 + 1 + 1 + 1 + 1 = 8  .

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