A spinner is mounted on a piece of cardboard divided into six areas of equal size. Each of the six areas is a different color (blue, yellow, red, green, white and orange). When the spinner is spun, each color should be selected by the spinner approximately $\frac { 1 } { 6 }$ of the time. A student suspects that a certain spinner is defective. The suspected spinner is spun 90 times. The results are shown below. Calculate the chi-square test statistic $\chi ^ { 2 }$ to test the student's claim. Use $\alpha = 0.10$.
$\begin{array}{l|cccccc}
\text { Color } & \text { Blue Yellow } & \text { Red Green } & \text { White Orange } \
\hline \text { Frequency } & 17 & 11 & 12 & 19 & 15 & 16
\end{array}$
A) $3.067$
B) $2.143$
C) $5.013$
D) $4.312$

Question

Quizplus · Accepted Answer

The chi-square test statistic is calculated as:$\chi ^ { 2 } = \sum \frac { (O - E) ^ { 2 } } { E }$where O is the observed frequency and E is the expected frequency.The expected frequency for each color is 90 / 6 = 15.Therefore, the chi-square test statistic is:$\chi ^ { 2 } = \frac { (17 - 15) ^ { 2 } } { 15 } + \frac { (11 - 15) ^ { 2 } } { 15 } + \frac { (12 - 15) ^ { 2 } } { 15 } + \frac { (19 - 15) ^ { 2 } } { 15 } + \frac { (15 - 15) ^ { 2 } } { 15 } + \frac { (16 - 15) ^ { 2 } } { 15 } = 3.067$The critical value for $\chi ^ { 2 }$ with 5 degrees of freedom and $\alpha = 0.10$ is 9.236.Since $\chi ^ { 2 } = 3.067 < 9.236$, we fail to reject the null hypothesis and conclude that the spinner is not defective.

A Spinner Is Mounted on a Piece of Cardboard Divided 16\frac { 1 } { 6 }61​

A Spinner Is Mounted on a Piece of Cardboard Divided $\frac { 1 } { 6 }$