Kortholts that fail to meet certain precise specifications must be reworked on the next day until they are within the desired specifications. A sample of one day's output of kortholts from the Melodic Kortholt Company showed the following frequencies: $\begin{array}{|c|c|c|c|}
\hline&\text { Plant A }&\text { Plant B }&\text { Row Total }\
\hline \text {Specification Met}&185 & 85&270 \
\hline \text { Specification Not Met }&15 & 15& 30\
\hline \text { Col Total }&200 & 100 &300\
\hline
\end{array}$ Find the chi-square test statistic for a hypothesis of independence.
A)7.22
B)4.17
C)5.13
D)6.08

Question

Kortholts that fail to meet certain precise specifications must be reworked on the next day until they are within the desired specifications. A sample of one day's output of kortholts from the Melodic Kortholt Company showed the following frequencies: $\begin{array}{|c|c|c|c|}
\hline&\text { Plant  A }&\text { Plant  B }&\text { Row Total }\
\hline \text {Specification Met}&185 & 85&270 \
\hline \text { Specification Not Met }&15 & 15& 30\
\hline \text { Col Total }&200 & 100 &300\
\hline
\end{array}$ Find the chi-square test statistic for a hypothesis of independence.
A)7.22
B)4.17
C)5.13
D)6.08

Quizplus · Accepted Answer

The chi-square test statistic is calculated using the formula $\chi^2 = \sum \frac{(O - E)^2}{E}$, where $O$ represents the observed frequencies and $E$ represents the expected frequencies under the hypothesis of independence. The expected frequency for each cell in a contingency table is calculated as $(\text{row total} \times \text{column total}) / \text{grand total}$.For Plant A, Specification Met: $E = (270 \times 200) / 300 = 180$For Plant A, Specification Not Met: $E = (30 \times 200) / 300 = 20$For Plant B, Specification Met: $E = (270 \times 100) / 300 = 90$For Plant B, Specification Not Met: $E = (30 \times 100) / 300 = 10$Calculating $\chi^2$:$\chi^2 = \frac{(185-180)^2}{180} + \frac{(15-20)^2}{20} + \frac{(85-90)^2}{90} + \frac{(15-10)^2}{10}$$\chi^2 = \frac{25}{180} + \frac{25}{20} + \frac{25}{90} + \frac{25}{10}$$\chi^2 = 0.1389 + 1.25 + 0.2778 + 2.5$$\chi^2 = 4.17$Therefore, the chi-square test statistic is 4.17.

Kortholts That Fail to Meet Certain Precise Specifications Must Be