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: Plant A Plant B Row Total Specification Met 85 35 120 Specification Not Met 15 25 40 Column Total 100 60 160 Find the chi-square test statistic for a hypothesis of independence. Multiple Choice 7.22 14.22 -0.18 14.70

The value of Chi-square test statistic for a hypothesis test of independence is 14.22.

Step-by-step explanation:

The data provided is for one day's output of Kortholt's from the Melodic Kortholt Company.

The formula to compute the chi-square test statistic for a hypothesis of independence is:

The formula to compute the expected frequencies (E) is:

Consider the Excel output attached.

Compute the value of Chi-square test statistic as follows:

Thus, the value of Chi-square test statistic for a hypothesis test of independence is 14.22.

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