As products come off a large assembly line, they are inspected for quality. They are placed into one of three categories: pass, fail, or need additional inspection. The manager claims that 80 percent of the products pass the inspection, 5 percent fail, and 15 percent need additional inspection. In an effort to increase the passing rate, the machines are recalibrated. After this recalibration, a random sample of 100 products is selected. It is discovered that 85 passed inspection, 2 failed, and 13 needed additional inspection. The manager would like to know if there is convincing evidence that the distribution has changed. What is the value of the chi-square test statistic and cap p -value? Use the chi-square distribution table.
Answer
A
chi squared is equal to 5 point 1 0 , cap p -value is less than 0 point 0 0 0 5
B
chi squared is equal to 2 point 3 8 , cap p -value is less than 0 point 0 0 0 5
C
chi squared is equal to 5 point 1 0 , cap p -value is greater than 0 point 2 5
D
chi squared is equal to 2 point 3 8 , cap p -value is greater than 0 point 2 5
Explanation
Under the manager’s claimed distribution, the expected counts in a sample of 100 are 80 pass, 5 fail, and 15 needing additional inspection. The chi-square statistic is (85−80)²/80 + (2−5)²/5 + (13−15)²/15 = 0.3125 + 1.8 + 0.2667 ≈ 2.38. With 3 categories, there are 2 degrees of freedom; the chi-square table shows this statistic has a p-value greater than 0.25, so choice D is correct.
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