session 6 13.1

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CaseStudy13.1AnswerGuide.docx

13.1: American Oil Company

The problem Chad faces is whether a particular component part has an average diameter that meets the company's needs. The supplier of the part states that the diameter is normally distributed with a mean of 0.16 inches and a standard deviation of 0.02 inches. Chad needs to determine whether the supplier's claim is valid or not.

The supplier has sent Chad the following information for 330 randomly selected components.

Diameter (inches)

Observed

Under 0.14

5

0.14 and under 0.15

70

0.15 and under 0.16

90

0.16 and under 0.17

105

0.17 and under 0.18

50

Over 0.18

10

Total

330

Chad can use a chi-square test to determine if the component supplier's claim is true. He would first need to determine the relevant probabilities for a normally distributed random variable having a mean of 0.16 and a standard deviation of 0.02. Using these probabilities, Chad can then determine the expected frequency of component parts that he would expect to see in a sample of 330 if the supplier's claim is true. The observed and expected frequencies can then be used to calculate a chi-square test statistic that can be compared to a critical chi-square value. Chad's computations are shown below.

Diameter (inches)

Observed

Probability

Expected

(O-E)2/E

Under 0.14

5

0.15865526

52.35623572

42.83373

0.14 and under 0.15

70

0.15 and under 0.16

90

0.16 and under 0.17

105

0.17 and under 0.18

50

0.149882273

49.46115005

0.00587

Over 0.18

10

Total

330

1.00000

330

The calculated value of the chi-square test statistic is ___________. The critical value for 5 degrees of freedom and a value of alpha = 0.01 is _________. Since the calculated value is __________ than the _______________ value, Chad should __________ the null hypothesis and conclude that the manufacturer's statement that the diameter of the component part is normally distributed with a mean of 0.16 inches and a standard deviation of 0.02 inches is ____________________(true or not true). In other words, the diameter of the component part is (or is not) normally distributed with the population mean and standard deviation parameters specified by the supplier.

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