3. The Bradfield Container Company makes \"cardboard\" boxes for commercial use.

Question

3. The Bradfield Container Company makes "cardboard" boxes for commercial use. One of the big issues for the company is the set-up time required to change over from one order to the next. At one particular machine, the set-up time is thought to be uniformly distributed. To test whether this is true or not, a random sample of 180 set-ups on this machine was selected with set-up time rounded to the nearest two-minute intervals. The following results occurred: Set-up Time Frequeney 10-11 minutes15 12-13 minutes 14-15 minutes 16-17 minutes 18-19 minutes 20-21 minutes 25 38 42 39 21 Using a significance level of 0.05, determine whether the distribution of setup times is uniform? (20 points) a. State the Null and Alternative Hypotheses of Interest: b· The critical value of the test statistic is: c. The calculated value of the test statistic is: d. Conclusion:

Explanation / Answer

Solution:

Here, we have to use Chi square test for goodness of fit.

Part a

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H0: Distribution of setup time is uniform.

Alternative hypothesis: Ha: Distribution of setup time is not uniform.

Part b

The level of significance for this test is given as 0.05.

Degrees of freedom = n – 1 = 6 – 1 = 5

Critical value = 11.0705

(by using Chi square table)

Part c

Now, we have to find the test statistic value. The formula for test statistic is given as below:

Chi square = [(O – E)^2/E]

Where, O is observed frequencies and E is expected frequencies.

Tables for calculations are summarized as below:

Set up Time

Observed Frequency

Expected frequency

10 to 11 minutes

15

30

12 to 13 minutes

25

30

14 to 15 minutes

38

30

16 to 17 minutes

42

30

18 to 19 minutes

39

30

20 to 21 minutes

21

30

Total

180

180

O

E

(O - E)

(O - E)^2

(O - E)^2/E

15

30

-15

225

7.5000

25

30

-5

25

0.8333

38

30

8

64

2.1333

42

30

12

144

4.8000

39

30

9

81

2.7000

21

30

-9

81

2.7000

Total

20.6667

Chi square = [(O – E)^2/E] = 20.6667

P-value = 0.0009

Part d

Here, we get Chi square test statistic value = 20.6667 is greater than Critical value = 11.0705

Chi square test statistic > Critical value

(P-value = 0.0009 < = 0.05)

So, we reject the null hypothesis that Distribution of setup time is uniform.

There is insufficient evidence to conclude that Distribution of setup time is uniform.

Set up Time

Observed Frequency

Expected frequency

10 to 11 minutes

15

30

12 to 13 minutes

25

30

14 to 15 minutes

38

30

16 to 17 minutes

42

30

18 to 19 minutes

39

30

20 to 21 minutes

21

30

Total

180

180

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