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An industrial engineer tests 4 different shop-floor layouts by having each of 6

ID: 3152605 • Letter: A

Question

An industrial engineer tests 4 different shop-floor layouts by having each of 6 work crews construct a subassembly and measuring the construction times (minutes) as follows:

Layout 1

Layout 2

Layout 3

Layout 4

Crew A

48.2

53.1

51.2

58.6

Crew B

49.5

52.9

50.0

60.1

Crew C

50.7

56.8

49.9

62.4

Crew D

48.6

50.6

47.5

57.5

Crew E

47.1

51.8

49.1

55.3

Crew F

52.4

57.2

53.5

61.7

Test at the 0.01 level of significance whether the 4 floors layouts produce different assembly times and whether some of the work crews are consistently faster in constructing this subassembly than the others.

{please use excel if needed for this problem, and provide steps to what you did to calculate any data found in excel. for everything else, please show all steps. thank you}

Layout 1

Layout 2

Layout 3

Layout 4

Crew A

48.2

53.1

51.2

58.6

Crew B

49.5

52.9

50.0

60.1

Crew C

50.7

56.8

49.9

62.4

Crew D

48.6

50.6

47.5

57.5

Crew E

47.1

51.8

49.1

55.3

Crew F

52.4

57.2

53.5

61.7

Explanation / Answer

Solution:

Here, we have to test whether the 4 floors layouts produce different assembly times and whether some of the work crews are consistently faster in constructing this subassembly than the others. For checking this claim we have to use the one way analysis of variance for finding the significant difference in the assembly times for given four floors layouts. The null and alternative hypothesis is given as below:

Null hypothesis: H0: There is no any significant difference in the average assembly times for four floors layouts.

Alternative hypothesis: Ha: There is a significant difference in the average assembly times for the four floors layouts.

We assume the level of significance for this test as alpha = 0.05.

The ANOVA table for this test is given as below:

Descriptives

Construction time

N

Mean

Std. Deviation

Std. Error

95% Confidence Interval for Mean

Minimum

Maximum

Lower Bound

Upper Bound

Layout 1

6

49.4167

1.90096

.77607

47.4217

51.4116

47.10

52.40

Layout 2

6

53.7333

2.68601

1.09656

50.9145

56.5521

50.60

57.20

Layout 3

6

50.2000

2.02583

.82704

48.0740

52.3260

47.50

53.50

Layout 4

6

59.2667

2.67333

1.09138

56.4612

62.0721

55.30

62.40

Total

24

53.1542

4.53374

.92545

51.2397

55.0686

47.10

62.40

ANOVA

Construction time

Sum of Squares

df

Mean Square

F

Sig.

Between Groups

362.365

3

120.788

21.883

.000

Within Groups

110.395

20

5.520

Total

472.760

23

For this ANOVA test, we get the p-value as 0.00 which is less than the given level of significance or alpha value 0.05, so we reject the null hypothesis that There is no any significant difference in the average assembly times for four floors layouts.

This means, There is a significant difference in the average assembly times for four floors layouts.

Descriptives

Construction time

N

Mean

Std. Deviation

Std. Error

95% Confidence Interval for Mean

Minimum

Maximum

Lower Bound

Upper Bound

Layout 1

6

49.4167

1.90096

.77607

47.4217

51.4116

47.10

52.40

Layout 2

6

53.7333

2.68601

1.09656

50.9145

56.5521

50.60

57.20

Layout 3

6

50.2000

2.02583

.82704

48.0740

52.3260

47.50

53.50

Layout 4

6

59.2667

2.67333

1.09138

56.4612

62.0721

55.30

62.40

Total

24

53.1542

4.53374

.92545

51.2397

55.0686

47.10

62.40