CASE: AIR FORCE TRAINING PROGRAM.

Submitted by Ashish Clifton

Application no – 60097

FSB 2

DATA

Current Proposed76 7476 7577 7774 7876 7474 8074 7377 7372 7878 7673 7678 7475 7780 6979 7672 7569 7279 7572 7270 7670 7281 7776 7378 7772 6982 7772 7573 7671 7470 7777 75

2

Current Proposed

78 78

73 72

79 77

82 78

65 78

77 76

79 75

73 76

76 76

81 75

69 76

75 80

75 77

77 76

79 75

76 73

78 77

76 77

76 77

73 79

77 75

84 75

74 72

74 82

69 76

79 76

66 74

70 72

74 78

72 71

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Question 1

Use appropriate descriptive statistics to summarize the training time data for each method. What similarities or differences do

you observe from sample data?

TABLE 1 x1 x2 TABLE 3 5

Number Summary

Mean 75.0656 75.4262 Current ProposedStandard Deviation

3.9449 2.5064 Lowest Value 65 69

Variance 15.5623 6.2820 First Quartile 72 74 Median 76 76

TABLE 2 x1 x2 Third Quartile 78 7765-69 5 2 Largest Value 84 8270-74 22 16 75-79 28 39 80-84 6 8

3%

25%

60%

12%

X2 proposed65-69 70-74 75-79 80-84

8%

36%

46%

10%

X1 current65-69 70-74 75-79 80-84

CONCLUSION

TABLE1-The mean of the two samples are similar. However, the standard deviation and variance of the two samples vary largely which denotes that the data is spread over a wider range from central tendency.

TABLE2-the second sample shows that the completion duration with the range of 75-84 has increased by 16% which means 16% candidates are taking more time to complete the course with the proposed training programme.

TABLE3- The 5-number summary shows very little variability between the two samples.

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Question 2

Use the methods of chapter 10 to comment on any difference between the populations means for the two methods. Discuss your findings.

Step 1: Developing Hypothesis

Null Hypothesis Hₒ: μ1-μ2 = 0

Alternative Hypothesis H1: μ1-μ2 ≠ 0

Null Hypothesis : There is no difference between the mean completion times for current and proposed method.

Alternate Hypothesis : There is difference between the mean completion times for current and proposed method.

Step 2: Level of Significance

Level of Significance Α = 0.05

Step 3: Calculation of test statistic

x1=75.0656x2= 75.4262

s12=∑ ( x1 i−x ).

2

n1−1 = 3.9499

s22=∑ ( x2 i−x ).

2

n2−1 = 2.5064

t = ¿¿¿¿ = - 0.5977

Test StatisticT =

-0.5977

Step 4

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Critical Value Approach

t (α/2) =2.00

Step 5: Determine whether to reject HₒRejection Rule: Critical value approach

Reject H0 if t <= -t(α/2) or if t >= t(α/2)

CONCLUSION

Do not reject H0 i.e. There is no significant difference between the mean completion times for current and proposed method. (α=0.05).

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Question 3

Compute the standard deviation and variance for the training method. Conduct a hypothesis test about the quality of the population variances for the two training methods. Discuss your findings.

CURRENT PROPOSEDSTANDARD DEVIATION

3.944907 2.506385VARIANCE

15.5623 6.281967

Step 1: Developing Hypothesis

Null Hypothesis Hₒ: σ1^2 = σ2^2Alternative Hypothesis H1: σ1^2 ≠ σ2^2

Step 2: Level of Significance

Level of Significance α = 0.05

Step 3: Calculation of test statistic

S12 = 15.5623 (from question 1)

S22 = 6.2820 (From question 1)

F =s1

2

s22 = 2.4773

Test Statistic F = 2.4773

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Step 4: Critical Value Approach

F (α/2) = 1.67

Step 5:Determine whether to reject Hₒ

F (α/2) FDECISIO

N1.67 < 2.4773 Reject

CONCLUSION

Reject Hₒ which means that we have sufficient evidence to conclude that the population variances of the two training methods is significantly different. (α =0.05)

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