Question

Question 2: The Office of the Superintendent of Bankruptcy of Canada (OSBC) is developing a new index to measure the vulnerability of firms in the new technology industry. The index is a ratio of current assets to current liabilities adjusted for various factors specific to this industry. The OSBC wants to compare the index among healthy and failed firms for validation purposes. They expect that failed firms should have a lower index than the healthy ones. Based on a Canadian business registry, they draw a random sample of 68 firms still in operations and another random sample of 33 firms which failed in the last 3 years. Index data for the sampled firms based on their latest financial statements are included in the Minitab file Bankruptcy_Index.mtw and in the Excel file Assign2.xlsx under the Bankruptcy_Index tab. Data is below:

a) Use Minitab or other appropriate software to produce boxplots of the index values for the two groups of firms and comment on their distribution. 2  

b) Use an appropriate statistical test to determine, at the 1% significance level, whether the data provide evidence of a higher average index for the healthy firms. Make sure you provide your manual calculations using the critical value approach.  

c) Calculate manually a 99% one-sided confidence interval for the difference in the average index of healthy and failed firms and compare your results with b) above.

d) Use Minitab or other appropriate software to perform the calculations in b) and c) above and comment on any differences

Here is the data for Bankruptcy_Index.mtw

Healthy	Failed
1.5	0.82
2.08	0.05
2.23	1.68
0.89	0.91
1.91	1.16
1.2	0.42
1.95	0.88
2.73	1.11
1.62	2.03
1.71	0.92
1.03	0.73
1.96	0.89
0.1	0.83
1.43	0.99
2.5	0.52
0.23	1.32
1.67	0.48
2.17	1.1
2.61	0.19
1.56	0.51
1.76	0.26
1.02	0.88
1.8	1.31
1.81	0.9
1.76	0.62
0.68	1.45
2.02	1.17
1.2	0.93
1.87	0.75
2.61	0.13
1.11	1.12
2.73	1.15
2.22	0.71
2.5
0.67
1.14
3.15
1.44
2.16
1.21
3.05
0.95
0.9
2.8
1.55
2.44
1.84
1.24
1.39
1.8
2.05
1.52
0.96
2.12
1.85
1.69
2.3
2.21
2.03
1.64
1.87
1.06
1.93
2.25
1.42
0.96
1.64
2.21

Answer 1

Question.a) Use Minitab or other appropriate software to produce boxplots of the index values for the two groups of firms and comment on their distribution.

Box plot from minitab is given below .

you can get the same from paste the data on minitab and use menu graph-boxplot and select the data and put in ggraph variables and click OK.

Interpretation of Distribution :

Key things to look .

• Median , ie. line in middle of dark box ,
• Interquarttilee range box (black box in middle represents 50%)
• Outlier -Failed cases show an outlier i.e datapoint which is far from other data values .
• Spread : Indicates variability - Variability is high in case of Healthy in comparision to Failed.

one can check the parameters for both

Median 1.78 Healthy 0.89 failed -- Q1 : 1.21 healthy 0.57 failed -Q3 2.17 healthy 1.14

• Skewness : Both the plots Median is not exactly on Middle , left side (i.e lower in the plot) is more than right , i.e distribution is bit skewed towards left or negatively skewed. indicates a deviation from Normal distribution. However its not very high , hence need to be confirmed by Normality test before conducting test of hypothesis.

Q2. Use an appropriate statistical test to determine, at the 1% significance level, whether the data provide evidence of a higher average index for the healthy firms. Make sure you provide your manual calculations using the critical value approach.  

Answer 2 sample t test is the appropriate test as data is continuous , data sets are independent and we want to comapre mean , population standard devation is not known,

Null Hypothesis H₀ : Mean Healthy Index(mu1) - Mean of Failed Index(mu2) =0

Alternate Hypothesi : H₁ : Mean Healthy Index(xbar) -Mean of Failed Index(ybar) >0

sample size n1 = 68 , n2 =33 so degrees of freedom =68+33-2 =99

Level of significance alpha =0.01 ,  

Criterion for rejecting Ho if t calculated > t 99, 0.01 (t critical ) = 2.365

tcalculated = (xbar - ybar)/Sp(1/n₁ +1/n₂)

whrere Sp = pooled standard deviation

Sp² =( (n₁-1) S₁² + (n₂-1) S₂² )/ (n₁ +n₂ -2)

Calculations are as follows

healthy failed

Mean	1.729559	0.876364
Std Deviation	0.635082	0.431906
variance	0.40333	0.186543
Sample Size	68	33

Sp = 0.5773 t critical : 6.97 ,

Hence Null Hypotheis is rejected , i.e Index for Healthy is higher than Failed  

c)  Calculate manually a 99% one-sided confidence interval for the difference in the average index of healthy and failed firms and compare your results with b) above.

One sided confidence interval for differeence of mean at 99% CL = (xbar-ybar) - (t 99, 0.01 (t critical ))*Sp*SQRT(1/n1 +1/n2)

= 0.853- 2.365*0.577*sqrt(0.015+0.030) = 0.853 -0.289 = 0.563 lowe bound

Output from Minitab (select stat-Basic stat-2 sample t ,select the samples in diffrent column put values in sample 1 and 2 , in options put ci 99% , test difference 0 and alternative greater than, select Assum equal variance)

Two-Sample T-Test and CI: Healthy, Failed

Two-sample T for Healthy vs Failed

N Mean StDev SE Mean
Healthy 68 1.730 0.635 0.077
Failed 33 0.876 0.432 0.075

Difference = mu (Healthy) - mu (Failed)
Estimate for difference: 0.853
99% lower bound for difference: 0.564
T-Test of difference = 0 (vs >): T-Value = 6.97 P-Value = 0.000 DF = 99
Both use Pooled StDev = 0.5773
There is no difference