Instructions. The Assignment has a total of 20 marks. Marks are awarded based on the quality of the work and on the steps taken to complete the task. You are encouraged to discuss the assignment with your peers, but must write your own solution. The dataset and the details of this assignment are given below. Please calculate your answer to two decimal places, unless otherwise specified below.
Table 1: Crediting Ratings for Clients of Bank X
Scenario. Recently, Bank X reassessed their outstanding loans and updated their credit ratings database. Similar to the old rating, the new rating is on a scale from 1 to 10, where 10 indicates the most reliable client and 1 indicates the least reliable client. You, as a data analyst of Bank X, are given a dataset which contains both the old credit ratings and the new credit ratings for 21 clients. The dataset is shown in Table 1. You have been told that the new credit rating is calculated in a different way from the old credit rating, thus the sample contains old credit rating in Table 1 is independent to the sample containing new credit rating in the same table. The loan system for Bank X is quite large and stored in a very complicate way. For this reason, you do not get access to information such as the variance of the distribution of the credit rating (both the old and the new credit ratings).
Your job is to analyse the given dataset and using appropriate statistical techniques to determine if there is a difference between the average of the old credit ratings and the average of the new credit ratings. You need to state any relevant assumptions that may be required to complete your task. A senior staff has given you a hint on how to complete your work. The hint is to consider two different cases and to perform relevant hypothesis testing.
Write a statistical report which includes all the steps that were required to complete this task and draw a conclusion based on your analysis. Note, if the degrees of freedom is calculated to be a decimal point then round up to the nearest integer.
Ho: μ1 = μ2
Ha: μ1 ≠ μ2
α = 0.05
Two-tailed Paired sample t-test (population variance unknown)
d bar = 1/n (Ʃi=1 n di) = -0.2
S= √1/(n-1) (Ʃi=1 n (di -
d)2) = 0.4690
d/S/√n = -1.9540
Critical Value Df =20 and α = 0.5 is: 2.09
Since t is not less than -2.09 or greater than 2.09, t is not in the critical region. We cannot reject the null hypothesis.
We are unable to reject the null hypothesis. Hence, we can conclude that at the assumed level of significance of 0.5, there is no statistically significant data to conclude that there is difference in average credit rating calculated basis the old method and the new method.
Critical Value Df =20 and α = 0.1 is: 1.73
Since t is less than -1.73, t (-1.9540) is in the critical region. We can reject the null hypothesis.
We can reject the null hypothesis. Hence, we can conclude that at the assumed level of significance of 0.1, there is statistically significant data to conclude that there is difference in average credit rating calculated basis the old method and the new method.