budgeting and variance analysis case study assignment

Question

Eastside laboratory has decided that varianes that are within two standard errors of the budgeted or expected value need not be investigated because they are quite likely to occur simply as a matter of chance. However if the variance is more than two standard errors, either plus or minus, there is a significant chance that a control problem exists. Investigation is required Assume that last year is considered to be a period during which particular effort was made to control ancillary usage. As a result, we are willing to accept last years, eight patients had a particular type of expensive lab test (this unrealistically low number of patients is used to simplify your calculations), and the number of units of resources consumed for each of the eight tests was as follows:

Patient #:        number of tests:

1                     30

2                     56

3                     42

4                     80

5                     66

6                     58

7                     44

8                     72

- Calculate the standard error based on this test group from last year. Suppose that in the first month of this year, the average number of resource units used for each test of this type was 70. Is the resulting deviation from the standard great enough to warrant investigation?

 

Homework template to fill in:

 

The first step in solving this problem is to calculate the standard error, as follows: (X-x ) #(X) Deviation from Mean Patient Number of Tests (x - x )2 1 30 -26 676 2 56 ____ ____ 3 42 ____ ____ 4 80 ____ ____ 5 66 ____ ____ 6 58 ____ ____ 7 44 ____ ____ 8 72 ____ ____ Total 448 ____ ____ Mean = x = Sum of observations Number of observations 448 8 = 56 Standard deviation = square root of: ( x - x )2 . (Number of observations - 1) Standard error = Standard deviation . Square root of Number of observations Two standard errors would be twice the ______standard error calculated above, or ______. In this problem, last month’s observed mean of 70 results in a deviation of 14 from the standard test mean of 56. Since 14 is greater than _______, the deviation is more than two standard errors. Two standard errors encompass all normal events with a 95 percent probability. That is, we are 95 percent confident that if the process of controlling ancillary usage is working properly, the observed mean for any month should not be more than two standard errors greater or fewer than the test mean. In this case, two standard errors is ______. There is only a 2.5 percent chance that we are controlling ancillary tests properly, but that patients require _______ more tests than the test mean of ____; and there is only a 2.5 percent chance that we are controlling ancillary tests properly, but that patients require ______ fewer tests than the test mean of _____. Thus, given that the mean last month of 70 is more than two standard errors from the test mean of ____, it is very unlikely that ancillary tests were being appropriately controlled last month. Investigation should take place.

Answer 1

Step 1: Calculate the mean

Total observations:

30+56+42+80+66+58+44+72=44830 + 56 + 42 + 80 + 66 + 58 + 44 + 72 = 44830+56+42+80+66+58+44+72=448Number of observations:

888Mean:

4488=56\frac{448}{8} = 568448​=56So the standard or expected mean is 56 resource units per test.

Step 2: Complete the deviation table

PatientNumber of Tests / Resource UnitsDeviation from MeanDeviation Squared
1	30	-26	676
2	56	0	0
3	42	-14	196
4	80	24	576
5	66	10	100
6	58	2	4
7	44	-12	144
8	72	16	256
Total	448	0	1,952

Step 3: Calculate the standard deviation

Formula:

Standard Deviation=∑(x−xˉ)2n−1\text{Standard Deviation} = \sqrt{\frac{\sum (x-\bar{x})^2}{n-1}}Standard Deviation=n−1∑(x−xˉ)2​​ =1,9528−1= \sqrt{\frac{1,952}{8-1}}=8−11,952​​ =1,9527= \sqrt{\frac{1,952}{7}}=71,952​​ =278.86= \sqrt{278.86}=278.86​ =16.70= 16.70=16.70So the standard deviation is 16.70.

Step 4: Calculate the standard error

Formula:

Standard Error=Standard Deviationn\text{Standard Error} = \frac{\text{Standard Deviation}}{\sqrt{n}}Standard Error=n​Standard Deviation​ =16.708= \frac{16.70}{\sqrt{8}}=8​16.70​ =16.702.828= \frac{16.70}{2.828}=2.82816.70​ =5.90= 5.90=5.90So the standard error is 5.90.

Step 5: Calculate two standard errors

2×5.90=11.802 \times 5.90 = 11.802×5.90=11.80So two standard errors = 11.80, approximately 11.81.

Step 6: Compare the current month’s result

This year’s first month average is 70.

Expected mean from last year:

565656Deviation:

70−56=1470 - 56 = 1470−56=14Now compare:

14>11.8114 > 11.8114>11.81Since the deviation of 14 is greater than two standard errors of 11.81, the variance is large enough to require investigation.

Final Answer

Yes, the deviation should be investigated.

The average usage this month was 70 resource units, compared with the expected average of 56 resource units. The difference is 14 units. Since this difference is greater than two standard errors, which is approximately 11.81 units, the variance is statistically significant.

This means the higher usage is unlikely to be due only to random chance. There may be a control problem in ancillary resource usage, so Eastside Laboratory should investigate.