Question

From time to time, unfortunately, Healthy Life employees have to deal with insurance fraud. Say, some people claim medical services that have never been provided or money they never paid. To that end, Healthy Life hired a number of investigators whose functions are not much different from those of police detectives. Doctor N.N. has been under suspicion for some time for deceiving both Healthy Life and his patients. Healthy Life approached the provincial authorities and they agreed to launch a formal investigation and open a case given a credible evidence of fraud is provided. The Healthy Life investigation department found a number of offences.These included “up-coding” or “upgrading,” which involved billing for more expensive treatments than those actually provided; providing and subsequently billing for treatments that were not medically necessary; scheduling extra visits for patients; referring patients to another physician when no further treatment was actually necessary; "phantom billing," or billing for services not rendered; and “ganging,” or billing for services to family members or other individuals who were accompanying the patient but who had not personally received any services. Jennifer Nguyen took part in this investigation together with Dr. Steinberg and Ian McGillivray, a former police detective and now a Healthy Life employee. At one point, Jennifer was asked to compare the amount doctor N.N. charged for a certain medical procedure with the province average. Jennifer randomly selected a sample of forty cases (see the Major Assignment Data file). Can we support at 1% significance level the doctor’s widely advertised claim that his average procedure fee is way below the population average $510? Use Data Analysis t-Test: Two-Sample Assuming Unequal Variances and “fool” Excel approach. Assume that the values are normally distributed.

Patient's #	Doctor's Fee
1	$450.00
2	$435.00
3	$375.00
4	$450.00
5	$450.00
6	$420.00
7	$580.00
8	$450.00
9	$500.00
10	$550.00
11	$500.00
12	$450.00
13	$550.00
14	$540.00
15	$550.00
16	$525.00
17	$525.00
18	$525.00
19	$550.00
20	$560.00
21	$495.00
22	$520.00
23	$500.00
24	$510.00
25	$525.00
26	$575.00
27	$475.00
28	$550.00
29	$500.00
30	$450.00
31	$550.00
32	$500.00
33	$550.00
34	$525.00
35	$525.00
36	$550.00
37	$495.00
38	$570.00
39	$485.00
40	$525.00

Answer 1

• Null Hypothesis (H_o) : average procedure fee is equal to the population average $510 that is μ = $510.
• Alternative Hypothesis (H₁) : average procedure fee is less than the population average $510 that is μ < $510.

Now for the computation of T-test we use excel, to do this follow the given steps:

1. Firstly you have to install Data analysis toolpak in excel, for this do the following steps:
2. Go to 'file' and click on 'Options'.
3. Then a dialog box will appear, click on add-ins
4. Then in the 'manage' option, select 'excel add-ins' and then click on go.
5. Then a dialog box will appear, tick on 'Analysis ToolPak' and press ok.

Now for T-test:

1. Firstly enter the data in the excel spreadsheet and create another column with dummy values equal to zero.
2. Then go to 'Data' option in the tool bar.
3. Click on 'Data Analysis', then a dialog box will appear.
4. In the dialog box, select t-Test: Two-Sample Assuming Unequal Variances and click on OK.
5. Then in 'Variable 1 range:' drag Doctor's Fee data and in 'Varible 2 range:' drag dummy data and put alpha = 0.01, and finally click on OK.

OUTPUT:

Now since our p-value (single tail) is greater than 0.01, So we may not reject our null hypothesis at 0.01 level of significance and conclude that the procedure fees for doctor is $510.