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

Based on the given data, a sample of 41 cases from the Major assignment data file:

Let $\mu$ denote the average fees charged by Dr. N.N. We need to compare this average to that of the population average, a hypothesized value of $\mu_{0}=$510$

We need to test:

$H_{0}:\mu \geq $510$ Vs $H_{0}:\mu<$510$

However, the data analysis tool pack of excel does not provide an easy approach for One sample t test, we need to resort to the available options, one of which, as mentioned in the problem is a t-test, Two sample assuming unequal variances,

But this would require two samples for comparison and at the same time we must maintain the average of the second sample at $510. Suppose, we create a hypothetical column of data with all values equal to $510 and run the test.

Thus, we would re-write the above hypothesis as:

$H_{0}:\mu_{1}-\mu_{2}=0$ Vs   $H_{a}:\mu_{1}-\mu_{2}<0$

Using excel, running a t-test, assuming unequal variances and that the data is normally distributed:

File Home Insert Page Layout Formulas Data Review View Show Detail Data Analysis Hide Detail ? Solver 21 22 21 Sort From Access From Web Connections Properties Refresh All GO Edit Links Connections From From Other Text Sources Get External Data Existing Connections Clear Reapply Filter Y Advanced Sort & Filter Group Ungroup Subtotal Text to Remove Data Consolidate What If Columns Duplicates Validation Analysis Data Tools Outline Analysis T24 foc D E F G H - J K L M N O P Q R S ? х OK Cancel Data Analysis Analysis Tools Histogram Moving Average Random Number Generation Rank and Percentile Regression Sampling t-Test: Paired Two Sample for Means t-Test: Two-Sample Assuming Equal Variances t-Test: Two-Sample Assuming Unequal Variances z-Test: Two Sample for Means Help A B C 1 Doctors's # Dr. N.N's Fee General Fee 2 1 $ 450 $ 510 3 2 $ 435 $ 510 4 3 $ 375 $ 510 5 4 $ 450 $ 510 6 5 $ 450 $ 510 7 6 $ 420 $ 510 8 7 $ 580 $ 510 9 8 $ 450 $ 510 10 9 $ 500 $ 510 11 10 $ 550 $ 510 12 11 $ 500 $ 510 13 12 $ 450 $ 510 14 13 $ 550 $ 510 14 on 15 10

T24 X K t-Test: Two-Sample Assuming Unequal Variances Input Variable 1 Range: $B$1:$B$42 Variable 2 Range: $C$1:$C$42 OK Cancel 3 4 Help Hypothesized Mean Difference: 0 UI US Labels Alpha: 0.011 А B C 1 Doctors's # Dr. N.N's Fee General Fee 1 $ 450 $ 510 2 $ 435 $ 510 3 $ 375 $ 510 4 $ 450 $ 510 5 $ 450 $ 510 6 $ 420 $ 510 7 $ 580 $ 510 8 $ 450 $ 510 20 9 $ 500 $ 510 1 10 $ 550 $ 510 2 11 $ 500 $ 510 -3 12 $ 450 $ 510 4 13 $ 550 $ 510 -5 14 $ 540 $ 510 -6 15 $ 550 $ 510 3 SE $2 Output options Output Range: New Worksheet Ply: New Workbook

We get the output:

We find that the p-value of the test p = 0.382 > 0.01 is not significant at 1% level (Also, the test statistic t = -0.30 > -2.42 does not lie in the rejection region (t < -t_0.01,40). We fail to reject the null hypothesis at 1% level of significance. We may conclude that the data does not provide sufficient evidence to support the Dr. N.N's claim that his average procedure fee is way below the population average of $510.

D E F G H t-Test: Two-Sample Assuming Unequal Variances Dr. N.N's Fee General Fee 507.80 510 2168.81 0 41 41 0 А B с Doctors's # Dr. N.N's Fee General Fee 1 $ 450 $ 510 2 $ 435 $ 510 3 $ 375 $ 510 4 $ 450 $ 510 5 $ 450 $ 510 6 $ 420 $ 510 7 $ 580 $ 510 8 $ 450 $ 510 9 $ 500 $ 510 10 $ 550 $ 510 11 $ 500 $ 510 12 $ 450 $ 510 13 $ 550 $ 510 14 $ 540 $ 510 15 $ 550 $ 510 16 $ 525 510 Mean Variance Observations Hypothesized Mean Difference df t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail 40 -0.30 0.382 2.42 0.764 2.70 $