Marina Petrenko, a Humber College Healthcare Management program graduate who always had only perfect marks in statistics, was hired by the famous Healthy Life medical insurance company. Marina is assigned to conduct statistical analysis of medical and financial data. As Marina is on probation, please help her to complete the following six tasks. In problems 2-6, state hypotheses H0 and H1 and provide detailed conclusions (based on P-values or critical values/test statistics) together with the Excel output. For your convenience the data are given in the Major Assignment Data file. You can also find useful information on the Blackboard in Excel Instructions folder.

1. Marina's manager Dr. Jonathan Stein, who has degrees in both mathematics and medicine, asked her to find estimates of the average dental claim reimbursement for 2018. As Healthy Life has many thousands of clients it is virtually impossible to calculate the population mean. Using the Excel Random Number Generator function, Marina found a random sample of 35 dental claims submitted to Healthy Life. The amounts covered by insurance you can see in the Major Assignment Data file. help Marina to construct 90%, 95%, and 99% confidence intervals for the true average reimbursement. Make sure that t-distribution is applicable: build a histogram with the bin values, for example, $100, $200, $300, $400, and $500, and check whether it is approximately symmetric and bell-shaped. Then, use Descriptive Statistics function from Data Analysis.

Dental Claim Number / Amount Covered 1 $192.75 2 $192.75 3 $350.25 4 $200.00 5 $225.00 6 $95.00 7 $375.50 8 $380.00 9 $192.75 10 $400.00 11 $230.00 12 $245.00 13 $750.00 14 $250.00 15 $250.00 16 $340.00 17 $225.50 18 $156.25 19 $300.00 20 $350.00 21 $435.00 22 $192.75 23 $192.75 24 $250.00 25 $225.00 26 $230.00 27 $245.00 28 $250.00 29 $250.00 30 $250.00 31 $350.00 32 $98.00 33 $405.00 34 $295.00 35 $205.00

2. Analyzing the most recent data, Marina's manager Dr. Stein found out that in 2018 the anti-cancer prescription drug claims submitted by Healthy Life clients contained amounts considerably greater than he anticipated. Specifically, he asked Marina to check whether patients with Oncotype DX scores over 30 claimed, on average, more than $1000 in 2018. Using the systematic sampling method Marina selected 60 patients from this category (see the Major Assignment Data file). help Marina to test the claim that the population mean annual amount of the anticancer prescription drug claims was over $1000 in 2018. Use Data Analysis t-Test: Two-Sample Assuming Unequal Variances and "fool" Excel approach. Use 5% significance level. Is it possible to reach the same conclusion at 1% significance level? As you know, you have to make sure that the distribution of data is symmetric and bell-shaped in order to use t-distribution. Otherwise you have to use nonparametric methods for data analysis. Please build a histogram for the data using bins $500, $1000, $1500, $2000, $2500.

Patient # / Monthly Claims 1 $1,254.17 2 $589.25 3 $1,047.12 4 $1,145.00 5 $1,055.25 6 $595.50 7 $1,050.00 8 $1,010.00 9 $1,290.90 10 $2,025.00 11 $750.00 12 $600.00 13 $1,250.00 14 $1,150.25 15 $1,350.00 16 $1,254.17 17 $1,150.00 18 $1,450.25 19 $1,190.90 20 $700.00 21 $1,250.00 22 $1,150.25 23 $1,150.25 24 $1,050.25 25 $1,150.25 26 $750.00 27 $1,550.00 28 $1,090.90 29 $1,254.17 30 $1,090.90 31 $1,254.17 32 $450.00 33 $470.00 34 $480.00 35 $2,190.90 36 $2,054.17 37 $1,150.25 38 $750.00 39 $1,150.25 40 $2,550.00 41 $600.00 42 $550.00 43 $550.00 44 $750.00 45 $750.00 46 $1,650.25 47 $1,550.25 48 $450.00 49 $450.00 50 $1,590.90 51 $1,554.17 52 $590.90 53 $1,554.17 54 $1,550.00 55 $1,550.00 56 $1,550.00 57 $590.90 58 $1,554.17 59 $1,650.25 60 $1,550.25

3. 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 and overcharging 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. Marina Petrenko took part in this investigation. At one point, she was asked to compare the amount doctor N.N. charged for a certain medical procedure with the province average. Marina randomly selected a sample of forty Ontario physicians of the same qualification and specialty in order to compare their fees with ones of doctor N.N. (see the Major Assignment Data file). Can we conclude at 0.5% significance level that the population average is significantly below $600 that N.N. charged for this procedure? After carefully reviewing the data Marina and her manager decided that variability in the population did not change over the last years so they assumed that the population standard deviation is still about $175 as it had been several years ago. Therefore Marina decided to use z-distribution. help Marina to test the claim. Use Z.TEST dialog box. (Please note that the test returns the one-tailed right P-value).

Doctor's # Doctor's Fee 1 $550.00 2 $535.00 3 $475.00 4 $550.00 5 $550.00 6 $520.00 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

4. It appears that not all people are equally vulnerable to cancer. People from some ethnic groups have higher or lower than average chances of developing certain types of cancer. Marina's manager Dr. Stein thinks that the company's insurance policy should reflect this fact and constantly raises this question in the meetings. To be more convincing, he asked Marina to take a random sample of those Healthy Life clients who belong to ethnic group X (at least on their maternal or paternal side) and a similar sample consisting of the clients without X ancestry. Then, Marina was asked to compare Y parameters indicating likelihood of developing W cancer. (As the matter is ticklish, we will not be naming the ethnic group, the type of cancer, the medical parameter, and the gender of sample members.) Then, it is known that in the general population Y parameters are distributed with the standard deviation of 3.0 (therefore the population variance is equal to 9.0). However in the X ethnic group the population standard deviation is about 2.1(and the population variance is equal to 4.41). As only 2-3% of Canadians have X ancestors, Marina and her manager assume that the standard deviation in the population of those who have no X ancestry is also around 3.0 as in the country in general. This assumption allows using z-distribution. The data are provided. Can we state at 1% level of significance that it matters whether a person belongs to X group or not in terms of the risk of developing W cancer? Or the chances are equal (on average Y parameters are the same)? Help Marina to conduct the test. Use Data Analysis z-Test: Two-Sample for Means.

# / Y-scores X Group/ Y-scores Not X Group 1 9.2 2.6 2 6.4 4.7 3 5.5 8.8 4 7.1 2.0 5 3.6 1.2 6 5.3 3.5 7 6.5 5.2 8 9.8 6.8 9 2.5 9.1 10 4.9 1.0 11 9.5 0.8 12 7.7 0.9 13 8.1 1.6 14 6.0 4.5 15 5.0 5.6 16 5.5 7.5 17 7.2 2.8 18 4.5 2.0 19 4.0 1.9 20 4.8 1.8 21 8.5 8.5 22 6.6 7.4 23 7.1 8.0 24 3.6 9.5 25 4.7 2.4 26 6.2 0.9 27 9.1 1.4 28 5.5 8.3 29 5.9 5.0 30 6.0 7.7 31 7.1 9.2 32 4.0 1.6 33 5.0 4.5 34 5.5 5.5 35 7.2 7.5 36 4.5 2.9 37 6.2 2.0 38 9.2 1.9 39 5.5 2.1 40 5.8 8.7

5. Smoking is a risk factor affecting patient's cardiovascular system (and not only cardiovascular system, of course). Marina was asked to take random samples of male Healthy Life members recently admitted to hospitals in GTA with heart attacks. She compiled two large separate files consisting of those male residents of Toronto who smoked and those who never smoked. The purpose of this research is to prove that the first heart attack occurs at an earlier age if a patient smoked. After carefully analyzing the data Marina came up with the hypothesis that male smokers who suffer a first heart attack are, on average, seven years younger than non-smoking males who suffer a first heart attack. Please calculate sample variances for both files and decide (depending on the sample variances) what function is most appropriate: Data Analysis t-Test: TwoSample Assuming Equal Variances or Data Analysis t-Test: Two-Sample Assuming Unequal Variances. help Marina to conduct the test at 1% significance level. As it is known that both samples come from normally distributed populations, no histograms are required.

# / Nonsmoking Patients / Smoking Patients 1 60.7 50.2 2 54.7 48.1 3 77.5 72.0 4 82.4 53.2 5 92.7 57.4 6 49.1 41.9 7 88.0 74.0 8 83.5 75.9 9 54.7 49.2 10 67.1 61.5 11 62.5 47.2 12 66.0 60.4 13 57.1 51.7 14 56.2 47.5 15 74.1 71.9 16 82.5 53.2 17 61.3 55.0 18 65.3 59.8 19 67.4 62.5 20 68.3 62.8 21 61.1 55.0 22 62.5 57.5 23 75.5 71.2 24 70.2 70.4 25 87.6 82.5 26 55.9 52.4 27 61.5 54.8 28 62.3 57.0 29 77.5 72.0 30 61.3 54.5 31 84.5 77.5 32 48.7 54.1 33 61.1 55.0 34 62.5 56.5 35 68.7 64.2 36 71.4 49.4 37 77.5 73.0 38 78.4 72.5 39 66.0 61.0 40 61.3 55.4 41 57.5 52.4 42 60.5 54.8 43 62.5 57.0 44 77.5 72.0 45 60.0 54.5 46 59.9 54.2 47 60.5 55.0 48 62.5 56.5 49 70.4 64.2 50 69.5 63.1 51 65.3 62.8 52 67.4 55.0 53 68.3 57.5 54 61.1 71.2 55 62.5 70.4 56 75.5 82.5 57 74.2 52.4 58 87.6 54.8 59 55.9 57.0 60 61.5 72.1 61 61.3 54.5 62 84.5 77.5 63 58.7 53.2 64 61.1 73.5 65 62.5 82.6 66 68.7 67 71.4 68 77.5 69 78.4 70 71.4

6. Marina's manager Dr. Jonathan Stein wonders whether Healthy Life members needed more chiropractic help in 2018 than in 2017, on average. Marina found a random sample of those who were treated by chiropractic doctors in both years. Data provided. help Marina to check whether annual expenses and number of visits increased, on average. For both tests use Data Analysis t-Test: Paired Two Sample for Means and 2% significance level. Marina and her manager know that in both cases differences are normally distributed, so here again there is no need to build histograms.

Patient # / [ (Number of visits) 2017 / 2018 ] / [ (Annual Expense) 2017 / 2018 ]

1 10 11 $340.00 $374.00

2 15 14 $560.00 $476.00

3 24 25 $816.00 $850.00

4 7 4 $238.00 $136.00

5 11 10 $374.00 $340.00

6 19 22 $646.00 $748.00

7 18 15 $612.00 $510.00

8 4 3 $136.00 $162.00

9 25 24 $850.00 $816.00

10 31 35 $1,054.00 $1,190.00

11 7 7 $298.00 $238.00

12 8 9 $272.00 $366.00

13 17 15 $578.00 $510.00

14 10 12 $340.00 $408.00

15 14 15 $476.00 $510.00

16 22 20 $748.00 $680.00

17 6 5 $204.00 $170.00

18 21 24 $774.00 $816.00

19 15 15 $510.00 $570.00

20 12 12 $408.00 $468.00

21 3 4 $140.00 $120.00

22 11 10 $450.00 $420.00

23 2 0 $90.00 $0.00

24 2 4 $80.00 $160.00

25 7 6 $300.00 $280.00