Jennifer’s manager Dr. Jonathan Steinberg, who has degrees and
publications in both mathematical statistics and medical science,
asked her to find estimates of the average dental claim
reimbursement for 2019. As Healthy Life has many thousands of
clients it is virtually impossible to calculate the population
mean. Using the Excel Random Number Generator function, Jennifer
found a random sample of 52
dental claims submitted to Healthy Life. The amounts covered by
insurance you can see in the Major Assignment Data file. Please
help Jennifer Nguyen 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. Constructing
confidence intervals, please round values to two decimal
places.
| 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 | $150,00 |
| 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 |
| 36 | $230,00 |
| 37 | $245,00 |
| 38 | $750,00 |
| 39 | $250,00 |
| 40 | $250,00 |
| 41 | $340,00 |
| 42 | $225,50 |
| 43 | $192,75 |
| 44 | $192,75 |
| 45 | $250,00 |
| 46 | $225,00 |
| 47 | $350,00 |
| 48 | $250,00 |
| 49 | $250,00 |
| 50 | $340,00 |
| 51 | $195,00 |
| 52 | $385,00 |
Program in minitab:
MTB > Histogram 'Amount Covered';
SUBC> Bar;
Descriptive Statistics: Amount Covered
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3
Maximum
Amount Covered 52 0 26722 1421 10250 9500 20125 25000 34000
75000
MTB > let c6(1)=26722+c3(1)*(10250/(52^0.5))
MTB > let c6(1)=26722+c3(1)*(10250/(52^0.5))
MTB > let c6(2)=26722+c3(2)*(10250/(52^0.5))
MTB > let c6(3)=26722+c3(3)*(10250/(52^0.5))
MTB > let c7(1)=26722-c3(1)*(10250/(52^0.5))
MTB > let c7(2)=26722-c3(2)*(10250/(52^0.5))
MTB > let c7(3)=26722-c3(3)*(10250/(52^0.5)
Confidence
Row Interval Upper bound Lower bound
1 90.00% 29095.8 24348.2
2 95.00% 29574.8 23869.2
3 99.00% 30524.3 22919.7
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