What would my average be?

Below I provided all details and info I've gathered so far but I'm having trouble finding the average.

Scenario:

You work at a hospital that has always been using "Trexatine" as a medication to treat autoimmune disease flare-ups. While Trexatine is effective in helping with the disease, it has a negative side effect...it raises Systolic Blood Pressure by an average of 12.9 points. This can be a problem for patients who have high blood pressure to begin with.

A new drug, "Florexitol," has been equally as effective as Trexatine in treating the autoimmune flare-ups, but the hospital is interested in seeing if Florexitol has a lower average impact on Systolic Blood Pressure than Trexatine.

A total of 136 patients have received this new drug, and their Systolic Blood Pressure increases were noted. Each patient was given an ID number from 1 to 136. A random number generator was used to produce 30 ID numbers, and you are to use those 30 patients in your sample. (The ID number tells you which patient to use...the value you will use in your sample is the Systolic Blood Pressure increase associated with that patient.)

30 random numbers

1. 028
2. 127
3. 002
4. 069
5. 038
6. 032
7. 016
8. 063
9. 056
10. 115
11. 036
12. 101
13. 116
14. 106
15. 136
16. 009
17. 048
18. 104
19. 107
20. 132
21. 022
22. 008
23. 085
24. 126
25. 039
26. 092
27. 006
28. 100
29. 047
30. 070

You are to use these 30 patients to determine if the hospital should switch to this new drug. In other words, you are to see if the average impact on Systolic Blood Pressure with Florexitol is less than 12.9 points. Since you will be doing a test on a medication, you will use a low significance level of .01.

Research

The data that was collected in this study was done through simple random sampling so that every possible sample from the sample size had the possibility to be chosen.

Shape: Normal Curve

Center: 66

Spread: Standard Deviation is 42.46

Hypothesis Test

Hypothesis:

• Null: Mu = 66
• Alternative: Mu > 66
• T-Score: 0.237
• P-Value: 0.593
• Conclusion: In this study, we fail to reject the null hypothesis because our p-value is higher than the alpha.