Question

QUESTION 6 20 points Save Answer A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is currently producing has a normal distribution with an mean of 7.4 minutes with a standard deviation of 1.2 minutes. The chemistry of the new anesthetic is such that the effective time should be normally distributed with the same standard deviation, but the mean effective time may be lower. If it is lower, the drug company will market the new anesthetic; otherwise, they will continue to produce the older one. A sample of size 36 results in a sample mean of 7.1. A hypothesis test will be done at the 10% level of significance to help make the decision. 1. W = 7.4 2. W27.4 I AL VI i ci प. प. - Step 1: The null hypothesis is Ho: su> 7.4 Step 1: The alternative hypothesis is H1: wiw N AV प. Step 2: The value of the test statistics is Step 3: The lower critical value is Note: If there is one critical value, one of the lower critical value" or "upper critical value" should be "none". Step 3: The upper critical value is Note: If there is one critical value, one of the "lower critical value" or "upper critical value should be "none" 7. -1.50 8. -1.28 9. 0.0668 10.1.28 Step 3: The p-value is 11.1.50 12. not reject Ho - Step 4: You should 13. reject Ho - Step 5: Your should conclude that 14. none 15. there is sufficient evidence that the mean effective time is less than 7.4 minutes. 16. there is not sufficient evidence that the mean effective time is less than 7.4 minutes 17. there is sufficient evidence that the mean effective time is not less than 7.4 minutes. 18. there is not sufficient evidence that the mean effective time is not less than 7.4 minutes

Answer 1

1) 6)

2)2

3) 7

4) 8

5) 14 none

6) 9

7) 13

8) 16

8 X="ZI., d=of. XNN ( u = 7o4 , 8=102), N=36, Ho: M 67.4 6.) Hao M7 7. 4.8.) 4 X-Me = 761 - 764 = 1 0150 102 1536 <=.-312= -4.5.7.) -63 46 162 = 12 Now the tail test, dr Lower critical value, A. oy=-1.28 8.). - upper value = None 14.) Pvalue: 05- Plosz €105] = .5- 04332 = .0668 9.). as I calculated=-1.5 Ž critical = -1.28 ² calculated L-fuitcal. -|65 < -1.28. 80 Ho is. Rejected. 13.) 1. Conclude that : Mean effective time is (7.4 Minutes. # No sufficient evidence that 160).